{"id":543675,"date":"2021-09-30T09:25:32","date_gmt":"2021-09-30T09:25:32","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=543675"},"modified":"2021-10-01T10:02:13","modified_gmt":"2021-10-01T10:02:13","slug":"rd-sharma-solutions-for-class-11-maths-chapter-23-the-straight-lines","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-23-the-straight-lines\/","title":{"rendered":"RD Sharma Solutions for Class 11 Maths Chapter 23\u2013The Straight Lines"},"content":{"rendered":"\n

Class 11: Maths Chapter 23 solutions. Complete Class 11 Maths Chapter 23 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 11 Maths Chapter 23\u2013The Straight Lines<\/h2>\n\n\n\n

RD Sharma 11th Maths Chapter 23, Class 11 Maths Chapter 23 solutions<\/p>\n\n\n\n

EXERCISE 23.1 PAGE NO: 23.12<\/h4>\n\n\n\n

1. Find the slopes of the lines which make the following angles with the positive direction of x \u2013 axis:<\/strong><\/p>\n\n\n\n

(i) \u2013 \u03c0\/4<\/strong><\/p>\n\n\n\n

(ii) 2\u03c0\/3<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>\u2013 \u03c0\/4<\/p>\n\n\n\n

Let the slope of the line be \u2018m\u2019<\/p>\n\n\n\n

Where, m = tan \u03b8<\/p>\n\n\n\n

So, the slope of Line is m = tan (\u2013 \u03c0\/4)<\/p>\n\n\n\n

= \u2013 1<\/p>\n\n\n\n

\u2234 The slope of the line is \u2013 1.<\/p>\n\n\n\n

(ii) <\/strong>2\u03c0\/3<\/p>\n\n\n\n

Let the slope of the line be \u2018m\u2019<\/p>\n\n\n\n

Where, m = tan \u03b8<\/p>\n\n\n\n

So, the slope of Line is m = tan (2\u03c0\/3)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u2234 The slope of the line is \u2013\u221a<\/strong>3<\/p>\n\n\n\n

2. Find the slopes of a line passing through the following points :
(i) (\u20133, 2) and (1, 4)<\/strong><\/p>\n\n\n\n

(ii) (at2<\/sup>1<\/sub>, 2at1<\/sub>) and (at2<\/sup>2<\/sub>, 2at2<\/sub>)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>(\u20133, 2) and (1, 4)<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u2234 The slope of the line is \u00bd.<\/p>\n\n\n\n

(ii) <\/strong>(at2<\/sup>1<\/sub>, 2at1<\/sub>) and (at2<\/sup>2<\/sub>, 2at2<\/sub>)<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

3. State whether the two lines in each of the following are parallel, perpendicular or neither:
(i) Through (5, 6) and (2, 3); through (9, \u20132) and (6, \u20135)<\/strong><\/p>\n\n\n\n

(ii) Through (9, 5) and (\u2013 1, 1); through (3, \u20135) and (8, \u20133)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>Through (5, 6) and (2, 3); through (9, \u20132) and (6, \u20135)<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

So, m2<\/sub> = 1<\/p>\n\n\n\n

Here, m1<\/sub> = m2 <\/sub>= 1<\/p>\n\n\n\n

\u2234 The lines are parallel to each other.<\/p>\n\n\n\n

(ii) <\/strong>Through (9, 5) and (\u2013 1, 1); through (3, \u20135) and (8, \u20133)<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

4. Find the slopes of a line
(i) which bisects the first quadrant angle
(ii) which makes an angle of 300<\/sup> with the positive direction of y \u2013 axis measured anticlockwise.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>Which bisects the first quadrant angle?<\/p>\n\n\n\n

Given: Line bisects the first quadrant<\/p>\n\n\n\n

We know that, if the line bisects in the first quadrant, then the angle must be between line and the positive direction of x \u2013 axis.<\/p>\n\n\n\n

Since, angle = 90\/2 = 45o<\/sup><\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The slope of the line, m = tan \u03b8<\/p>\n\n\n\n

The slope of the line for a given angle is m = tan 45o<\/sup><\/p>\n\n\n\n

So, m = 1<\/p>\n\n\n\n

\u2234 The slope of the line is 1.<\/p>\n\n\n\n

(ii)<\/strong> Which makes an angle of 300<\/sup> with the positive direction of y \u2013 axis measured anticlockwise?<\/p>\n\n\n\n

Given: The line makes an angle of 30o<\/sup> with the positive direction of y \u2013 axis.<\/p>\n\n\n\n

We know that, angle between line and positive side of axis => 90o<\/sup> + 30o<\/sup> = 120o<\/sup><\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The slope of the line, m = tan \u03b8<\/p>\n\n\n\n

The slope of the line for a given angle is m = tan 120o<\/sup><\/p>\n\n\n\n

So, m = \u2013 \u221a3<\/p>\n\n\n\n

\u2234 The slope of the line is \u2013 \u221a3.<\/p>\n\n\n\n

5. Using the method of slopes show that the following points are collinear:
(i) A (4, 8), B (5, 12), C (9, 28)<\/strong><\/p>\n\n\n\n

(ii) A(16, \u2013 18), B(3, \u2013 6), C(\u2013 10, 6)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>A (4, 8), B (5, 12), C (9, 28)<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The slope of the line = [y2<\/sub> \u2013 y1<\/sub>] \/ [x2<\/sub> \u2013 x1<\/sub>]<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

The slope of line AB = [12 \u2013 8] \/ [5 \u2013 4]<\/p>\n\n\n\n

= 4 \/ 1<\/p>\n\n\n\n

The slope of line BC = [28 \u2013 12] \/ [9 \u2013 5]<\/p>\n\n\n\n

= 16 \/ 4<\/p>\n\n\n\n

= 4<\/p>\n\n\n\n

The slope of line CA = [8 \u2013 28] \/ [4 \u2013 9]<\/p>\n\n\n\n

= -20 \/ -5<\/p>\n\n\n\n

= 4<\/p>\n\n\n\n

Here, AB = BC = CA<\/p>\n\n\n\n

\u2234 The Given points are collinear.<\/p>\n\n\n\n

(ii) <\/strong>A(16, \u2013 18), B(3, \u2013 6), C(\u2013 10, 6)<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The slope of the line = [y2<\/sub> \u2013 y1<\/sub>] \/ [x2<\/sub> \u2013 x1<\/sub>]<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

The slope of line AB = [-6 \u2013 (-18)] \/ [3 \u2013 16]<\/p>\n\n\n\n

= 12 \/ -13<\/p>\n\n\n\n

The slope of line BC = [6 \u2013 (-6)] \/ [-10 \u2013 3]<\/p>\n\n\n\n

= 12 \/ -13<\/p>\n\n\n\n

The slope of line CA = [6 \u2013 (-18)] \/ [-10 \u2013 16]<\/p>\n\n\n\n

= 12 \/ -13<\/p>\n\n\n\n

= 4<\/p>\n\n\n\n

Here, AB = BC = CA<\/p>\n\n\n\n

\u2234 The Given points are collinear.<\/p>\n\n\n\n

EXERCISE 23.2 PAGE NO: 23.17<\/h4>\n\n\n\n

1. Find the equation of the parallel to x\u2013axis and passing through (3, \u20135).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: A line which is parallel to x\u2013axis and passing through (3, \u20135)<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of line: [y \u2013 y1<\/sub> = m(x \u2013 x1<\/sub>)]<\/p>\n\n\n\n

We know that the parallel lines have equal slopes<\/p>\n\n\n\n

And, the slope of x\u2013axis is always 0<\/p>\n\n\n\n

Then<\/p>\n\n\n\n

The slope of line, m = 0<\/p>\n\n\n\n

Coordinates of line are (x1<\/sub>, y1<\/sub>) = (3, \u20135)<\/p>\n\n\n\n

The equation of line = y \u2013 y1<\/sub> = m(x \u2013 x1<\/sub>)<\/p>\n\n\n\n

Now, substitute the values, we get<\/p>\n\n\n\n

y \u2013 (\u2013 5) = 0(x \u2013 3)<\/p>\n\n\n\n

y + 5 = 0<\/p>\n\n\n\n

\u2234 The equation of line is y + 5 = 0<\/p>\n\n\n\n

2. Find the equation of the line perpendicular to x\u2013axis and having intercept \u2013 2 on x\u2013axis.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: A line which is perpendicular to x\u2013axis and having intercept \u20132<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of line: [y \u2013 y1<\/sub> = m(x \u2013 x1<\/sub>)]<\/p>\n\n\n\n

We know that, the line is perpendicular to the x\u2013axis, then x is 0 and y is \u20131.<\/p>\n\n\n\n

The slope of line is, m = y\/x<\/p>\n\n\n\n

= -1\/0<\/p>\n\n\n\n

It is given that x\u2013intercept is \u20132, so, y is 0.<\/p>\n\n\n\n

Coordinates of line are (x1<\/sub>, y1<\/sub>) = (\u20132, 0)<\/p>\n\n\n\n

The equation of line = y \u2013 y1<\/sub> = m(x \u2013 x1<\/sub>)<\/p>\n\n\n\n

Now, substitute the values, we get<\/p>\n\n\n\n

y \u2013 0 = (-1\/0) (x \u2013 (\u2013 2))<\/p>\n\n\n\n

x + 2 = 0<\/p>\n\n\n\n

\u2234 The equation of line is x + 2 = 0<\/p>\n\n\n\n

3. Find the equation of the line parallel to x\u2013axis and having intercept \u2013 2 on y \u2013 axis.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: A line which is parallel to x\u2013axis and having intercept \u20132 on y \u2013 axis<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of line: [y \u2013 y1<\/sub> = m(x \u2013 x1<\/sub>)]<\/p>\n\n\n\n

The parallel lines have equal slopes,<\/p>\n\n\n\n

And, the slope of x\u2013axis is always 0<\/p>\n\n\n\n

Then<\/p>\n\n\n\n

The slope of line, m = 0<\/p>\n\n\n\n

It is given that intercept is \u20132, on y \u2013 axis then<\/p>\n\n\n\n

Coordinates of line are (x1<\/sub>, y1<\/sub>) = (0, \u2013 2)<\/p>\n\n\n\n

The equation of line is y \u2013 y1<\/sub> = m(x \u2013 x1<\/sub>)<\/p>\n\n\n\n

Now, substitute the values, we get<\/p>\n\n\n\n

y \u2013 (\u2013 2) = 0 (x \u2013 0)<\/p>\n\n\n\n

y + 2 = 0<\/p>\n\n\n\n

\u2234 The equation of line is y + 2 = 0<\/p>\n\n\n\n

4. Draw the lines x = \u20133, x = 2, y = \u20132, y = 3 and write the coordinates of the vertices of the square so formed.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: x = \u20133, x = 2, y = \u20132 and y = 3<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u2234 The Coordinates of the square are: A(2, 3), B(2, \u20132), C(\u20133, 3), and D(\u20133, \u20132).<\/p>\n\n\n\n

5. Find the equations of the straight lines which pass through (4, 3) and are respectively parallel and perpendicular to the x\u2013axis.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: A line which is perpendicular and parallel to x\u2013axis respectively and passing through (4, 3)<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of line: [y \u2013 y1<\/sub> = m(x \u2013 x1<\/sub>)]<\/p>\n\n\n\n

Let us consider,<\/p>\n\n\n\n

Case 1: When Line is parallel to x\u2013axis<\/p>\n\n\n\n

The parallel lines have equal slopes,<\/p>\n\n\n\n

And, the slope of x\u2013axis is always 0, then<\/p>\n\n\n\n

The slope of line, m = 0<\/p>\n\n\n\n

Coordinates of line are (x1<\/sub>, y1<\/sub>) = (4, 3)<\/p>\n\n\n\n

The equation of line is y \u2013 y1<\/sub> = m(x \u2013 x1<\/sub>)<\/p>\n\n\n\n

Now substitute the values, we get<\/p>\n\n\n\n

y \u2013 (3) = 0(x \u2013 4)<\/p>\n\n\n\n

y \u2013 3 = 0<\/p>\n\n\n\n

Case 2: When line is perpendicular to x\u2013axis<\/p>\n\n\n\n

The line is perpendicular to the x\u2013axis, then x is 0 and y is \u2013 1.<\/p>\n\n\n\n

The slope of the line is, m = y\/x<\/p>\n\n\n\n

= -1\/0<\/p>\n\n\n\n

Coordinates of line are (x1<\/sub>, y1<\/sub>) = (4, 3)<\/p>\n\n\n\n

The equation of line = y \u2013 y1<\/sub> = m(x \u2013 x1<\/sub>)<\/p>\n\n\n\n

Now substitute the values, we get<\/p>\n\n\n\n

y \u2013 3 = (-1\/0) (x \u2013 4)<\/p>\n\n\n\n

x = 4<\/p>\n\n\n\n

\u2234 The equation of line when it is parallel to x \u2013 axis is y = 3 and it is perpendicular is x = 4.<\/p>\n\n\n\n

EXERCISE 23.3 PAGE NO: 23.21<\/h4>\n\n\n\n

1. Find the equation of a line making an angle of 150\u00b0 with the x\u2013axis and cutting off an intercept 2 from y\u2013axis.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: A line which makes an angle of 150o<\/sup> with the x\u2013axis and cutting off an intercept at 2<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of a line is y = mx + c<\/p>\n\n\n\n

We know that angle, \u03b8 = 150o<\/sup><\/p>\n\n\n\n

The slope of the line, m = tan \u03b8<\/p>\n\n\n\n

Where, m = tan 150o<\/sup><\/p>\n\n\n\n

= -1\/ \u221a<\/strong>3<\/p>\n\n\n\n

Coordinate of y\u2013intercept is (0, 2)<\/p>\n\n\n\n

The required equation of the line is y = mx + c<\/p>\n\n\n\n

Now substitute the values, we get<\/p>\n\n\n\n

y = -x\/\u221a<\/strong>3 + 2<\/p>\n\n\n\n

\u221a<\/strong>3y \u2013 2\u221a<\/strong>3 + x = 0<\/p>\n\n\n\n

x + \u221a<\/strong>3y = 2\u221a<\/strong>3<\/p>\n\n\n\n

\u2234 The equation of line is x + \u221a<\/strong>3y = 2\u221a<\/strong>3<\/p>\n\n\n\n

2. Find the equation of a straight line:
(i) with slope 2 and y \u2013 intercept 3;
(ii) with slope \u2013 1\/ 3 and y \u2013 intercept \u2013 4.
(iii) with slope \u2013 2 and intersecting the x\u2013axis at a distance of 3 units to the left of origin.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>With slope 2 and y \u2013 intercept 3<\/p>\n\n\n\n

The slope is 2 and the coordinates are (0, 3)<\/p>\n\n\n\n

Now, the required equation of line is<\/p>\n\n\n\n

y = mx + c<\/p>\n\n\n\n

Substitute the values, we get<\/p>\n\n\n\n

y = 2x + 3<\/p>\n\n\n\n

(ii) <\/strong>With slope \u2013 1\/ 3 and y \u2013 intercept \u2013 4<\/p>\n\n\n\n

The slope is \u2013 1\/3 and the coordinates are (0, \u2013 4)<\/p>\n\n\n\n

Now, the required equation of line is<\/p>\n\n\n\n

y = mx + c<\/p>\n\n\n\n

Substitute the values, we get<\/p>\n\n\n\n

y = -1\/3x \u2013 4<\/p>\n\n\n\n

3y + x = \u2013 12<\/p>\n\n\n\n

(iii) <\/strong>With slope \u2013 2 and intersecting the x\u2013axis at a distance of 3 units to the left of origin<\/p>\n\n\n\n

The slope is \u2013 2 and the coordinates are (\u2013 3, 0)<\/p>\n\n\n\n

Now, the required equation of line is y \u2013 y1<\/sub> = m (x \u2013 x1<\/sub>)<\/p>\n\n\n\n

Substitute the values, we get<\/p>\n\n\n\n

y \u2013 0 = \u2013 2(x + 3)<\/p>\n\n\n\n

y = \u2013 2x \u2013 6<\/p>\n\n\n\n

2x + y + 6 = 0<\/p>\n\n\n\n

3. Find the equations of the bisectors of the angles between the coordinate axes.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

There are two bisectors of the coordinate axes.<\/p>\n\n\n\n

Their inclinations with the positive x-axis are 45o<\/sup> and 135o<\/sup><\/p>\n\n\n\n

The slope of the bisector is m = tan 45o<\/sup> or m = tan 135o<\/sup><\/p>\n\n\n\n

i.e., m = 1 or m = -1, c = 0<\/p>\n\n\n\n

By using the formula, y = mx + c<\/p>\n\n\n\n

Now, substitute the values of m and c, we get<\/p>\n\n\n\n

y = x + 0<\/p>\n\n\n\n

x \u2013 y = 0 or y = -x + 0<\/p>\n\n\n\n

x + y = 0<\/p>\n\n\n\n

\u2234 The equation of the bisector is x \u00b1 <\/strong>y = 0<\/p>\n\n\n\n

4. Find the equation of a line which makes an angle of tan \u2013 1<\/sup> (3) with the x\u2013axis and cuts off an intercept of 4 units on the negative direction of y\u2013axis.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equation which makes an angle of tan \u2013 1<\/sup>(3) with the x\u2013axis and cuts off an intercept of 4 units on the negative direction of y\u2013axis<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of the line is y = mx + c<\/p>\n\n\n\n

Here, angle \u03b8 = tan \u2013 1<\/sup>(3)<\/p>\n\n\n\n

So, tan \u03b8 = 3<\/p>\n\n\n\n

The slope of the line is, m = 3<\/p>\n\n\n\n

And, Intercept in the negative direction of y\u2013axis is (0, -4)<\/p>\n\n\n\n

The required equation of the line is y = mx + c<\/p>\n\n\n\n

Now, substitute the values, we get<\/p>\n\n\n\n

y = 3x \u2013 4<\/p>\n\n\n\n

\u2234 The equation of the line is y = 3x \u2013 4.<\/p>\n\n\n\n

5. Find the equation of a line that has y \u2013 intercept \u2013 4 and is parallel to the line joining (2, \u20135) and (1, 2).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

A line segment joining (2, \u2013 5) and (1, 2) if it cuts off an intercept \u2013 4 from y\u2013axis<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of line is y = mx + C<\/p>\n\n\n\n

It is given that, c = \u2013 4<\/p>\n\n\n\n

Slope of line joining (x1<\/sub> \u2013 x2<\/sub>) and (y1<\/sub> \u2013 y2<\/sub>),<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

So, Slope of line joining (2, \u2013 5) and (1, 2),<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

m = \u2013 7<\/p>\n\n\n\n

The equation of line is y = mx + c<\/p>\n\n\n\n

Now, substitute the values, we get<\/p>\n\n\n\n

y = \u20137x \u2013 4<\/p>\n\n\n\n

y + 7x + 4 = 0<\/p>\n\n\n\n

\u2234 The equation of line is y + 7x + 4 = 0.<\/p>\n\n\n\n

EXERCISE 23.4 PAGE NO: 23.29<\/p>\n\n\n\n

1. Find the equation of the straight line passing through the point (6, 2) and having slope \u2013 3.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given, A straight line passing through the point (6, 2) and the slope is \u2013 3<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of line is [y \u2013 y1<\/sub> = m(x \u2013 x1<\/sub>)]<\/p>\n\n\n\n

Here, the line is passing through (6, 2)<\/p>\n\n\n\n

It is given that, the slope of line, m = \u20133<\/p>\n\n\n\n

Coordinates of line are (x1<\/sub>, y1<\/sub>) = (6,2)<\/p>\n\n\n\n

The equation of line = y \u2013 y1<\/sub> = m(x \u2013 x1<\/sub>)<\/p>\n\n\n\n

Now, substitute the values, we get<\/p>\n\n\n\n

y \u2013 2 = \u2013 3(x \u2013 6)<\/p>\n\n\n\n

y \u2013 2 = \u2013 3x + 18<\/p>\n\n\n\n

y + 3x \u2013 20 = 0<\/p>\n\n\n\n

\u2234 The equation of line is 3x + y \u2013 20 = 0<\/p>\n\n\n\n

2. Find the equation of the straight line passing through (\u20132, 3) and indicated at an angle of 45\u00b0 with the x \u2013 axis.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

A line which is passing through (\u20132, 3), the angle is 45o<\/sup>.<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of line is [y \u2013 y1<\/sub> = m(x \u2013 x1<\/sub>)]<\/p>\n\n\n\n

Here, angle, \u03b8 = 45o<\/sup><\/p>\n\n\n\n

The slope of the line, m = tan \u03b8<\/p>\n\n\n\n

m = tan 45o<\/sup><\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

The line passing through (x1<\/sub>, y1<\/sub>) = (\u20132, 3)<\/p>\n\n\n\n

The required equation of line is y \u2013 y1<\/sub> = m(x \u2013 x1<\/sub>)<\/p>\n\n\n\n

Now, substitute the values, we get<\/p>\n\n\n\n

y \u2013 3 = 1(x \u2013 (\u2013 2))<\/p>\n\n\n\n

y \u2013 3 = x + 2<\/p>\n\n\n\n

x \u2013 y + 5 = 0<\/p>\n\n\n\n

\u2234The equation of line is x \u2013 y + 5 = 0<\/p>\n\n\n\n

3. Find the equation of the line passing through (0, 0) with slope m<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

A straight line passing through the point (0, 0) and slope is m.<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of line is [y \u2013 y1<\/sub> = m(x \u2013 x1<\/sub>)]<\/p>\n\n\n\n

It is given that, the line is passing through (0, 0) and the slope of line, m = m<\/p>\n\n\n\n

Coordinates of line are (x1<\/sub>, y1<\/sub>) = (0, 0)<\/p>\n\n\n\n

The equation of line = y \u2013 y1<\/sub> = m(x \u2013 x1<\/sub>)<\/p>\n\n\n\n

Now, substitute the values, we get<\/p>\n\n\n\n

y \u2013 0 = m(x \u2013 0)<\/p>\n\n\n\n

y = mx<\/p>\n\n\n\n

\u2234 The equation of line is y = mx.<\/p>\n\n\n\n

4. Find the equation of the line passing through (2, 2\u221a3) and inclined with x \u2013 axis at an angle of 75o<\/sup>.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

A line which is passing through (2, 2\u221a3), the angle is 75o<\/sup>.<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of line is [y \u2013 y1<\/sub> = m(x \u2013 x1<\/sub>)]<\/p>\n\n\n\n

Here, angle, \u03b8 = 75o<\/sup><\/p>\n\n\n\n

The slope of the line, m = tan \u03b8<\/p>\n\n\n\n

m = tan 75o<\/sup><\/p>\n\n\n\n

= 3.73 = 2 + \u221a3<\/p>\n\n\n\n

The line passing through (x1<\/sub>, y1<\/sub>) = (2, 2\u221a3)<\/p>\n\n\n\n

The required equation of the line is y \u2013 y1<\/sub> = m(x \u2013 x1<\/sub>)<\/p>\n\n\n\n

Now, substitute the values, we get<\/p>\n\n\n\n

y \u2013 2\u221a3 = 2 + \u221a3 (x \u2013 2)<\/p>\n\n\n\n

y \u2013 2\u221a3 = (2 + \u221a3)x \u2013 7.46<\/p>\n\n\n\n

(2 + \u221a3)x \u2013 y \u2013 4 = 0<\/p>\n\n\n\n

\u2234 The equation of the line is (2 + \u221a3)x \u2013 y \u2013 4 = 0<\/p>\n\n\n\n

5. Find the equation of the straight line which passes through the point (1, 2) and makes such an angle with the positive direction of x \u2013 axis whose sine is 3\/5.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

A line which is passing through (1, 2)<\/p>\n\n\n\n

To Find: The equation of a straight line.<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of line is [y \u2013 y1<\/sub> = m(x \u2013 x1<\/sub>)]<\/p>\n\n\n\n

Here, sin \u03b8 = 3\/5<\/p>\n\n\n\n

We know, sin \u03b8 = perpendicular\/hypotenuse<\/p>\n\n\n\n

= 3\/5<\/p>\n\n\n\n

So, according to Pythagoras theorem,<\/p>\n\n\n\n

(Hypotenuse)2<\/sup> = (Base)2<\/sup> + (Perpendicular)2<\/sup><\/p>\n\n\n\n

(5)2<\/sup> = (Base)2<\/sup> + (3)2<\/sup><\/p>\n\n\n\n

(Base) = \u221a(25 \u2013 9)<\/p>\n\n\n\n

(Base)2<\/sup> = \u221a16<\/p>\n\n\n\n

Base = 4<\/p>\n\n\n\n

Hence, tan \u03b8 = perpendicular\/base<\/p>\n\n\n\n

= 3\/4<\/p>\n\n\n\n

The slope of the line, m = tan \u03b8<\/p>\n\n\n\n

= 3\/4<\/p>\n\n\n\n

The line passing through (x1<\/sub>,y1<\/sub>) = (1,2)<\/p>\n\n\n\n

The required equation of line is y \u2013 y1<\/sub> = m(x \u2013 x1<\/sub>)<\/p>\n\n\n\n

Now, substitute the values, we get<\/p>\n\n\n\n

y \u2013 2 = (\u00be) (x \u2013 1)<\/p>\n\n\n\n

4y \u2013 8 = 3x \u2013 3<\/p>\n\n\n\n

3x \u2013 4y + 5 = 0<\/p>\n\n\n\n

\u2234 The equation of line is 3x \u2013 4y + 5 = 0<\/p>\n\n\n\n

EXERCISE 23.5 PAGE NO: 23.35<\/h4>\n\n\n\n

1. Find the equation of the straight lines passing through the following pair of points:
(i) (0, 0) and (2, -2)<\/strong><\/p>\n\n\n\n

(ii) (a, b) and (a + c sin \u03b1, b + c cos \u03b1)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>(0, 0) and (2, -2)<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

(x1<\/sub>, y1<\/sub>) = (0, 0), (x2<\/sub>, y2<\/sub>) = (2, -2)<\/p>\n\n\n\n

The equation of the line passing through the two points (0, 0) and (2, \u22122) is<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

y = \u2013 x<\/p>\n\n\n\n

\u2234 The equation of line is y = -x<\/p>\n\n\n\n

(ii) <\/strong>(a, b) and (a + c sin \u03b1, b + c cos \u03b1)<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

(x1<\/sub>, y1<\/sub>) = (a, b), (x2<\/sub>, y2<\/sub>) = (a + c sin \u03b1, b + c cos \u03b1)<\/p>\n\n\n\n

So, the equation of the line passing through the two points (0, 0) and (2, \u22122) is<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

y \u2013 b = cot \u03b1 (x \u2013 a)<\/p>\n\n\n\n

\u2234 The equation of line is y \u2013 b = cot \u03b1 (x \u2013 a)<\/p>\n\n\n\n

2. Find the equations to the sides of the triangles the coordinates of whose angular points are respectively:
(i) (1, 4), (2, -3) and (-1, -2)<\/strong><\/p>\n\n\n\n

(ii) (0, 1), (2, 0) and (-1, -2)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>(1, 4), (2, -3) and (-1, -2)<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Points A (1, 4), B (2, -3) and C (-1, -2).<\/p>\n\n\n\n

Let us assume,<\/p>\n\n\n\n

m1,<\/sub> m2,<\/sub> and m3<\/sub> be the slope of the sides AB, BC and CA, respectively.<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

The equation of the line passing through the two points (x1<\/sub>, y1<\/sub>) and (x2<\/sub>, y2<\/sub>).<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

m1 =<\/sub> -7, m2<\/sub> = -1\/3 and m3<\/sub> = 3<\/p>\n\n\n\n

So, the equation of the sides AB, BC and CA are<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

y \u2013 y1<\/sub>= m (x \u2013 x1<\/sub>)<\/p>\n\n\n\n

=> y \u2013 4 = -7 (x \u2013 1)<\/p>\n\n\n\n

y \u2013 4 = -7x + 7<\/p>\n\n\n\n

y + 7x = 11,<\/p>\n\n\n\n

=> y + 3 = (-1\/3) (x \u2013 2)<\/p>\n\n\n\n

3y + 9 = -x + 2<\/p>\n\n\n\n

3y + x = \u2013 7<\/p>\n\n\n\n

x + 3y + 7 = 0 and<\/p>\n\n\n\n

=> y + 2 = 3(x+1)<\/p>\n\n\n\n

y + 2 = 3x + 3<\/p>\n\n\n\n

y \u2013 3x = 1<\/p>\n\n\n\n

So, we get<\/p>\n\n\n\n

y + 7x =11, x+ 3y + 7 =0 and y \u2013 3x = 1<\/p>\n\n\n\n

\u2234 The equation of sides are y + 7x =11, x+ 3y + 7 =0 and y \u2013 3x = 1<\/p>\n\n\n\n

(ii) <\/strong>(0, 1), (2, 0) and (-1, -2)<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Points A (0, 1), B (2, 0) and C (-1, -2).<\/p>\n\n\n\n

Let us assume,<\/p>\n\n\n\n

m1,<\/sub> m2,<\/sub> and m3<\/sub> be the slope of the sides AB, BC and CA, respectively.<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

The equation of the line passing through the two points (x1<\/sub>, y1<\/sub>) and (x2<\/sub>, y2<\/sub>).<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

m1<\/sub> = -1\/2, m2<\/sub> = -2\/3 and m3<\/sub>= 3<\/p>\n\n\n\n

So, the equation of the sides AB, BC and CA are<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

y \u2013 y1<\/sub>= m (x \u2013 x1<\/sub>)<\/p>\n\n\n\n

=> y \u2013 1 = (-1\/2) (x \u2013 0)<\/p>\n\n\n\n

2y \u2013 2 = -x<\/p>\n\n\n\n

x + 2y = 2<\/p>\n\n\n\n

=> y \u2013 0 = (-2\/3) (x \u2013 2)<\/p>\n\n\n\n

3y = -2x + 4<\/p>\n\n\n\n

2x \u2013 3y = 4<\/p>\n\n\n\n

=> y + 2 = 3(x+1)<\/p>\n\n\n\n

y + 2 = 3x + 3<\/p>\n\n\n\n

y \u2013 3x = 1<\/p>\n\n\n\n

So, we get<\/p>\n\n\n\n

x + 2y = 2, 2x \u2013 3y =4 and y \u2013 3x = 1<\/p>\n\n\n\n

\u2234 The equation of sides are x + 2y = 2, 2x \u2013 3y =4 and y \u2013 3x = 1<\/p>\n\n\n\n

3. Find the equations of the medians of a triangle, the coordinates of whose vertices are (-1, 6), (-3,-9) and (5, -8).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

A (\u22121, 6), B (\u22123, \u22129) and C (5, \u22128) be the coordinates of the given triangle.<\/p>\n\n\n\n

Let us assume: D, E, and F be midpoints of BC, CA and AB, respectively. So, the coordinates of D, E and F are<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Median AD passes through A (-1, 6) and D (1, -17\/2)<\/p>\n\n\n\n

So, by using the formula,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

4y \u2013 24 = -29x \u2013 29<\/p>\n\n\n\n

29x + 4y + 5 = 0<\/p>\n\n\n\n

Similarly, Median BE passes through B (-3,-9) and E (2,-1)<\/p>\n\n\n\n

So, by using the formula,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

5y + 45 = 8x + 24<\/p>\n\n\n\n

8x \u2013 5y \u2013 21=0<\/p>\n\n\n\n

Similarly, Median CF passes through C (5,-8) and F(-2,-3\/2)<\/p>\n\n\n\n

So, by using the formula,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

-14y \u2013 112 = 13x \u2013 65<\/p>\n\n\n\n

13x + 14y + 47 = 0<\/p>\n\n\n\n

\u2234 The equation of lines are: 29x + 4y + 5 = 0, 8x \u2013 5y \u2013 21=0 and 13x + 14y + 47 = 0<\/p>\n\n\n\n

4. Find the equations to the diagonals of the rectangle the equations of whose sides are x = a, x = a\u2019, y = b and y = b\u2019.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The rectangle formed by the lines x = a, x = a\u2019, y = b and y = b\u2019<\/p>\n\n\n\n

It is clear that, the vertices of the rectangle are A (a, b), B (a\u2019, b), C (a\u2019, b\u2019) and D (a, b\u2019) .<\/p>\n\n\n\n

The diagonal passing through A (a, b) and C (a\u2019, b\u2019) is<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(a\u2019 \u2013 a)y \u2013 b(a\u2019 \u2013 a) = (b\u2019 \u2013 b)x \u2013 a(b\u2019 \u2013 b)<\/p>\n\n\n\n

(a\u2019 \u2013 a) \u2013 (b\u2019 \u2013 b)x = ba\u2019 \u2013 ab\u2019<\/p>\n\n\n\n

Similarly, the diagonal passing through B (a\u2019, b) and D (a, b\u2019) is<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(a\u2019 \u2013 a)y \u2013 b(a \u2013 a\u2019) = (b\u2019 \u2013 b)x \u2013 a\u2019 (b\u2019 \u2013 b)<\/p>\n\n\n\n

(a\u2019 \u2013 a) + (b\u2019 \u2013 b)x = a\u2019b\u2019 \u2013 ab<\/p>\n\n\n\n

\u2234 The equation of diagonals are y(a\u2019 \u2013 a) \u2013 x(b\u2019 \u2013 b) = a\u2019b \u2013 ab\u2019 and<\/p>\n\n\n\n

y(a\u2019 \u2013 a) + x(b\u2019 \u2013 b) = a\u2019b\u2019 \u2013 ab<\/p>\n\n\n\n

5. Find the equation of the side BC of the triangle ABC whose vertices are A (-1, -2), B (0, 1) and C (2, 0) respectively. Also, find the equation of the median through A (-1, -2).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The vertices of triangle ABC are A (-1, -2), B(0, 1) and C(2, 0).<\/p>\n\n\n\n

Let us find the equation of median through A.<\/p>\n\n\n\n

So, the equation of BC is<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

x + 2y \u2013 2 = 0<\/p>\n\n\n\n

Let D be the midpoint of median AD,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

4y + 8 = 5x + 5<\/p>\n\n\n\n

5x \u2013 4y \u2013 3 = 0<\/p>\n\n\n\n

\u2234 The equation of line BC is x + 2y \u2013 2 = 0<\/p>\n\n\n\n

The equation of median is 5x \u2013 4y \u2013 3 = 0<\/p>\n\n\n\n

EXERCISE 23.6 PAGE NO: 23.46<\/h4>\n\n\n\n

1. Find the equation to the straight line
(i) cutting off intercepts 3 and 2 from the axes.<\/strong><\/p>\n\n\n\n

(ii) cutting off intercepts -5 and 6 from the axes.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>Cutting off intercepts 3 and 2 from the axes.<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

a = 3, b = 2<\/p>\n\n\n\n

Let us find the equation of line cutoff intercepts from the axes.<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of the line is x\/a + y\/b = 1<\/p>\n\n\n\n

x\/3 + y\/2 = 1<\/p>\n\n\n\n

By taking LCM,<\/p>\n\n\n\n

2x + 3y = 6<\/p>\n\n\n\n

\u2234 The equation of line cut off intercepts 3 and 2 from the axes is 2x + 3y = 6<\/p>\n\n\n\n

(ii) <\/strong>Cutting off intercepts -5 and 6 from the axes.<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

a = -5, b = 6<\/p>\n\n\n\n

Let us find the equation of line cutoff intercepts from the axes.<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of the line is x\/a + y\/b = 1<\/p>\n\n\n\n

x\/-5 + y\/6 = 1<\/p>\n\n\n\n

By taking LCM,<\/p>\n\n\n\n

6x \u2013 5y = -30<\/p>\n\n\n\n

\u2234 The equation of line cut off intercepts 3 and 2 from the axes is 6x \u2013 5y = -30<\/p>\n\n\n\n

2. Find the equation of the straight line which passes through (1, -2) and cuts off equal intercepts on the axes.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

A line passing through (1, -2)<\/p>\n\n\n\n

Let us assume, the equation of the line cutting equal intercepts at coordinates of length \u2018a\u2019 is<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of the line is x\/a + y\/b = 1<\/p>\n\n\n\n

x\/a + y\/a = 1<\/p>\n\n\n\n

x + y = a<\/p>\n\n\n\n

The line x + y = a passes through (1, -2)<\/p>\n\n\n\n

Hence, the point satisfies the equation.<\/p>\n\n\n\n

1 -2 = a<\/p>\n\n\n\n

a = -1<\/p>\n\n\n\n

\u2234 The equation of the line is x+ y = -1<\/p>\n\n\n\n

3. Find the equation to the straight line which passes through the point (5, 6) and has intercepts on the axes
(i) Equal in magnitude and both positive<\/strong><\/p>\n\n\n\n

(ii) Equal in magnitude but opposite in sign<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>Equal in magnitude and both positive<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

a = b<\/p>\n\n\n\n

Let us find the equation of line cutoff intercepts from the axes.<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of the line is x\/a + y\/b = 1<\/p>\n\n\n\n

x\/a + y\/a = 1<\/p>\n\n\n\n

x + y = a<\/p>\n\n\n\n

The line passes through the point (5, 6)<\/p>\n\n\n\n

Hence, the equation satisfies the points.<\/p>\n\n\n\n

5 + 6 = a<\/p>\n\n\n\n

a = 11<\/p>\n\n\n\n

\u2234 The equation of the line is x + y = 11<\/p>\n\n\n\n

(ii) <\/strong>Equal in magnitude but opposite in sign<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

b = -a<\/p>\n\n\n\n

Let us find the equation of line cutoff intercepts from the axes.<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of the line is x\/a + y\/b = 1<\/p>\n\n\n\n

x\/a + y\/-a = 1<\/p>\n\n\n\n

x \u2013 y = a<\/p>\n\n\n\n

The line passes through the point (5, 6)<\/p>\n\n\n\n

Hence, the equation satisfies the points.<\/p>\n\n\n\n

5 \u2013 6 = a<\/p>\n\n\n\n

a = -1<\/p>\n\n\n\n

\u2234 The equation of the line is x \u2013 y = -1<\/p>\n\n\n\n

4. For what values of a and b the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x \u2013 3y + 6 = 0 on the axes.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Intercepts cut off on the coordinate axes by the line ax + by +8 = 0 \u2026\u2026 (i)<\/p>\n\n\n\n

And are equal in length but opposite in sign to those cut off by the line<\/p>\n\n\n\n

2x \u2013 3y +6 = 0 \u2026\u2026(ii)<\/p>\n\n\n\n

We know that, the slope of two lines is equal<\/p>\n\n\n\n

The slope of the line (i) is \u2013a\/b<\/p>\n\n\n\n

The slope of the line (ii) is 2\/3<\/p>\n\n\n\n

So let us equate,<\/p>\n\n\n\n

-a\/b = 2\/3<\/p>\n\n\n\n

a = -2b\/3<\/p>\n\n\n\n

The length of the perpendicular from the origin to the line (i) is<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The length of the perpendicular from the origin to the line (ii) is<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

It is given that, d1<\/sub> = d2<\/sub><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

b = 4<\/p>\n\n\n\n

So, a = -2b\/3<\/p>\n\n\n\n

= -8\/3<\/p>\n\n\n\n

\u2234 The value of a is -8\/3 and b is 4.<\/p>\n\n\n\n

5. Find the equation to the straight line which cuts off equal positive intercepts on the axes and their product is 25.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

a = b and ab = 25<\/p>\n\n\n\n

Let us find the equation of the line which cutoff intercepts on the axes.<\/p>\n\n\n\n

\u2234 a2<\/sup> = 25<\/p>\n\n\n\n

a = 5 [considering only positive value of intercepts]<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of the line with intercepts a and b is x\/a + y\/b = 1<\/p>\n\n\n\n

x\/5 + y\/5 = 1<\/p>\n\n\n\n

By taking LCM<\/p>\n\n\n\n

x + y = 5<\/p>\n\n\n\n

\u2234 The equation of line is x + y = 5<\/p>\n\n\n\n

EXERCISE 23.7 PAGE NO: 23.53<\/h4>\n\n\n\n

1. Find the equation of a line for which<\/strong><\/p>\n\n\n\n

(i) p = 5, \u03b1 = 60\u00b0<\/strong><\/p>\n\n\n\n

(ii) p = 4, \u03b1 = 150\u00b0<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>p = 5, \u03b1 = 60\u00b0<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

p = 5, \u03b1 = 60\u00b0<\/p>\n\n\n\n

The equation of the line in normal form is given by<\/p>\n\n\n\n

Using the formula,<\/p>\n\n\n\n

x cos \u03b1 + y sin \u03b1 = p<\/p>\n\n\n\n

Now, substitute the values, we get<\/p>\n\n\n\n

x cos 60\u00b0 + y sin 60\u00b0 = 5<\/p>\n\n\n\n

x\/2 + \u221a<\/strong>3y\/2 = 5<\/p>\n\n\n\n

x + \u221a3y = 10<\/p>\n\n\n\n

\u2234 The equation of line in normal form is x + \u221a3y = 10.<\/p>\n\n\n\n

(ii) <\/strong>p = 4, \u03b1 = 150\u00b0<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

p = 4, \u03b1 = 150\u00b0<\/p>\n\n\n\n

The equation of the line in normal form is given by<\/p>\n\n\n\n

Using the formula,<\/p>\n\n\n\n

x cos \u03b1 + y sin \u03b1 = p<\/p>\n\n\n\n

Now, substitute the values, we get<\/p>\n\n\n\n

x cos 150\u00b0 + y sin 150\u00b0 = 4<\/p>\n\n\n\n

cos (180\u00b0 \u2013 \u03b8) = \u2013 cos \u03b8 , sin (180\u00b0 \u2013 \u03b8) = sin \u03b8<\/p>\n\n\n\n

x cos(180\u00b0 \u2013 30\u00b0) + y sin(180\u00b0 \u2013 30\u00b0) = 4<\/p>\n\n\n\n

\u2013 x cos 30\u00b0 + y sin 30\u00b0 = 4<\/p>\n\n\n\n

\u2013\u221a<\/strong>3x\/2 + y\/2 = 4<\/p>\n\n\n\n

-\u221a<\/strong>3x + y = 8<\/p>\n\n\n\n

\u2234 The equation of line in normal form is -\u221a<\/strong>3x + y = 8.<\/p>\n\n\n\n

2. Find the equation of the line on which the length of the perpendicular segment from the origin to the line is 4 and the inclination of the perpendicular segment with the positive direction of x\u2013axis is 30\u00b0.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

p = 4, \u03b1 = 30\u00b0<\/p>\n\n\n\n

The equation of the line in normal form is given by<\/p>\n\n\n\n

Using the formula,<\/p>\n\n\n\n

x cos \u03b1 + y sin \u03b1 = p<\/p>\n\n\n\n

Now, substitute the values, we get<\/p>\n\n\n\n

x cos 30\u00b0 + y sin 30\u00b0 = 4<\/p>\n\n\n\n

x\u221a<\/strong>3\/2 + y1\/2 = 4<\/p>\n\n\n\n

\u221a3x + y = 8<\/p>\n\n\n\n

\u2234 The equation of line in normal form is \u221a3x + y = 8.<\/p>\n\n\n\n

3. Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with the positive direction of x\u2013axis is 15\u00b0.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

p = 4, \u03b1 = 15\u00b0<\/p>\n\n\n\n

The equation of the line in normal form is given by<\/p>\n\n\n\n

We know that, cos 15\u00b0 = cos (45\u00b0 \u2013 30\u00b0) = cos45\u00b0cos30\u00b0 + sin45\u00b0sin30\u00b0<\/p>\n\n\n\n

Cos (A \u2013 B) = cos A cos B + sin A sin B<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

And sin 15 = sin (45\u00b0 \u2013 30\u00b0) = sin 45\u00b0 cos 30\u00b0 \u2013 cos 45\u00b0 sin 30\u00b0<\/p>\n\n\n\n

Sin (A \u2013 B) = sin A cos B \u2013 cos A sin B<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Now, by using the formula,<\/p>\n\n\n\n

x cos \u03b1 + y sin \u03b1 = p<\/p>\n\n\n\n

Now, substitute the values, we get<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(\u221a3+1)x +(\u221a3-1) y = 8\u221a2<\/p>\n\n\n\n

\u2234 The equation of line in normal form is (\u221a3+1)x +(\u221a3-1) y = 8\u221a2.<\/p>\n\n\n\n

4. Find the equation of the straight line at a distance of 3 units from the origin such that the perpendicular from the origin to the line makes an angle \u03b1 given by tan \u03b1 = 5\/12 with the positive direction of x\u2013axis.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

p = 3, \u03b1 = tan-1<\/sup> (5\/12)<\/p>\n\n\n\n

So, tan \u03b1 = 5\/12<\/p>\n\n\n\n

sin \u03b1 = 5\/13<\/p>\n\n\n\n

cos \u03b1 = 12\/13<\/p>\n\n\n\n

The equation of the line in normal form is given by<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

x cos \u03b1 + y sin \u03b1 = p<\/p>\n\n\n\n

Now, substitute the values, we get<\/p>\n\n\n\n

12x\/13 + 5y\/13 = 3<\/p>\n\n\n\n

12x + 5y = 39<\/p>\n\n\n\n

\u2234 The equation of line in normal form is 12x + 5y = 39.<\/p>\n\n\n\n

5. Find the equation of the straight line on which the length of the perpendicular from the origin is 2 and the perpendicular makes an angle \u03b1 with x\u2013axis such that sin \u03b1 = 1\/3.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

p = 2, sin \u03b1 = 1\/3<\/p>\n\n\n\n

We know that cos \u03b1 = \u221a(1 \u2013 sin2<\/sup> \u03b1)<\/p>\n\n\n\n

= \u221a(1 \u2013 1\/9)<\/p>\n\n\n\n

= 2\u221a2\/3<\/p>\n\n\n\n

The equation of the line in normal form is given by<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

x cos \u03b1 + y sin \u03b1 = p<\/p>\n\n\n\n

Now, substitute the values, we get<\/p>\n\n\n\n

x2\u221a2\/3 + y\/3 = 2<\/p>\n\n\n\n

2\u221a2x + y = 6<\/p>\n\n\n\n

\u2234 The equation of line in normal form is 2\u221a2x + y = 6.<\/p>\n\n\n\n

EXERCISE 23.8 PAGE NO: 23.65<\/h4>\n\n\n\n

1. A line passes through a point A (1, 2) and makes an angle of 600<\/sup> with the x\u2013axis and intercepts the line x + y = 6 at the point P. Find AP.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

(x1<\/sub>, y1<\/sub>) = A (1, 2), \u03b8 = 60\u00b0<\/p>\n\n\n\n

Let us find the distance AP.<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of the line is given by:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Here, r represents the distance of any point on the line from point A (1, 2).<\/p>\n\n\n\n

The coordinate of any point P on this line are (1 + r\/2, 2 + \u221a3r\/2)<\/p>\n\n\n\n

It is clear that, P lies on the line x + y = 6<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u2234 The value of AP is 3(\u221a3 \u2013 1)<\/p>\n\n\n\n

2. If the straight line through the point P(3, 4) makes an angle \u03c0\/6 with the x\u2013axis and meets the line 12x + 5y + 10 = 0 at Q, find the length PQ.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

(x1<\/sub>, y1<\/sub>) = A (3, 4), \u03b8 = \u03c0\/6 = 30\u00b0<\/p>\n\n\n\n

Let us find the length PQ.<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of the line is given by:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

x \u2013 \u221a3 y + 4\u221a3 \u2013 3 = 0<\/p>\n\n\n\n

Let PQ = r<\/p>\n\n\n\n

Then, the coordinate of Q are given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

3. A straight line drawn through the point A (2, 1) making an angle \u03c0\/4 with positive x\u2013axis intersects another line x + 2y + 1 = 0 in the point B. Find length AB.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

(x1<\/sub>, y1<\/sub>) = A (2, 1), \u03b8 = \u03c0\/4 = 45\u00b0<\/p>\n\n\n\n

Let us find the length AB.<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of the line is given by:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

x \u2013 y \u2013 1 = 0<\/p>\n\n\n\n

Let AB = r<\/p>\n\n\n\n

Then, the coordinate of B is given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

4. A line a drawn through A (4, \u2013 1) parallel to the line 3x \u2013 4y + 1 = 0. Find the coordinates of the two points on this line which are at a distance of 5 units from A.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

(x1<\/sub>, y1<\/sub>) = A (4, -1)<\/p>\n\n\n\n

Let us find Coordinates of the two points on this line which are at a distance of 5 units from A.<\/p>\n\n\n\n

Given: Line 3x \u2013 4y + 1 = 0<\/p>\n\n\n\n

4y = 3x + 1<\/p>\n\n\n\n

y = 3x\/4 + 1\/4<\/p>\n\n\n\n

Slope tan \u03b8 = 3\/4<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

Sin \u03b8 = 3\/5<\/p>\n\n\n\n

Cos \u03b8 = 4\/5<\/p>\n\n\n\n

The equation of the line passing through A (4, \u22121) and having slope \u00be is<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of the line is given by:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

3x \u2013 4y = 16<\/p>\n\n\n\n

Here, AP = r = \u00b1 5<\/p>\n\n\n\n

Thus, the coordinates of P are given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

x
\"https:\/\/gradeup-question-images.grdp.co\/liveData\/PROJ26150\/1549601523027278.png\"= \u00b14 + 4 and y = \u00b13 \u20131<\/p>\n\n\n\n

x = 8, 0 and y = 2, \u2013 4<\/p>\n\n\n\n

\u2234 The coordinates of the two points at a distance of 5 units from A are (8, 2) and (0, \u22124).<\/p>\n\n\n\n

5. The straight line through P(x1<\/sub>, y1<\/sub>) inclined at an angle \u03b8 with the x\u2013axis meets the line ax + by + c = 0 in Q. Find the length of PQ.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equation of the line that passes through P(x1<\/sub>, y1<\/sub>) and makes an angle of \u03b8 with the x\u2013axis.<\/p>\n\n\n\n

Let us find the length of PQ.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

The equation of the line is given by:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

EXERCISE 23.9 PAGE NO: 23.72<\/h4>\n\n\n\n

1. Reduce the equation \u221a3x + y + 2 = 0 to:
(i) slope \u2013 intercept form and find slope and y \u2013 intercept;
(ii) Intercept form and find intercept on the axes
(iii) The normal form and find p and \u03b1.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i)<\/strong> Given:<\/p>\n\n\n\n

\u221a3x + y + 2 = 0<\/p>\n\n\n\n

y = \u2013 \u221a3x \u2013 2<\/p>\n\n\n\n

This is the slope intercept form of the given line.<\/p>\n\n\n\n

\u2234 The slope = \u2013 \u221a3 and y \u2013 intercept = -2<\/p>\n\n\n\n

(ii)<\/strong> Given:<\/p>\n\n\n\n

\u221a3x + y + 2 = 0<\/p>\n\n\n\n

\u221a3x + y = -2<\/p>\n\n\n\n

Divide both sides by -2, we get<\/p>\n\n\n\n

\u221a3x\/-2 + y\/-2 = 1<\/p>\n\n\n\n

\u2234 The intercept form of the given line. Here, x \u2013 intercept = \u2013 2\/\u221a3 and y \u2013 intercept = -2<\/p>\n\n\n\n

(iii)<\/strong> Given:<\/p>\n\n\n\n

\u221a3x + y + 2 = 0<\/p>\n\n\n\n

-\u221a3x \u2013 y = 2<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

2. Reduce the following equations to the normal form and find p and \u03b1 in each case:
(i) x + \u221a3y \u2013 4 = 0
(ii) x + y + \u221a2 = 0<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>x + \u221a3y \u2013 4 = 0<\/p>\n\n\n\n

x + \u221a3y = 4<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The normal form of the given line, where p = 2, cos \u03b1 = 1\/2 and sin \u03b1 = \u221a3\/2<\/p>\n\n\n\n

\u2234 p = 2 and \u03b1 = \u03c0\/3<\/p>\n\n\n\n

(ii) <\/strong>x + y + \u221a2 = 0<\/p>\n\n\n\n

-x \u2013 y = \u221a2<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The normal form of the given line, where p = 1, cos \u03b1 = -1\/\u221a2 and sin \u03b1 = -1\/\u221a2<\/p>\n\n\n\n

\u2234 p = 1 and \u03b1 = 225o<\/sup><\/p>\n\n\n\n

3. Put the equation x\/a + y\/b = 1 the slope intercept form and find its slope and y \u2013 intercept.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: the equation is x\/a + y\/b = 1 <\/strong><\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

General equation of line y = mx + c.<\/p>\n\n\n\n

bx + ay = ab<\/p>\n\n\n\n

ay = \u2013 bx + ab<\/p>\n\n\n\n

y = -bx\/a + b<\/p>\n\n\n\n

The slope intercept form of the given line.<\/p>\n\n\n\n

\u2234 Slope = \u2013 b\/a and y \u2013 intercept = b<\/p>\n\n\n\n

4. Reduce the lines 3x \u2013 4y + 4 = 0 and 2x + 4y \u2013 5 = 0 to the normal form and hence find which line is nearer to the origin.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The normal forms of the lines 3x \u2212 4y + 4 = 0 and 2x + 4y \u2212 5 = 0.<\/p>\n\n\n\n

Let us find, in given normal form of a line, which is nearer to the origin.<\/p>\n\n\n\n

-3x + 4y = 4<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Now 2x + 4y = \u2013 5<\/p>\n\n\n\n

-2x \u2013 4y = 5<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

From equations (1) and (2):<\/p>\n\n\n\n

45 < 525<\/p>\n\n\n\n

\u2234 The line 3x \u2212 4y + 4 = 0 is nearer to the origin.<\/p>\n\n\n\n

5. Show that the origin is equidistant from the lines 4x + 3y + 10 = 0; 5x \u2013 12y + 26 = 0 and 7x + 24y = 50.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The lines 4x + 3y + 10 = 0; 5x \u2013 12y + 26 = 0 and 7x + 24y = 50.<\/p>\n\n\n\n

We need to prove that, the origin is equidistant from the lines 4x + 3y + 10 = 0; 5x \u2013 12y + 26 = 0 and 7x + 24y = 50.<\/p>\n\n\n\n

Let us write down the normal forms of the given lines.<\/p>\n\n\n\n

First line: 4x + 3y + 10 = 0<\/p>\n\n\n\n

-4x \u2013 3y = 10<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

So, p = 2<\/p>\n\n\n\n

Second line: 5x \u2212 12y + 26 = 0<\/p>\n\n\n\n

-5x + 12y = 26<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

So, p = 2<\/p>\n\n\n\n

Third line: 7x + 24y = 50<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

So, p = 2<\/p>\n\n\n\n

\u2234 The origin is equidistant from the given lines.<\/p>\n\n\n\n

EXERCISE 23.10 PAGE NO: 23.77<\/h4>\n\n\n\n

1. Find the point of intersection of the following pairs of lines:
(i) 2x \u2013 y + 3 = 0 and x + y \u2013 5 = 0<\/strong><\/p>\n\n\n\n

(ii) bx + ay = ab and ax + by = ab<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>2x \u2013 y + 3 = 0 and x + y \u2013 5 = 0<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equations of the lines are as follows:<\/p>\n\n\n\n

2x \u2212 y + 3 = 0 \u2026 (1)<\/p>\n\n\n\n

x + y \u2212 5 = 0 \u2026 (2)<\/p>\n\n\n\n

Let us find the point of intersection of pair of lines.<\/p>\n\n\n\n

By solving (1) and (2) using cross \u2013 multiplication method, we get<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

x = 2\/3 and y = 13\/3<\/p>\n\n\n\n

\u2234 The point of intersection is (2\/3, 13\/3)<\/p>\n\n\n\n

(ii) <\/strong>bx + ay = ab and ax + by = ab<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equations of the lines are as follows:<\/p>\n\n\n\n

bx + ay \u2212 ab = 0\u2026 (1)<\/p>\n\n\n\n

ax + by = ab \u21d2 ax + by \u2212 ab = 0 \u2026 (2)<\/p>\n\n\n\n

Let us find the point of intersection of pair of lines.<\/p>\n\n\n\n

By solving (1) and (2) using cross \u2013 multiplication method, we get<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u2234 The point of intersection is (ab\/a+b, ab\/a+b)<\/p>\n\n\n\n

2. Find the coordinates of the vertices of a triangle, the equations of whose sides are: (i) x + y \u2013 4 = 0, 2x \u2013 y + 3 0 and x \u2013 3y + 2 = 0<\/strong><\/p>\n\n\n\n

(ii) y (t1<\/sub> + t2<\/sub>) = 2x + 2at1<\/sub>t2<\/sub>, y (t2<\/sub> + t3<\/sub>) = 2x + 2at2<\/sub>t3<\/sub> and, y(t3<\/sub> + t1<\/sub>) = 2x + 2at1<\/sub>t3<\/sub>.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>x + y \u2013 4 = 0, 2x \u2013 y + 3 0 and x \u2013 3y + 2 = 0<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

x + y \u2212 4 = 0, 2x \u2212 y + 3 = 0 and x \u2212 3y + 2 = 0<\/p>\n\n\n\n

Let us find the point of intersection of pair of lines.<\/p>\n\n\n\n

x + y \u2212 4 = 0 \u2026 (1)<\/p>\n\n\n\n

2x \u2212 y + 3 = 0 \u2026 (2)<\/p>\n\n\n\n

x \u2212 3y + 2 = 0 \u2026 (3)<\/p>\n\n\n\n

By solving (1) and (2) using cross \u2013 multiplication method, we get<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

x = 1\/3, y = 11\/3<\/p>\n\n\n\n

Solving (1) and (3) using cross \u2013 multiplication method, we get<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

x = 5\/2, y = 3\/2<\/p>\n\n\n\n

Similarly, solving (2) and (3) using cross \u2013 multiplication method, we get<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

x = \u2013 7\/5, y = 1\/5<\/p>\n\n\n\n

\u2234 The coordinates of the vertices of the triangle are (1\/3, 11\/3), (5\/2, 3\/2) and (-7\/5, 1\/5)<\/p>\n\n\n\n

(ii) <\/strong>y (t1<\/sub> + t2<\/sub>) = 2x + 2at1<\/sub>t2<\/sub>, y (t2<\/sub> + t3<\/sub>) = 2x + 2at2<\/sub>t3<\/sub> and, y(t3<\/sub> + t1<\/sub>) = 2x + 2at1<\/sub>t3<\/sub>.<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

y (t1<\/sub> + t2<\/sub>) = 2x + 2a t1<\/sub>t2<\/sub>, y (t2<\/sub> + t3<\/sub>) = 2x + 2a t2<\/sub>t3<\/sub> and y (t3<\/sub> + t1<\/sub>) = 2x + 2a t1<\/sub>t3<\/sub><\/p>\n\n\n\n

Let us find the point of intersection of pair of lines.<\/p>\n\n\n\n

2x \u2212 y (t1<\/sub> + t2<\/sub>) + 2a t1<\/sub>t2<\/sub> = 0 \u2026 (1)<\/p>\n\n\n\n

2x \u2212 y (t2<\/sub> + t3<\/sub>) + 2a t2<\/sub>t3<\/sub> = 0 \u2026 (2)<\/p>\n\n\n\n

2x \u2212 y (t3<\/sub> + t1<\/sub>) + 2a t1<\/sub>t3<\/sub> = 0 \u2026 (3)<\/p>\n\n\n\n

By solving (1) and (2) using cross \u2013 multiplication method, we get<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving (1) and (3) using cross \u2013 multiplication method, we get<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving (2) and (3) using cross \u2013 multiplication method, we get<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u2234 The coordinates of the vertices of the triangle are (at2<\/sup>1<\/sub>, 2at1<\/sub>), (at2<\/sup>2<\/sub>, 2at2<\/sub>) and (at2<\/sup>3<\/sub>, 2at3<\/sub>).<\/p>\n\n\n\n

3. Find the area of the triangle formed by the lines
y = m1<\/sub>x + c1<\/sub>, y = m2<\/sub>x + c2<\/sub> and x = 0<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

y = m1<\/sub>x + c1<\/sub> \u2026 (1)<\/p>\n\n\n\n

y = m2<\/sub>x + c2<\/sub> \u2026 (2)<\/p>\n\n\n\n

x = 0 \u2026 (3)<\/p>\n\n\n\n

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.<\/p>\n\n\n\n

Solving (1) and (2), we get<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving (1) and (3):<\/p>\n\n\n\n

x = 0, y = c1<\/sub><\/p>\n\n\n\n

Thus, AB and CA intersect at A 0,c1<\/sub>.<\/p>\n\n\n\n

Similarly, solving (2) and (3):<\/p>\n\n\n\n

x = 0, y = c2<\/sub><\/p>\n\n\n\n

Thus, BC and CA intersect at C 0,c2<\/sub>.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

4. Find the equations of the medians of a triangle, the equations of whose sides are:
3x + 2y + 6 = 0, 2x \u2013 5y + 4 = 0 and x \u2013 3y \u2013 6 = 0<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

3x + 2y + 6 = 0 \u2026 (1)<\/p>\n\n\n\n

2x \u2212 5y + 4 = 0 \u2026 (2)<\/p>\n\n\n\n

x \u2212 3y \u2212 6 = 0 \u2026 (3)<\/p>\n\n\n\n

Let us assume, in triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.<\/p>\n\n\n\n

Solving equations (1) and (2), we get<\/p>\n\n\n\n

x = \u22122, y = 0<\/p>\n\n\n\n

Thus, AB and BC intersect at B (\u22122, 0).<\/p>\n\n\n\n

Now, solving (1) and (3), we get<\/p>\n\n\n\n

x = \u2013 6\/11, y = \u2013 24\/11<\/p>\n\n\n\n

Thus, AB and CA intersect at A (-6\/11, -24\/11)<\/p>\n\n\n\n

Similarly, solving (2) and (3), we get<\/p>\n\n\n\n

x = \u221242, y = \u221216<\/p>\n\n\n\n

Thus, BC and CA intersect at C (\u221242, \u221216).<\/p>\n\n\n\n

Now, let D, E and F be the midpoints the sides BC, CA and AB, respectively.<\/p>\n\n\n\n

Then, we have:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

\u2234 The equations of the medians of a triangle are: 41x \u2013 112y \u2013 70 = 0,<\/p>\n\n\n\n

16x \u2013 59y \u2013 120 = 0, 25x \u2013 53y + 50 = 0<\/p>\n\n\n\n

5. Prove that the lines y = \u221a3x + 1, y = 4 and y = -\u221a3x + 2 form an equilateral triangle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

y = \u221a3x + 1\u2026\u2026 (1)<\/p>\n\n\n\n

y = 4 \u2026\u2026. (2)<\/p>\n\n\n\n

y = \u2013 \u221a3x + 2\u2026\u2026. (3)<\/p>\n\n\n\n

Let us assume in triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.<\/p>\n\n\n\n

By solving equations (1) and (2), we get<\/p>\n\n\n\n

x = \u221a3, y = 4<\/p>\n\n\n\n

Thus, AB and BC intersect at B(\u221a3,4)<\/p>\n\n\n\n

Now, solving equations (1) and (3), we get<\/p>\n\n\n\n

x = 1\/2\u221a3, y = 3\/2<\/p>\n\n\n\n

Thus, AB and CA intersect at A (1\/2\u221a3, 3\/2)<\/p>\n\n\n\n

Similarly, solving equations (2) and (3), we get<\/p>\n\n\n\n

x = -2\/\u221a3, y = 4<\/p>\n\n\n\n

Thus, BC and AC intersect at C (-2\/\u221a3,4)<\/p>\n\n\n\n

Now, we have:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Hence proved, the given lines form an equilateral triangle.<\/p>\n\n\n\n

EXERCISE 23.11 PAGE NO: 23.83<\/h4>\n\n\n\n

1. Prove that the following sets of three lines are concurrent:
(i) 15x \u2013 18y + 1 = 0, 12x + 10y \u2013 3 = 0 and 6x + 66y \u2013 11 = 0<\/strong><\/p>\n\n\n\n

(ii) 3x \u2013 5y \u2013 11 = 0, 5x + 3y \u2013 7 = 0 and x + 2y = 0<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>15x \u2013 18y + 1 = 0, 12x + 10y \u2013 3 = 0 and 6x + 66y \u2013 11 = 0<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

15x \u2013 18y + 1 = 0 \u2026\u2026 (i)<\/p>\n\n\n\n

12x + 10y \u2013 3 = 0 \u2026\u2026 (ii)<\/p>\n\n\n\n

6x + 66y \u2013 11 = 0 \u2026\u2026 (iii)<\/p>\n\n\n\n

Now, consider the following determinant:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

=> 1320 \u2013 2052 + 732 = 0<\/p>\n\n\n\n

Hence proved, the given lines are concurrent.<\/p>\n\n\n\n

(ii) <\/strong>3x \u2013 5y \u2013 11 = 0, 5x + 3y \u2013 7 = 0 and x + 2y = 0<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

3x \u2212 5y \u2212 11 = 0 \u2026\u2026 (i)<\/p>\n\n\n\n

5x + 3y \u2212 7 = 0 \u2026\u2026 (ii)<\/p>\n\n\n\n

x + 2y = 0 \u2026\u2026 (iii)<\/p>\n\n\n\n

Now, consider the following determinant:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Hence, the given lines are concurrent.<\/p>\n\n\n\n

2. For what value of \u03bb are the three lines 2x \u2013 5y + 3 = 0, 5x \u2013 9y + \u03bb = 0 and x \u2013 2y + 1 = 0 concurrent?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

2x \u2212 5y + 3 = 0 \u2026 (1)<\/p>\n\n\n\n

5x \u2212 9y + \u03bb = 0 \u2026 (2)<\/p>\n\n\n\n

x \u2212 2y + 1 = 0 \u2026 (3)<\/p>\n\n\n\n

It is given that the three lines are concurrent.<\/p>\n\n\n\n

Now, consider the following determinant:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

2(-9 + 2\u03bb) + 5(5 \u2013 \u03bb) + 3(-10 + 9) = 0<\/p>\n\n\n\n

-18 + 4\u03bb + 25 \u2013 5\u03bb \u2013 3 = 0<\/p>\n\n\n\n

\u03bb = 4<\/p>\n\n\n\n

\u2234 The value of \u03bb is 4.<\/p>\n\n\n\n

3. Find the conditions that the straight lines y = m1<\/sub>x + c1<\/sub>, y = m2<\/sub>x + c2<\/sub> and y = m3<\/sub>x + c3<\/sub> may meet in a point.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

m1<\/sub>x \u2013 y + c1<\/sub> = 0 \u2026 (1)<\/p>\n\n\n\n

m2<\/sub>x \u2013 y + c2<\/sub> = 0 \u2026 (2)<\/p>\n\n\n\n

m3<\/sub>x \u2013 y + c3<\/sub> = 0 \u2026 (3)<\/p>\n\n\n\n

It is given that the three lines are concurrent.<\/p>\n\n\n\n

Now, consider the following determinant:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

m1<\/sub>(-c3<\/sub> + c2<\/sub>) + 1(m2<\/sub>c3<\/sub>-m3<\/sub>c2<\/sub>) + c1<\/sub>(-m2<\/sub> + m3<\/sub>) = 0<\/p>\n\n\n\n

m1<\/sub>(c2<\/sub>-c3<\/sub>) + m2<\/sub>(c3<\/sub>-c1<\/sub>) + m3<\/sub>(c1<\/sub>-c2<\/sub>) = 0<\/p>\n\n\n\n

\u2234 The required condition is m1<\/sub>(c2<\/sub>-c3<\/sub>) + m2<\/sub>(c3<\/sub>-c1<\/sub>) + m3<\/sub>(c1<\/sub>-c2<\/sub>) = 0<\/p>\n\n\n\n

4. If the lines p1<\/sub>x + q1<\/sub>y = 1, p2<\/sub>x + q2<\/sub>y = 1 and p3<\/sub>x + q3<\/sub>y = 1 be concurrent, show that the points (p1<\/sub>, q1<\/sub>), (p2<\/sub>, q2<\/sub>) and (p3<\/sub>, q3<\/sub>) are collinear.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

p1<\/sub>x + q1<\/sub>y = 1<\/p>\n\n\n\n

p2<\/sub>x + q2<\/sub>y = 1<\/p>\n\n\n\n

p3<\/sub>x + q3<\/sub>y = 1<\/p>\n\n\n\n

The given lines can be written as follows:<\/p>\n\n\n\n

p1<\/sub> x + q1<\/sub> y \u2013 1 = 0 \u2026 (1)<\/p>\n\n\n\n

p2<\/sub> x + q2<\/sub> y \u2013 1 = 0 \u2026 (2)<\/p>\n\n\n\n

p3<\/sub> x + q3<\/sub> y \u2013 1 = 0 \u2026 (3)<\/p>\n\n\n\n

It is given that the three lines are concurrent.<\/p>\n\n\n\n

Now, consider the following determinant:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Hence proved, the given three points, (p1<\/sub>, q1<\/sub>), (p2<\/sub>, q2<\/sub>) and (p3<\/sub>, q3<\/sub>) are collinear.<\/p>\n\n\n\n

5. Show that the straight lines L1<\/sub> = (b + c)x + ay + 1 = 0, L2<\/sub> = (c + a)x + by + 1 = 0 and L3<\/sub> = (a + b)x + cy + 1 = 0 are concurrent.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

L1<\/sub> = (b + c)x + ay + 1 = 0<\/p>\n\n\n\n

L2<\/sub> = (c + a)x + by + 1 = 0<\/p>\n\n\n\n

L3<\/sub> = (a + b)x + cy + 1 = 0<\/p>\n\n\n\n

The given lines can be written as follows:<\/p>\n\n\n\n

(b + c) x + ay + 1 = 0 \u2026 (1)<\/p>\n\n\n\n

(c + a) x + by + 1 = 0 \u2026 (2)<\/p>\n\n\n\n

(a + b) x + cy + 1 = 0 \u2026 (3)<\/p>\n\n\n\n

Consider the following determinant.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Hence proved, the given lines are concurrent.<\/p>\n\n\n\n

EXERCISE 23.12 PAGE NO: 23.92<\/h4>\n\n\n\n

1. Find the equation of a line passing through the point (2, 3) and parallel to the line 3x\u2013 4y + 5 = 0.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equation is parallel to 3x \u2212 4y + 5 = 0 and pass through (2, 3)<\/p>\n\n\n\n

The equation of the line parallel to 3x \u2212 4y + 5 = 0 is<\/p>\n\n\n\n

3x \u2013 4y + \u03bb = 0,<\/p>\n\n\n\n

Where, \u03bb is a constant.<\/p>\n\n\n\n

It passes through (2, 3).<\/p>\n\n\n\n

Substitute the values in above equation, we get<\/p>\n\n\n\n

3 (2) \u2013 4 (3) + \u03bb = 0<\/p>\n\n\n\n

6 \u2013 12 + \u03bb = 0<\/p>\n\n\n\n

\u03bb = 6<\/p>\n\n\n\n

Now, substitute the value of \u03bb = 6 in 3x \u2013 4y + \u03bb = 0, we get<\/p>\n\n\n\n

3x \u2212 4y + 6<\/p>\n\n\n\n

\u2234 The required line is 3x \u2212 4y + 6 = 0.<\/p>\n\n\n\n

2. Find the equation of a line passing through (3, -2) and perpendicular to the line x \u2013 3y + 5 = 0.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equation is perpendicular to x \u2013 3y + 5 = 0 and passes through (3,-2)<\/p>\n\n\n\n

The equation of the line perpendicular to x \u2212 3y + 5 = 0 is<\/p>\n\n\n\n

3x + y + \u03bb = 0,<\/p>\n\n\n\n

Where, \u03bb is a constant.<\/p>\n\n\n\n

It passes through (3, \u2212 2).<\/p>\n\n\n\n

Substitute the values in above equation, we get<\/p>\n\n\n\n

3 (3) + (-2) + \u03bb = 0<\/p>\n\n\n\n

9 \u2013 2 + \u03bb = 0<\/p>\n\n\n\n

\u03bb = \u2013 7<\/p>\n\n\n\n

Now, substitute the value of \u03bb = \u2212 7 in 3x + y + \u03bb = 0, we get<\/p>\n\n\n\n

3x + y \u2013 7 = 0<\/p>\n\n\n\n

\u2234 The required line is 3x + y \u2013 7 = 0.<\/p>\n\n\n\n

3. Find the equation of the perpendicular bisector of the line joining the points (1, 3) and (3, 1).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

A (1, 3) and B (3, 1) be the points joining the perpendicular bisector<\/p>\n\n\n\n

Let C be the midpoint of AB.<\/p>\n\n\n\n

So, coordinates of C = [(1+3)\/2, (3+1)\/2]<\/p>\n\n\n\n

= (2, 2)<\/p>\n\n\n\n

Slope of AB = [(1-3) \/ (3-1)]<\/p>\n\n\n\n

= -1<\/p>\n\n\n\n

Slope of the perpendicular bisector of AB = 1<\/p>\n\n\n\n

Thus, the equation of the perpendicular bisector of AB is given as,<\/p>\n\n\n\n

y \u2013 2 = 1(x \u2013 2)<\/p>\n\n\n\n

y = x<\/p>\n\n\n\n

x \u2013 y = 0<\/p>\n\n\n\n

\u2234 The required equation is y = x.<\/p>\n\n\n\n

4. Find the equations of the altitudes of a \u0394ABC whose vertices are A (1, 4), B (-3, 2) and C (-5, -3).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The vertices of \u2206ABC are A (1, 4), B (\u2212 3, 2) and C (\u2212 5, \u2212 3).<\/p>\n\n\n\n

Now let us find the slopes of \u2206ABC.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Slope of AB = [(2 \u2013 4) \/ (-3-1)]<\/p>\n\n\n\n

= \u00bd<\/p>\n\n\n\n

Slope of BC = [(-3 \u2013 2) \/ (-5+3)]<\/p>\n\n\n\n

= 5\/2<\/p>\n\n\n\n

Slope of CA = [(4 + 3) \/ (1 + 5)]<\/p>\n\n\n\n

= 7\/6<\/p>\n\n\n\n

Thus, we have:<\/p>\n\n\n\n

Slope of CF = -2<\/p>\n\n\n\n

Slope of AD = -2\/5<\/p>\n\n\n\n

Slope of BE = -6\/7<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Equation of CF is:<\/p>\n\n\n\n

y + 3 = -2(x + 5)<\/p>\n\n\n\n

y + 3 = -2x \u2013 10<\/p>\n\n\n\n

2x + y + 13 = 0<\/p>\n\n\n\n

Equation of AD is:<\/p>\n\n\n\n

y \u2013 4 = (-2\/5) (x \u2013 1)<\/p>\n\n\n\n

5y \u2013 20 = -2x + 2<\/p>\n\n\n\n

2x + 5y \u2013 22 = 0<\/p>\n\n\n\n

Equation of BE is:<\/p>\n\n\n\n

y \u2013 2 = (-6\/7) (x + 3)<\/p>\n\n\n\n

7y \u2013 14 = -6x \u2013 18<\/p>\n\n\n\n

6x + 7y + 4 = 0<\/p>\n\n\n\n

\u2234 The required equations are 2x + y + 13 = 0, 2x + 5y \u2013 22 = 0, 6x + 7y + 4 = 0.<\/p>\n\n\n\n

5. Find the equation of a line which is perpendicular to the line\u221a3x \u2013 y + 5 = 0 and which cuts off an intercept of 4 units with the negative direction of y-axis.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equation is perpendicular to \u221a3x \u2013 y + 5 = 0 equation and cuts off an intercept of 4 units with the negative direction of y-axis.<\/p>\n\n\n\n

The line perpendicular to \u221a3x \u2013 y + 5 = 0 is x + \u221a3y + \u03bb = 0<\/p>\n\n\n\n

It is given that the line x + \u221a3y + \u03bb = 0 cuts off an intercept of 4 units with the negative direction of the y-axis.<\/p>\n\n\n\n

This means that the line passes through (0,-4).<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

Let us substitute the values in the equation x + \u221a3y + \u03bb = 0, we get<\/p>\n\n\n\n

0 \u2013 \u221a3 (4) + \u03bb = 0<\/p>\n\n\n\n

\u03bb = 4\u221a3<\/p>\n\n\n\n

Now, substitute the value of \u03bb back, we get<\/p>\n\n\n\n

x + \u221a3y + 4\u221a3 = 0<\/p>\n\n\n\n

\u2234 The required equation of line is x + \u221a3y + 4\u221a3 = 0.<\/p>\n\n\n\n

EXERCISE 23.13 PAGE NO: 23.99<\/h4>\n\n\n\n

1. Find the angles between each of the following pairs of straight lines:
(i) 3x + y + 12 = 0 and x + 2y \u2013 1 = 0<\/strong><\/p>\n\n\n\n

(ii) 3x \u2013 y + 5 = 0 and x \u2013 3y + 1 = 0<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>3x + y + 12 = 0 and x + 2y \u2013 1 = 0<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equations of the lines are<\/p>\n\n\n\n

3x + y + 12 = 0 \u2026 (1)<\/p>\n\n\n\n

x + 2y \u2212 1 = 0 \u2026 (2)<\/p>\n\n\n\n

Let m1<\/sub> and m2<\/sub> be the slopes of these lines.<\/p>\n\n\n\n

m1<\/sub> = -3, m2<\/sub> = -1\/2<\/p>\n\n\n\n

Let \u03b8 be the angle between the lines.<\/p>\n\n\n\n

Then, by using the formula<\/p>\n\n\n\n

tan \u03b8 = [(m1<\/sub> \u2013 m2<\/sub>) \/ (1 + m1<\/sub>m2<\/sub>)]<\/p>\n\n\n\n

= [(-3 + 1\/2) \/ (1 + 3\/2)]<\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

\u03b8 = \u03c0\/4 or 45o<\/sup><\/p>\n\n\n\n

\u2234 The acute angle between the lines is 45\u00b0<\/p>\n\n\n\n

(ii) <\/strong>3x \u2013 y + 5 = 0 and x \u2013 3y + 1 = 0<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equations of the lines are<\/p>\n\n\n\n

3x \u2212 y + 5 = 0 \u2026 (1)<\/p>\n\n\n\n

x \u2212 3y + 1 = 0 \u2026 (2)<\/p>\n\n\n\n

Let m1<\/sub> and m2<\/sub> be the slopes of these lines.<\/p>\n\n\n\n

m1<\/sub> = 3, m2<\/sub> = 1\/3<\/p>\n\n\n\n

Let \u03b8 be the angle between the lines.<\/p>\n\n\n\n

Then, by using the formula<\/p>\n\n\n\n

tan \u03b8 = [(m1<\/sub> \u2013 m2<\/sub>) \/ (1 + m1<\/sub>m2<\/sub>)]<\/p>\n\n\n\n

= [(3 + 1\/3) \/ (1 + 1)]<\/p>\n\n\n\n

= 4\/3<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

\u03b8 = tan-1<\/sup> (4\/3)<\/p>\n\n\n\n

\u2234 The acute angle between the lines is tan-1<\/sup> (4\/3).<\/p>\n\n\n\n

2. Find the acute angle between the lines 2x \u2013 y + 3 = 0 and x + y + 2 = 0.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equations of the lines are<\/p>\n\n\n\n

2x \u2212 y + 3 = 0 \u2026 (1)<\/p>\n\n\n\n

x + y + 2 = 0 \u2026 (2)<\/p>\n\n\n\n

Let m1<\/sub> and m2<\/sub> be the slopes of these lines.<\/p>\n\n\n\n

m1<\/sub> = 2, m2<\/sub> = -1<\/p>\n\n\n\n

Let \u03b8 be the angle between the lines.<\/p>\n\n\n\n

Then, by using the formula<\/p>\n\n\n\n

tan \u03b8 = [(m1<\/sub> \u2013 m2<\/sub>) \/ (1 + m1<\/sub>m2<\/sub>)]<\/p>\n\n\n\n

= [(2 + 1) \/ (1 + 2)]<\/p>\n\n\n\n

= 3<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

\u03b8 = tan-1<\/sup> (3)<\/p>\n\n\n\n

\u2234 The acute angle between the lines is tan-1<\/sup> (3).<\/p>\n\n\n\n

3. Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

To prove:<\/p>\n\n\n\n

The points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram<\/p>\n\n\n\n

Let us assume the points, A (2, \u2212 1), B (0, 2), C (2, 3) and D (4, 0) be the vertices.<\/p>\n\n\n\n

Now, let us find the slopes<\/p>\n\n\n\n

Slope of AB = [(2+1) \/ (0-2)]<\/p>\n\n\n\n

= -3\/2<\/p>\n\n\n\n

Slope of BC = [(3-2) \/ (2-0)]<\/p>\n\n\n\n

= \u00bd<\/p>\n\n\n\n

Slope of CD = [(0-3) \/ (4-2)]<\/p>\n\n\n\n

= -3\/2<\/p>\n\n\n\n

Slope of DA = [(-1-0) \/ (2-4)]<\/p>\n\n\n\n

= \u00bd<\/p>\n\n\n\n

Thus, AB is parallel to CD and BC is parallel to DA.<\/p>\n\n\n\n

Hence proved, the given points are the vertices of a parallelogram.<\/p>\n\n\n\n

Now, let us find the angle between the diagonals AC and BD.<\/p>\n\n\n\n

Let m1<\/sub> and m2<\/sub> be the slopes of AC and BD, respectively.<\/p>\n\n\n\n

m1<\/sub> = [(3+1) \/ (2-2)]<\/p>\n\n\n\n

= \u221e<\/p>\n\n\n\n

m2<\/sub> = [(0-2) \/ (4-0)]<\/p>\n\n\n\n

= -1\/2<\/p>\n\n\n\n

Thus, the diagonal AC is parallel to the y-axis.<\/p>\n\n\n\n

\u2220ODB = tan-1<\/sup> (1\/2)<\/p>\n\n\n\n

In triangle MND,<\/p>\n\n\n\n

\u2220DMN = \u03c0\/2 \u2013 tan-1<\/sup> (1\/2)<\/p>\n\n\n\n

\u2234 The angle between the diagonals is \u03c0\/2 \u2013 tan-1<\/sup> (1\/2).<\/p>\n\n\n\n

4. Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Points (2, 0), (0, 3) and the line x + y = 1.<\/p>\n\n\n\n

Let us assume A (2, 0), B (0, 3) be the given points.<\/p>\n\n\n\n

Now, let us find the slopes<\/p>\n\n\n\n

Slope of AB = m1<\/sub><\/p>\n\n\n\n

= [(3-0) \/ (0-2)]<\/p>\n\n\n\n

= -3\/2<\/p>\n\n\n\n

Slope of the line x + y = 1 is -1<\/p>\n\n\n\n

\u2234 m2<\/sub> = -1<\/p>\n\n\n\n

Let \u03b8 be the angle between the line joining the points (2, 0), (0, 3) and the line x + y =<\/p>\n\n\n\n

tan \u03b8 = |[(m1<\/sub> \u2013 m2<\/sub>) \/ (1 + m1<\/sub>m2<\/sub>)]|<\/p>\n\n\n\n

= [(-3\/2 + 1) \/ (1 + 3\/2)]<\/p>\n\n\n\n

= 1\/5<\/p>\n\n\n\n

\u03b8 = tan-1<\/sup> (1\/5)<\/p>\n\n\n\n

\u2234 The acute angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1 is tan-1<\/sup> (1\/5).<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We need to prove:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Let us assume A (x1<\/sub>, y1<\/sub>) and B (x2<\/sub>, y2<\/sub>) be the given points and O be the origin.<\/p>\n\n\n\n

Slope of OA = m1<\/sub> = y1\u00d71<\/sub><\/p>\n\n\n\n

Slope of OB = m2<\/sub> = y2\u00d72<\/sub><\/p>\n\n\n\n

It is given that \u03b8 is the angle between lines OA and OB.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

EXERCISE 23.14 PAGE NO: 23.102<\/h4>\n\n\n\n

1. Find the values of \u03b1 so that the point P(\u03b1 2<\/sup>, \u03b1) lies inside or on the triangle formed by the lines x \u2013 5y + 6 = 0, x \u2013 3y + 2 = 0 and x \u2013 2y \u2013 3 = 0.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

x \u2013 5y + 6 = 0, x \u2013 3y + 2 = 0 and x \u2013 2y \u2013 3 = 0 forming a triangle and point P(\u03b12<\/sup>, \u03b1) lies inside or on the triangle<\/p>\n\n\n\n

Let ABC be the triangle of sides AB, BC and CA whose equations are x \u2212 5y + 6 = 0, x \u2212 3y + 2 = 0 and x \u2212 2y \u2212 3 = 0, respectively.<\/p>\n\n\n\n

On solving the equations, we get A (9, 3), B (4, 2) and C (13, 5) as the coordinates of the vertices.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

It is given that point P (\u03b12<\/sup>, \u03b1) lies either inside or on the triangle. The three conditions are given below.<\/p>\n\n\n\n

(i) A and P must lie on the same side of BC.<\/p>\n\n\n\n

(ii) B and P must lie on the same side of AC.<\/p>\n\n\n\n

(iii) C and P must lie on the same side of AB.<\/p>\n\n\n\n

If A and P lie on the same side of BC, then<\/p>\n\n\n\n

(9 \u2013 9 + 2)(\u03b12<\/sup> \u2013 3\u03b1 + 2) \u22650<\/p>\n\n\n\n

(\u03b1 \u2013 2)(\u03b1 \u2013 1) \u2265 0<\/p>\n\n\n\n

\u03b1 \u2208 (- \u221e, 1 ] \u222a [ 2, \u221e) \u2026 (1)<\/p>\n\n\n\n

If B and P lie on the same side of AC, then<\/p>\n\n\n\n

(4 \u2013 4 \u2013 3) (\u03b12<\/sup> \u2013 2\u03b1 \u2013 3) \u2265 0<\/p>\n\n\n\n

(\u03b1 \u2013 3)(\u03b1 + 1) \u2264 0<\/p>\n\n\n\n

\u03b1 \u2208 [- 1, 3] \u2026 (2)<\/p>\n\n\n\n

If C and P lie on the same side of AB, then<\/p>\n\n\n\n

(13 \u2013 25 + 6)(\u03b12<\/sup> \u2013 5\u03b1 + 6) \u22650<\/p>\n\n\n\n

(\u03b1 \u2013 3)(\u03b1 \u2013 2) \u2264 0<\/p>\n\n\n\n

\u03b1 \u2208 [ 2, 3] \u2026 (3)<\/p>\n\n\n\n

From equations (1), (2) and (3), we get<\/p>\n\n\n\n

\u03b1\u2208 [2, 3]<\/p>\n\n\n\n

\u2234 \u03b1\u2208 [2, 3]<\/p>\n\n\n\n

2. Find the values of the parameter a so that the point (a, 2) is an interior point of the triangle formed by the lines x + y \u2013 4 = 0, 3x \u2013 7y \u2013 8 = 0 and 4x \u2013 y \u2013 31 = 0.<\/strong><\/p>\n\n\n\n

Solutions:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

x + y \u2013 4 = 0, 3x \u2013 7y \u2013 8 = 0 and 4x \u2013 y \u2013 31 = 0 forming a triangle and point (a, 2)is an interior point of the triangle<\/p>\n\n\n\n

Let ABC be the triangle of sides AB, BC and CA whose equations are x + y \u2212 4 = 0, 3x \u2212 7y \u2212 8 = 0 and 4x \u2212 y \u2212 31 = 0, respectively.<\/p>\n\n\n\n

On solving them, we get A (7, \u2013 3), B (18\/5, 2\/5) and C (209\/25, 61\/25) as the coordinates of the vertices.<\/p>\n\n\n\n

Let P (a, 2) be the given point.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

It is given that point P (a, 2) lies inside the triangle. So, we have the following:<\/p>\n\n\n\n

(i) A and P must lie on the same side of BC.<\/p>\n\n\n\n

(ii) B and P must lie on the same side of AC.<\/p>\n\n\n\n

(iii) C and P must lie on the same side of AB.<\/p>\n\n\n\n

Thus, if A and P lie on the same side of BC, then<\/p>\n\n\n\n

21 + 21 \u2013 8 \u2013 3a \u2013 14 \u2013 8 > 0<\/p>\n\n\n\n

a > 22\/3 \u2026 (1)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

From (1), (2) and (3), we get:<\/p>\n\n\n\n

A \u2208 (22\/3, 33\/4)<\/p>\n\n\n\n

\u2234 A \u2208 (22\/3, 33\/4)<\/p>\n\n\n\n

3. Determine whether the point (-3, 2) lies inside or outside the triangle whose sides are given by the equations x + y \u2013 4 = 0, 3x \u2013 7y + 8 = 0, 4x \u2013 y \u2013 31 = 0.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

x + y \u2013 4 = 0, 3x \u2013 7y + 8 = 0, 4x \u2013 y \u2013 31 = 0 forming a triangle and point (-3, 2)<\/p>\n\n\n\n

Let ABC be the triangle of sides AB, BC and CA, whose equations x + y \u2212 4 = 0, 3x \u2212 7y + 8 = 0 and 4x \u2212 y \u2212 31 = 0, respectively.<\/p>\n\n\n\n

On solving them, we get A (7, \u2013 3), B (2, 2) and C (9, 5) as the coordinates of the vertices.<\/p>\n\n\n\n

Let P (\u2212 3, 2) be the given point.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The given point P (\u2212 3, 2) will lie inside the triangle ABC, if<\/p>\n\n\n\n

(i) A and P lies on the same side of BC<\/p>\n\n\n\n

(ii) B and P lies on the same side of AC<\/p>\n\n\n\n

(iii) C and P lies on the same side of AB<\/p>\n\n\n\n

Thus, if A and P lie on the same side of BC, then<\/p>\n\n\n\n

21 + 21 + 8 \u2013 9 \u2013 14 + 8 > 0<\/p>\n\n\n\n

50 \u00d7 \u2013 15 > 0<\/p>\n\n\n\n

-750 > 0,<\/p>\n\n\n\n

This is false<\/p>\n\n\n\n

\u2234 The point (\u22123, 2) lies outside triangle ABC.<\/p>\n\n\n\n

EXERCISE 23.15 PAGE NO: 23.107<\/h4>\n\n\n\n

1. Find the distance of the point (4, 5) from the straight line 3x \u2013 5y + 7 = 0.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The line: 3x \u2013 5y + 7 = 0<\/p>\n\n\n\n

Comparing ax + by + c = 0 and 3x \u2212 5y + 7 = 0, we get:<\/p>\n\n\n\n

a = 3, b = \u2212 5 and c = 7<\/p>\n\n\n\n

So, the distance of the point (4, 5) from the straight line 3x \u2212 5y + 7 = 0 is<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u2234 The required distance is 6\/\u221a<\/strong>34<\/p>\n\n\n\n

2. Find the perpendicular distance of the line joining the points (cos \u03b8, sin \u03b8) and (cos \u03d5, sin \u03d5) from the origin.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The points (cos \u03b8, sin \u03b8) and (cos \u03d5, sin \u03d5) from the origin.<\/p>\n\n\n\n

The equation of the line joining the points (cos \u03b8, sin \u03b8) and (cos \u03d5, sin \u03d5) is given below:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

3. Find the length of the perpendicular from the origin to the straight line joining the two points whose coordinates are (a cos \u03b1, a sin \u03b1) and (a cos \u03b2, a sin \u03b2).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Coordinates are (a cos \u03b1, a sin \u03b1) and (a cos \u03b2, a sin \u03b2).<\/p>\n\n\n\n

Equation of the line passing through (a cos \u03b1, a sin \u03b1) and (a cos \u03b2, a sin \u03b2) is<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

4. Show that the perpendicular let fall from any point on the straight line 2x + 11y \u2013 5 = 0 upon the two straight lines 24x + 7y = 20 and 4x \u2013 3y \u2013 2 = 0 are equal to each other.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The lines 24x + 7y = 20 and 4x \u2013 3y \u2013 2 = 0<\/p>\n\n\n\n

Let us assume, P(a, b) be any point on 2x + 11y \u2212 5 = 0<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

2a + 11b \u2212 5 = 0<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

5. Find the distance of the point of intersection of the lines 2x + 3y = 21 and 3x \u2013 4y + 11 = 0 from the line 8x + 6y + 5 = 0.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The lines 2x + 3y = 21 and 3x \u2013 4y + 11 = 0<\/p>\n\n\n\n

Solving the lines 2x + 3y = 21 and 3x \u2212 4y + 11 = 0 we get:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

x = 3, y = 5<\/p>\n\n\n\n

So, the point of intersection of 2x + 3y = 21 and 3x \u2212 4y + 11 = 0 is (3, 5).<\/p>\n\n\n\n

Now, the perpendicular distance d of the line 8x + 6y + 5 = 0 from the point (3, 5) is<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u2234 The distance is 59\/10.<\/p>\n\n\n\n

EXERCISE 23.16 PAGE NO: 23.114<\/h4>\n\n\n\n

1. Determine the distance between the following pair of parallel lines:
(i) 4x \u2013 3y \u2013 9 = 0 and 4x \u2013 3y \u2013 24 = 0<\/strong><\/p>\n\n\n\n

(ii) 8x + 15y \u2013 34 = 0 and 8x + 15y + 31 = 0<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>4x \u2013 3y \u2013 9 = 0 and 4x \u2013 3y \u2013 24 = 0<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The parallel lines are<\/p>\n\n\n\n

4x \u2212 3y \u2212 9 = 0 \u2026 (1)<\/p>\n\n\n\n

4x \u2212 3y \u2212 24 = 0 \u2026 (2)<\/p>\n\n\n\n

Let d be the distance between the given lines.<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u2234 The distance between givens parallel line is 3units.<\/p>\n\n\n\n

(ii) <\/strong>8x + 15y \u2013 34 = 0 and 8x + 15y + 31 = 0<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The parallel lines are<\/p>\n\n\n\n

8x + 15y \u2212 34 = 0 \u2026 (1)<\/p>\n\n\n\n

8x + 15y + 31 = 0 \u2026 (2)<\/p>\n\n\n\n

Let d be the distance between the given lines.<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u2234 The distance between givens parallel line is 65\/17 units.<\/p>\n\n\n\n

2. The equations of two sides of a square are 5x \u2013 12y \u2013 65 = 0 and 5x \u2013 12y + 26 = 0. Find the area of the square.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Two side of square are 5x \u2013 12y \u2013 65 = 0 and 5x \u2013 12y + 26 = 0<\/p>\n\n\n\n

The sides of a square are<\/p>\n\n\n\n

5x \u2212 12y \u2212 65 = 0 \u2026 (1)<\/p>\n\n\n\n

5x \u2212 12y + 26 = 0 \u2026 (2)<\/p>\n\n\n\n

We observe that lines (1) and (2) are parallel.<\/p>\n\n\n\n

So, the distance between them will give the length of the side of the square.<\/p>\n\n\n\n

Let d be the distance between the given lines.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u2234 Area of the square = 72<\/sup> = 49 square units<\/p>\n\n\n\n

3. Find the equation of two straight lines which are parallel to x + 7y + 2 = 0 and at unit distance from the point (1, -1).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equation is parallel to x + 7y + 2 = 0 and at unit distance from the point (1, -1)<\/p>\n\n\n\n

The equation of given line is<\/p>\n\n\n\n

x + 7y + 2 = 0 \u2026 (1)<\/p>\n\n\n\n

The equation of a line parallel to line x + 7y + 2 = 0 is given below:<\/p>\n\n\n\n

x + 7y + \u03bb = 0 \u2026 (2)<\/p>\n\n\n\n

The line x + 7y + \u03bb = 0 is at a unit distance from the point (1, \u2212 1).<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

1 =<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u03bb \u2013 6 = \u00b1<\/strong> 5\u221a<\/strong>2<\/p>\n\n\n\n

\u03bb = 6 + 5\u221a<\/strong>2, 6 \u2013 5\u221a<\/strong>2<\/p>\n\n\n\n

now, substitute the value of \u03bb back in equation x + 7y + \u03bb = 0, we get<\/p>\n\n\n\n

x + 7y + 6 + 5\u221a<\/strong>2 = 0 and x + 7y + 6 \u2013 5\u221a<\/strong>2<\/p>\n\n\n\n

\u2234 The required lines:<\/p>\n\n\n\n

x + 7y + 6 + 5\u221a<\/strong>2 = 0 and x + 7y + 6 \u2013 5\u221a<\/strong>2<\/p>\n\n\n\n

4. Prove that the lines 2x + 3y = 19 and 2x + 3y + 7 = 0 are equidistant from the line 2x + 3y = 6.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The lines A, 2x + 3y = 19 and B, 2x + 3y + 7 = 0 also a line C, 2x + 3y = 6.<\/p>\n\n\n\n

Let d1<\/sub> be the distance between lines 2x + 3y = 19 and 2x + 3y = 6,<\/p>\n\n\n\n

While d2<\/sub> is the distance between lines 2x + 3y + 7 = 0 and 2x + 3y = 6<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Hence proved, the lines 2x + 3y = 19 and 2x + 3y + 7 = 0 are equidistant from the line 2x + 3y = 6<\/p>\n\n\n\n

5. Find the equation of the line mid-way between the parallel lines 9x + 6y \u2013 7 = 0 and 3x + 2y + 6 = 0.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

9x + 6y \u2013 7 = 0 and 3x + 2y + 6 = 0 are parallel lines<\/p>\n\n\n\n

The given equations of the lines can be written as:<\/p>\n\n\n\n

3x + 2y \u2013 7\/3 = 0 \u2026 (1)<\/p>\n\n\n\n

3x + 2y + 6 = 0 \u2026 (2)<\/p>\n\n\n\n

Let the equation of the line midway between the parallel lines (1) and (2) be<\/p>\n\n\n\n

3x + 2y + \u03bb = 0 \u2026 (3)<\/p>\n\n\n\n

The distance between (1) and (3) and the distance between (2) and (3) are equal.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Now substitute the value of \u03bb back in equation 3x + 2y + \u03bb = 0, we get<\/p>\n\n\n\n

3x + 2y + 11\/6 = 0<\/p>\n\n\n\n

By taking LCM<\/p>\n\n\n\n

18x + 12y + 11 = 0<\/p>\n\n\n\n

\u2234 The required equation of line is 18x + 12y + 11 = 0<\/p>\n\n\n\n

EXERCISE 23.17 PAGE NO: 23.117<\/h4>\n\n\n\n

1. Prove that the area of the parallelogram formed by the lines
a1<\/sub>x + b1<\/sub>y + c1<\/sub> = 0, a1<\/sub>x + b1<\/sub>y + d1<\/sub> = 0, a2<\/sub>x + b2<\/sub>y + c2<\/sub> = 0, a2<\/sub>x + b2<\/sub>y + d2<\/sub> = 0 is  sq. units.
Deduce the condition for these lines to form a rhombus.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The given lines are<\/p>\n\n\n\n

a1<\/sub>x + b1<\/sub>y + c1<\/sub> = 0 \u2026 (1)<\/p>\n\n\n\n

a1<\/sub>x + b1<\/sub>y + d1<\/sub> = 0 \u2026 (2)<\/p>\n\n\n\n

a2<\/sub>x + b2<\/sub>y + c2<\/sub> = 0 \u2026 (3)<\/p>\n\n\n\n

a2<\/sub>x + b2<\/sub>y + d2<\/sub> = 0 \u2026 (4)<\/p>\n\n\n\n

Let us prove, the area of the parallelogram formed by the lines a1<\/sub>x + b1<\/sub>y + c1<\/sub> = 0, a1<\/sub>x + b1<\/sub>y + d1<\/sub> = 0, a2<\/sub>x + b2<\/sub>y + c2<\/sub> = 0, a2<\/sub>x + b2<\/sub>y + d2<\/sub> = 0 is
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

 sq. units.<\/p>\n\n\n\n

The area of the parallelogram formed by the lines a1<\/sub>x + b1<\/sub>y + c1<\/sub> = 0, a1<\/sub>x + b1<\/sub>y + d1<\/sub> = 0, a2<\/sub>x + b2<\/sub>y + c2<\/sub> = 0 and a2<\/sub>x + b2<\/sub>y + d2<\/sub> = 0 is given below:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

2. Prove that the area of the parallelogram formed by the lines 3x \u2013 4y + a = 0, 3x \u20134y + 3a = 0, 4x \u2013 3y \u2013 a = 0 and 4x \u2013 3y \u2013 2a = 0 is 2a2<\/sup>\/7 sq. units.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The given lines are<\/p>\n\n\n\n

3x \u2212 4y + a = 0 \u2026 (1)<\/p>\n\n\n\n

3x \u2212 4y + 3a = 0 \u2026 (2)<\/p>\n\n\n\n

4x \u2212 3y \u2212 a = 0 \u2026 (3)<\/p>\n\n\n\n

4x \u2212 3y \u2212 2a = 0 \u2026 (4)<\/p>\n\n\n\n

Let us prove, the area of the parallelogram formed by the lines 3x \u2013 4y + a = 0, 3x \u2013 4y + 3a = 0, 4x \u2013 3y \u2013 a = 0 and 4x \u2013 3y \u2013 2a = 0 is 2a2<\/sup>\/7 sq. units.<\/p>\n\n\n\n

From above solution, we know that<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

3. Show that the diagonals of the parallelogram whose sides are lx + my + n = 0, lx + my + n\u2019 = 0, mx + ly + n = 0 and mx + ly + n\u2019 = 0 include an angle \u03c0\/2.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The given lines are<\/p>\n\n\n\n

lx + my + n = 0 \u2026 (1)<\/p>\n\n\n\n

mx + ly + n\u2019 = 0 \u2026 (2)<\/p>\n\n\n\n

lx + my + n\u2019 = 0 \u2026 (3)<\/p>\n\n\n\n

mx + ly + n = 0 \u2026 (4)<\/p>\n\n\n\n

Let us prove, the diagonals of the parallelogram whose sides are lx + my + n = 0, lx + my + n\u2019 = 0, mx + ly + n = 0 and mx + ly + n\u2019 = 0 include an angle \u03c0\/2.<\/p>\n\n\n\n

By solving (1) and (2), we get<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

\u2234 m1<\/sub>m2<\/sub> = -1<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

EXERCISE 23.18 PAGE NO: 23.124<\/h4>\n\n\n\n

1. Find the equation of the straight lines passing through the origin and making an angle of 45o<\/sup> with the straight line \u221a3x + y = 11.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Equation passes through (0, 0) and make an angle of 45\u00b0 with the line \u221a3x + y = 11.<\/p>\n\n\n\n

We know that, the equations of two lines passing through a point x1,y1 and making an angle \u03b1 with the given line y = mx + c are<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

2. Find the equations to the straight lines which pass through the origin and are inclined at an angle of 75o<\/sup> to the straight line x + y + \u221a3(y \u2013 x) = a.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equation passes through (0,0) and make an angle of 75\u00b0 with the line x + y + \u221a3(y \u2013 x) = a.<\/p>\n\n\n\n

We know that the equations of two lines passing through a point x1,y1 and making an angle \u03b1 with the given line y = mx + c are<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

3. Find the equations of straight lines passing through (2, -1) and making an angle of 45o<\/sup> with the line 6x + 5y \u2013 8 = 0.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equation passes through (2,-1) and make an angle of 45\u00b0 with the line 6x + 5y \u2013 8 = 0<\/p>\n\n\n\n

We know that the equations of two lines passing through a point x1<\/sub>, y1<\/sub> and making an angle \u03b1 with the given line y = mx + c are<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Here, equation of the given line is,<\/p>\n\n\n\n

6x + 5y \u2013 8 = 0<\/p>\n\n\n\n

5y = \u2013 6x + 8<\/p>\n\n\n\n

y = -6x\/5 + 8\/5<\/p>\n\n\n\n

Comparing this equation with y = mx + c<\/p>\n\n\n\n

We get, m = -6\/5<\/p>\n\n\n\n

Where, x1<\/sub> = 2, y1<\/sub> = \u2013 1, \u03b1 = 45\u00b0, m = -6\/5<\/p>\n\n\n\n

So, the equations of the required lines are<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

x + 11y + 9 = 0 and 11x \u2013 y \u2013 23 = 0<\/p>\n\n\n\n

\u2234 The equation of given line is x + 11y + 9 = 0 and 11x \u2013 y \u2013 23 = 0<\/p>\n\n\n\n

4. Find the equations to the straight lines which pass through the point (h, k) and are inclined at angle tan-1<\/sup> m to the straight line y = mx + c.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equation passes through (h, k) and make an angle of tan-1<\/sup> m with the line y = mx + c<\/p>\n\n\n\n

We know that the equations of two lines passing through a point x1<\/sub>, y1<\/sub> and making an angle \u03b1 with the given line y = mx + c are<\/p>\n\n\n\n

m\u2032 = m<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Here,<\/p>\n\n\n\n

x1<\/sub> = h, y1<\/sub> = k, \u03b1 = tan-1<\/sup> m, m\u2032 = m.<\/p>\n\n\n\n

So, the equations of the required lines are<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

5. Find the equations to the straight lines passing through the point (2, 3) and inclined at an angle of 450<\/sup> to the lines 3x + y \u2013 5 = 0.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equation passes through (2, 3) and make an angle of 450<\/sup>with the line 3x + y \u2013 5 = 0.<\/p>\n\n\n\n

We know that the equations of two lines passing through a point x1,y1 and making an angle \u03b1 with the given line y = mx + c are<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Here,<\/p>\n\n\n\n

Equation of the given line is,<\/p>\n\n\n\n

3x + y \u2013 5 = 0<\/p>\n\n\n\n

y = \u2013 3x + 5<\/p>\n\n\n\n

Comparing this equation with y = mx + c we get, m = \u2013 3<\/p>\n\n\n\n

x1<\/sub> = 2, y1<\/sub> = 3, \u03b1 = 45\u2218, m = \u2013 3.<\/p>\n\n\n\n

So, the equations of the required lines are<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

x + 2y \u2013 8 = 0 and 2x \u2013 y \u2013 1 = 0<\/p>\n\n\n\n

\u2234 The equation of given line is x + 2y \u2013 8 = 0 and 2x \u2013 y \u2013 1 = 0<\/p>\n\n\n\n

EXERCISE 23.19 PAGE NO: 23.124<\/h4>\n\n\n\n

1. Find the equation of a straight line through the point of intersection of the lines 4x \u2013 3y = 0 and 2x \u2013 5y + 3 = 0 and parallel to 4x + 5 y + 6 = 0.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Lines 4x \u2013 3y = 0 and 2x \u2013 5y + 3 = 0 and parallel to 4x + 5 y + 6 = 0<\/p>\n\n\n\n

The equation of the straight line passing through the points of intersection of 4x \u2212 3y = 0 and 2x \u2212 5y + 3 = 0 is given below:<\/p>\n\n\n\n

4x \u2212 3y + \u03bb (2x \u2212 5y + 3) = 0<\/p>\n\n\n\n

(4 + 2\u03bb)x + (\u2212 3 \u2212 5\u03bb)y + 3\u03bb = 0<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The required line is parallel to 4x + 5y + 6 = 0 or, y = -4x\/5 \u2013 6\/5<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u03bb = -16\/15<\/p>\n\n\n\n

\u2234 The required equation is<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

28x + 35y \u2013 48 = 0<\/p>\n\n\n\n

2. Find the equation of a straight line passing through the point of intersection of x + 2y + 3 = 0 and 3x + 4y + 7 = 0 and perpendicular to the straight line x \u2013 y + 9 = 0.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

x + 2y + 3 = 0 and 3x + 4y + 7 = 0<\/p>\n\n\n\n

The equation of the straight line passing through the points of intersection of x + 2y + 3 = 0 and 3x + 4y + 7 = 0 is<\/p>\n\n\n\n

x + 2y + 3 + \u03bb(3x + 4y + 7) = 0<\/p>\n\n\n\n

(1 + 3\u03bb)x + (2 + 4\u03bb)y + 3 + 7\u03bb = 0<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The required line is perpendicular to x \u2212 y + 9 = 0 or, y = x + 9<\/p>\n\n\n\n

3. Find the equation of the line passing through the point of intersection of 2x \u2013 7y + 11 = 0 and x + 3y \u2013 8 = 0 and is parallel to (i) x = axis (ii) y-axis.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equations, 2x \u2013 7y + 11 = 0 and x + 3y \u2013 8 = 0<\/p>\n\n\n\n

The equation of the straight line passing through the points of intersection of 2x \u2212 7y + 11 = 0 and x + 3y \u2212 8 = 0 is given below:<\/p>\n\n\n\n

2x \u2212 7y + 11 + \u03bb(x + 3y \u2212 8) = 0<\/p>\n\n\n\n

(2 + \u03bb)x + (\u2212 7 + 3\u03bb)y + 11 \u2212 8\u03bb = 0<\/p>\n\n\n\n

(i)<\/strong> The required line is parallel to the x-axis. So, the coefficient of x should be zero.<\/p>\n\n\n\n

2 + \u03bb = 0<\/p>\n\n\n\n

\u03bb = -2<\/p>\n\n\n\n

Now, substitute the value of \u03bb back in equation, we get<\/p>\n\n\n\n

0 + (\u2212 7 \u2212 6)y + 11 + 16 = 0<\/p>\n\n\n\n

13y \u2212 27 = 0<\/p>\n\n\n\n

\u2234 The equation of the required line is 13y \u2212 27 = 0<\/p>\n\n\n\n

(ii)<\/strong> The required line is parallel to the y-axis. So, the coefficient of y should be zero.<\/p>\n\n\n\n

-7 + 3\u03bb = 0<\/p>\n\n\n\n

\u03bb = 7\/3<\/p>\n\n\n\n

Now, substitute the value of \u03bb back in equation, we get<\/p>\n\n\n\n

(2 + 7\/3)x + 0 + 11 \u2013 8(7\/3) = 0<\/p>\n\n\n\n

13x \u2013 23 = 0<\/p>\n\n\n\n

\u2234 The equation of the required line is 13x \u2013 23 = 0<\/p>\n\n\n\n

4. Find the equation of the straight line passing through the point of intersection of 2x + 3y + 1 = 0 and 3x \u2013 5y \u2013 5 = 0 and equally inclined to the axes.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The equations, 2x + 3y + 1 = 0 and 3x \u2013 5y \u2013 5 = 0<\/p>\n\n\n\n

The equation of the straight line passing through the points of intersection of 2x + 3y + 1 = 0 and 3x \u2212 5y \u2212 5 = 0 is<\/p>\n\n\n\n

2x + 3y + 1 + \u03bb(3x \u2212 5y \u2212 5) = 0<\/p>\n\n\n\n

(2 + 3\u03bb)x + (3 \u2212 5\u03bb)y + 1 \u2212 5\u03bb = 0<\/p>\n\n\n\n

y = \u2013 [(2 + 3\u03bb) \/ (3 \u2013 5\u03bb)] \u2013 [(1 \u2013 5\u03bb) \/ (3 \u2013 5\u03bb)]<\/p>\n\n\n\n

The required line is equally inclined to the axes. So, the slope of the required line is either 1 or \u2212 1.<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

\u2013 [(2 + 3\u03bb) \/ (3 \u2013 5\u03bb)] = 1 and \u2013 [(2 + 3\u03bb) \/ (3 \u2013 5\u03bb)] = -1<\/p>\n\n\n\n

-2 \u2013 3\u03bb = 3 \u2013 5\u03bb and 2 + 3\u03bb = 3 \u2013 5\u03bb<\/p>\n\n\n\n

\u03bb = 5\/2 and 1\/8<\/p>\n\n\n\n

Now, substitute the values of \u03bb in (2 + 3\u03bb)x + (3 \u2212 5\u03bb)y + 1 \u2212 5\u03bb = 0, we get the equations of the required lines as:<\/p>\n\n\n\n

(2 + 15\/2)x + (3 \u2013 25\/2)y + 1 \u2013 25\/2 = 0 and (2 + 3\/8)x + (3 \u2013 5\/8)y + 1 \u2013 5\/8 = 0<\/p>\n\n\n\n

19x \u2013 19y \u2013 23 = 0 and 19x + 19y + 3 = 0<\/p>\n\n\n\n

\u2234 The required equation is 19x \u2013 19y \u2013 23 = 0 and 19x + 19y + 3 = 0<\/p>\n\n\n\n

5. Find the equation of the straight line drawn through the point of intersection of the lines x + y = 4 and 2x \u2013 3y = 1 and perpendicular to the line cutting off intercepts 5, 6 on the axes.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

The lines x + y = 4 and 2x \u2013 3y = 1<\/p>\n\n\n\n

The equation of the straight line passing through the point of intersection of x + y = 4 and 2x \u2212 3y = 1 is<\/p>\n\n\n\n

x + y \u2212 4 + \u03bb(2x \u2212 3y \u2212 1) = 0<\/p>\n\n\n\n

(1 + 2\u03bb)x + (1 \u2212 3\u03bb)y \u2212 4 \u2212 \u03bb = 0 \u2026 (1)<\/p>\n\n\n\n

y = \u2013 [(1 + 2\u03bb) \/ (1 \u2013 3\u03bb)]x + [(4 + \u03bb) \/ (1 \u2013 3\u03bb)]<\/p>\n\n\n\n

The equation of the line with intercepts 5 and 6 on the axis is<\/p>\n\n\n\n

x\/5 + y\/6 = 1 \u2026. (2)<\/p>\n\n\n\n

So, the slope of this line is -6\/5<\/p>\n\n\n\n

The lines (1) and (2) are perpendicular.<\/p>\n\n\n\n

\u2234 -6\/5 \u00d7 [(-1+2\u03bb) \/ (1 \u2013 3\u03bb)] = -1<\/p>\n\n\n\n

\u03bb = 11\/3<\/p>\n\n\n\n

Now, substitute the values of \u03bb in (1), we get the equation of the required line.<\/p>\n\n\n\n

(1 + 2(11\/3))x + (1 \u2013 3(11\/3))y \u2212 4 \u2013 11\/3 = 0<\/p>\n\n\n\n

(1 + 22\/3)x + (1 \u2013 11)y \u2013 4 \u2013 11\/3 = 0<\/p>\n\n\n\n

25x \u2013 30y \u2013 23 = 0<\/p>\n\n\n\n

\u2234 The required equation is 25x \u2013 30y \u2013 23 = 0<\/p>\n\n\n\n

RD Sharma Solutions for Class 11 Maths Chapter 23: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 11 Maths Chapter 23\u2013The Straight Lines<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 11 Maths Chapter 23\u2013The Straight Lines PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 11 Maths :<\/strong><\/h2>\n\n\n\n