{"id":600293,"date":"2022-05-11T04:48:22","date_gmt":"2022-05-11T04:48:22","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=600293"},"modified":"2022-05-12T10:24:13","modified_gmt":"2022-05-12T10:24:13","slug":"ml-aggarwal-solutions-for-class-10-maths-chapter-11-section-formula","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ml-aggarwal-solutions-for-class-10-maths-chapter-11-section-formula\/","title":{"rendered":"ML Aggarwal Solutions for Class 10 Maths Chapter 11- Section Formula"},"content":{"rendered":"\n

Class 10: Maths Chapter 11 solutions. Complete Class 10 Maths Chapter 11 Notes.<\/p>\n\n\n\n

ML Aggarwal Solutions for Class 10 Maths Chapter 11- Section Formula<\/h2>\n\n\n\n

ML Aggarwal 10th Maths Chapter 11, Class 10 Maths Chapter 11 solutions<\/p>\n\n\n\n

1. Find the co-ordinates of the mid-point of the line segments joining the following pairs of points:<\/strong><\/p>\n\n\n\n

(i) (2, \u2013 3), ( \u2013 6, 7)<\/strong>

(ii) (5, \u2013 11), (4, 3)<\/strong>

(iii) (a + 3, 5b), (2a \u2013 1, 3b + 4)<\/strong><\/p>\n\n\n\n

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

Co-ordinates of midpoint of line joining the points (x1<\/sub>,y1<\/sub>) and (x2<\/sub>,y2<\/sub>) = {(x1<\/sub>+x2<\/sub>)\/2 ,(y1<\/sub>+y2<\/sub>)\/2}<\/strong><\/p>\n\n\n\n

(i) Co-ordinates of midpoint of line joining the points (2, -3) and (-6,7) = {(2+-6)\/2, (-3+7)\/2}<\/strong><\/p>\n\n\n\n

= (-4\/2, 4\/2)<\/strong><\/p>\n\n\n\n

= (-2, 2)<\/strong><\/p>\n\n\n\n

Hence the co-ordinates of midpoint of line joining the points (2, -3) and (-6,7) is (-2, 2).<\/strong><\/p>\n\n\n\n

(ii) Co-ordinates of midpoint of line joining the points (x1<\/sub>,y1<\/sub>) and (x2<\/sub>,y2<\/sub>) = {(x1<\/sub>+x2<\/sub>)\/2 ,(y1<\/sub>+y2<\/sub>)\/2}<\/strong><\/p>\n\n\n\n

Co-ordinates of midpoint of line joining the points (5, -11) and (4,3) = {(5+4)\/2, (-11+3)\/2}<\/strong><\/p>\n\n\n\n

= (9\/2, -8\/2)<\/strong><\/p>\n\n\n\n

= (9\/2, -4)<\/strong><\/p>\n\n\n\n

Hence the co-ordinates of midpoint of line joining the points (5, -11) and (4,3) is (9\/2, -4).<\/strong><\/p>\n\n\n\n

(iii) Co-ordinates of midpoint of line joining the points (x1<\/sub>,y1<\/sub>) and (x2<\/sub>,y2<\/sub>) = {(x1<\/sub>+x2<\/sub>)\/2 ,(y1<\/sub>+y2<\/sub>)\/2}<\/strong><\/p>\n\n\n\n

Co-ordinates of midpoint of line joining the points (a+3, 5b) and (2a-1,3b+4) = {(a+3+2a-1)\/2, (5b+3b+4)\/2}<\/strong><\/p>\n\n\n\n

= {(3a+2)\/2, (8b+4)\/2}<\/strong><\/p>\n\n\n\n

= {(3a+2)\/2, (4b+2)}<\/strong><\/p>\n\n\n\n

Hence the co-ordinates of midpoint of line joining the points (a+3, 5b) and (2a-1,3b+4) are {(3a+2)\/2, (4b+2)}.<\/strong><\/p>\n\n\n\n

2.<\/strong> The co-ordinates of two points A and B are ( \u2013 3, 3) and (12, \u2013 7) respectively. P is a point on the line segment AB such that AP : PB = 2 : 3. Find the co-ordinates of P.<\/strong><\/p>\n\n\n\n

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

Let the co-ordinates of P(x, y) divides AB in the ratio m:n.<\/p>\n\n\n\n

A(-3,3) <\/strong>and B(12,-7) <\/strong>are the given points.<\/p>\n\n\n\n

Given m:n = 2:3<\/p>\n\n\n\n

x1<\/sub> = -3 , y1<\/sub> = 3 , x2<\/sub> = 12 , y2<\/sub> = -7 , m = 2 and n = 3<\/p>\n\n\n\n

By Section formula x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

x = (2\u00d712+3\u00d7-3)\/(2+3)<\/p>\n\n\n\n

x = (24-9)\/5<\/p>\n\n\n\n

x = 15\/5<\/p>\n\n\n\n

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

By Section formula y = (my2<\/sub>+ny1<\/sub>)\/(m+n)<\/p>\n\n\n\n

y = (2\u00d7-7+3\u00d73)\/5<\/p>\n\n\n\n

y = (-14+9)\/5<\/p>\n\n\n\n

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

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

Hence the co-ordinate of point P are (3,-1).<\/p>\n\n\n\n

3. P divides the distance between A ( \u2013 2, 1) and B (1, 4) in the ratio of 2 : 1. Calculate the co-ordinates of the point P.<\/strong><\/p>\n\n\n\n

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

Let the co-ordinates of P(x, y) divides AB in the ratio m:n.<\/p>\n\n\n\n

A(-2,1) <\/strong>and B(1,4) <\/strong>are the given points.<\/p>\n\n\n\n

Given m:n = 2:1<\/p>\n\n\n\n

x1<\/sub> = -2 , y1<\/sub> = 1 , x2<\/sub> = 1 , y2<\/sub> = 4 , m = 2 and n = 1<\/p>\n\n\n\n

By Section formula x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

x = (2\u00d71+1\u00d7-2)\/(2+1)<\/p>\n\n\n\n

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

x = 0\/3<\/p>\n\n\n\n

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

By Section formula y = (my2<\/sub>+ny1<\/sub>)\/(m+n)<\/p>\n\n\n\n

y = (2\u00d74+1\u00d71)\/(2+1)<\/p>\n\n\n\n

y = (8+1)\/3<\/p>\n\n\n\n

y = 9\/3<\/p>\n\n\n\n

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

Hence the co-ordinate of point P are (0,3).<\/p>\n\n\n\n

4. (i) Find the co-ordinates of the points of trisection of the line segment joining the point (3, \u2013 3)<\/strong><\/p>\n\n\n\n

and (6, 9).<\/strong><\/p>\n\n\n\n

(ii) The line segment joining the points (3, \u2013 4) and (1, 2) is trisected at the points P and Q. If the coordinates of P and Q are (p, \u2013 2) and (5\/3, q) respectively, find the values of p and q.<\/strong><\/p>\n\n\n\n

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

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

Let P and Q be the points of trisection of AB<\/p>\n\n\n\n

i.e., AP = PQ = QB<\/p>\n\n\n\n

Given A(3,-3) and B(6,9)<\/p>\n\n\n\n

x1<\/sub> = 3, y1<\/sub> = -3, x2<\/sub> = 6, y2<\/sub> = 9<\/p>\n\n\n\n

P(x, y) divides AB internally in the ratio 1 : 2.<\/p>\n\n\n\n

m:n = 1:2<\/p>\n\n\n\n

By applying the section formula, the coordinates of P are as follows.<\/p>\n\n\n\n

By Section formula x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

x = (1\u00d76+2\u00d73)\/(1+2)<\/p>\n\n\n\n

x = (6+6)\/3<\/p>\n\n\n\n

x = 12\/3<\/p>\n\n\n\n

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

By Section formula y = (my2<\/sub>+ny1<\/sub>)\/(m+n)<\/p>\n\n\n\n

y = (1\u00d79+2\u00d7-3)\/(2+1)<\/p>\n\n\n\n

y = (9-6)\/3<\/p>\n\n\n\n

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

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

Hence the co-ordinate of point P are (4,1).<\/p>\n\n\n\n

Now, Q also divides AB internally in the ratio 2 : 1.<\/p>\n\n\n\n

m:n = 2:1<\/p>\n\n\n\n

By applying the section formula, the coordinates of P are as follows.<\/p>\n\n\n\n

By Section formula x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

x = (2\u00d76+1\u00d73)\/(1+2)<\/p>\n\n\n\n

x = (12+3)\/3<\/p>\n\n\n\n

x = 15\/3<\/p>\n\n\n\n

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

By Section formula y = (my2<\/sub>+ny1<\/sub>)\/(m+n)<\/p>\n\n\n\n

y = (2\u00d79+1\u00d7-3)\/(2+1)<\/p>\n\n\n\n

y = (18-3)\/3<\/p>\n\n\n\n

y = 15\/3<\/p>\n\n\n\n

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

Hence the co-ordinate of point Q are (5,5).<\/p>\n\n\n\n

(ii) Let P(p,-2) and Q(5\/3, q) be the points of trisection of AB<\/p>\n\n\n\n

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

i.e., AP = PQ = QB<\/p>\n\n\n\n

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

x1<\/sub> = 3, y1<\/sub> = -4, x2<\/sub> = 1, y2<\/sub> = 2<\/p>\n\n\n\n

P(p, -2) divides AB internally in the ratio 1 : 2.<\/p>\n\n\n\n

By Section formula x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

p = (1\u00d71+2\u00d73)\/(1+2)<\/p>\n\n\n\n

p = (1+6)\/3<\/p>\n\n\n\n

p = 7\/3<\/p>\n\n\n\n

Now, Q also divides AB internally in the ratio 2 : 1.<\/p>\n\n\n\n

m:n = 2:1<\/p>\n\n\n\n

Q(5\/3, q) divides AB internally in the ratio 2 : 1.<\/p>\n\n\n\n

By Section formula y = (my2<\/sub>+ny1<\/sub>)\/(m+n)<\/p>\n\n\n\n

q = (2\u00d72+1\u00d7-4)\/(2+1)<\/p>\n\n\n\n

q = (4-4)\/3<\/p>\n\n\n\n

q = 0\/3<\/p>\n\n\n\n

q = 0<\/p>\n\n\n\n

Hence the value of p and q are 7\/3 and 0 respectively.<\/p>\n\n\n\n

5.<\/strong> (i) The line segment joining the points A (3, 2) and B (5, 1) is divided at the point P in the ratio 1 : 2 and it lies on the line 3x \u2013 18y + k = 0. Find the value of k.<\/strong>

(ii) A point P divides the line segment joining the points A (3, \u2013 5) and B ( \u2013 4, 8) such that AP\/PB = k\/1  If P lies on the line x + y = 0, then find the value of k.<\/strong><\/p>\n\n\n\n

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

(i) Let the co-ordinates of P(x, y) divides AB in the ratio m:n.<\/p>\n\n\n\n

A(3,2) <\/strong>and B(5,1) <\/strong>are the given points.<\/p>\n\n\n\n

Given m:n = 1:2<\/p>\n\n\n\n

x1<\/sub> = 3 , y1<\/sub> = 2 , x2<\/sub> = 5 , y2<\/sub> = 1 , m = 1 and n = 2<\/p>\n\n\n\n

By Section formula x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

x = (1\u00d75+2\u00d73)\/(1+2)<\/p>\n\n\n\n

x = (5+6)\/3<\/p>\n\n\n\n

x = 11\/3<\/p>\n\n\n\n

By Section formula y = (my2<\/sub>+ny1<\/sub>)\/(m+n)<\/p>\n\n\n\n

y = (1\u00d71+2\u00d72)\/(1+2)<\/p>\n\n\n\n

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

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

Given P lies on the line 3x-18y+k = 0<\/p>\n\n\n\n

Substitute x and y in above equation<\/p>\n\n\n\n

3\u00d7(11\/3)-18\u00d7(5\/3)+k = 0<\/p>\n\n\n\n

11-30+k = 0<\/p>\n\n\n\n

-19+k = 0<\/p>\n\n\n\n

k = 19<\/p>\n\n\n\n

Hence the value of k is 19.<\/p>\n\n\n\n

(ii) Let the co-ordinates of P(x, y) divides AB in the ratio m:n.<\/p>\n\n\n\n

A(3,-5) <\/strong>and B(-4,8) <\/strong>are the given points.<\/p>\n\n\n\n

Given AP\/PB = k\/1<\/p>\n\n\n\n

m:n = k:1<\/p>\n\n\n\n

x1<\/sub> = 3 , y1<\/sub> = -5 , x2<\/sub> = -4 , y2<\/sub> = 8 , m = k and n = 1<\/p>\n\n\n\n

By Section formula x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

x = (k\u00d7-4+1\u00d73)\/(k+1)<\/p>\n\n\n\n

x = (-4k+3)\/(k+1)<\/p>\n\n\n\n

x = (-4k+3)\/(k+1)<\/p>\n\n\n\n

By Section formula y = (my2<\/sub>+ny1<\/sub>)\/(m+n)<\/p>\n\n\n\n

y = (k\u00d78+1\u00d7-5)\/(k+1)<\/p>\n\n\n\n

y = (-4k+3)\/(k+1)<\/p>\n\n\n\n

Co-ordinate of P is ((-4k+3)\/(k+1), (8k-5)\/(k+1))<\/p>\n\n\n\n

Given P lies on line x+y = 0<\/p>\n\n\n\n

Substitute value of x and y in above equation<\/p>\n\n\n\n

(-4k+3)\/(k+1) + (8k-5)\/(k+1) = 0<\/p>\n\n\n\n

(-4k+3) + (8k-5) = 0<\/p>\n\n\n\n

4k-2 = 0<\/p>\n\n\n\n

4k = 2<\/p>\n\n\n\n

k = 2\/4 = \u00bd<\/p>\n\n\n\n

Hence the value of k is \u00bd .<\/p>\n\n\n\n

6.<\/strong> Find the coordinates of the point which is three-fourth of the way from A (3, 1) to B ( \u2013 2, 5).<\/strong><\/p>\n\n\n\n

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

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

Let P be the point which is three-fourth of the way from A(3,1) to B(-2,5).<\/p>\n\n\n\n

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

AB = AP+PB<\/p>\n\n\n\n

AP\/AB = AP\/(AP+PB) = 3\/4<\/p>\n\n\n\n

4AP = 3AP+3PB<\/p>\n\n\n\n

4AP-3AP = 3PB<\/p>\n\n\n\n

AP = 3PB<\/p>\n\n\n\n

AP\/PB = 3\/1<\/p>\n\n\n\n

The ratio m:n = 3:1<\/p>\n\n\n\n

x1<\/sub> = 3 , y1<\/sub> = 1 , x2<\/sub> = -2, y2<\/sub> = 5<\/p>\n\n\n\n

By Section formula x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

x = (3\u00d7-2+1\u00d73)\/(3+1)<\/p>\n\n\n\n

x = (-6+3)\/4<\/p>\n\n\n\n

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

By Section formula y = (my2<\/sub>+ny1<\/sub>)\/(m+n)<\/p>\n\n\n\n

y = (3\u00d75+1\u00d71)\/(3+1)<\/p>\n\n\n\n

y = (15+1)\/4<\/p>\n\n\n\n

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

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

Hence the co-ordinates of P are (-3\/4, 4).<\/p>\n\n\n\n

7.<\/strong> Point P (3, \u2013 5) is reflected to P\u2019 in the x- axis. Also P on reflection in the y-axis is mapped as P\u201d.<\/strong>

(i) Find the co-ordinates of P\u2019 and P\u201d.<\/strong>

(ii) Compute the distance P\u2019 P\u201d.<\/strong>

(iii) Find the middle point of the line segment P\u2019 P\u201d.<\/strong>

(iv) On which co-ordinate axis does the middle point of the line segment P P\u201d lie ?<\/strong><\/p>\n\n\n\n

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

(i) The image of P(3,-5) when reflected in X-axis will be (3,5).<\/strong><\/p>\n\n\n\n

When you reflect a point across the X-axis, the x-coordinate remains the same,<\/p>\n\n\n\n

but the y-coordinate is transformed into its opposite (its sign is changed).<\/p>\n\n\n\n

Co-ordinates of P\u2019 = (3,5)<\/strong><\/p>\n\n\n\n

Image of P(3,-5) when reflected in Y axis will be (-3,-5).<\/strong><\/p>\n\n\n\n

When you reflect a point across the Y-axis, the y-coordinate remains the same,<\/p>\n\n\n\n

but the x-coordinate is transformed into its opposite (its sign is changed)<\/p>\n\n\n\n

Co-ordinates of P\u2019\u2019 = (-3,-5)<\/strong><\/p>\n\n\n\n

(ii)Let P\u2019(x1<\/sub>, y1<\/sub>) and P\u2019\u2019(x2<\/sub> , y2<\/sub>) be the given points<\/p>\n\n\n\n

By distance formula d(P\u2019,P\u2019\u2019) = \u221a[(x2<\/sub>-x1<\/sub>)2<\/sup>+(y2<\/sub>-y1<\/sub>)2<\/sup>]<\/p>\n\n\n\n

Co-ordinates of P\u2019 = (3,5)<\/strong><\/p>\n\n\n\n

Co-ordinates of P\u2019\u2019 = (-3,-5)<\/strong><\/p>\n\n\n\n

Here x1<\/sub> = 3, y1<\/sub> = 5 , x2<\/sub> = -3, y2<\/sub> = -5<\/p>\n\n\n\n

d(P\u2019,P\u2019\u2019) = \u221a[(-3-3)2<\/sup>+(-5-5)2<\/sup>]<\/p>\n\n\n\n

= \u221a[(-6)2<\/sup>+(-10)2<\/sup>]<\/p>\n\n\n\n

= \u221a(36+100)<\/p>\n\n\n\n

= \u221a136<\/p>\n\n\n\n

= \u221a(4\u00d734)<\/p>\n\n\n\n

= 2\u221a34<\/p>\n\n\n\n

Hence the distance between P\u2019 and P\u2019\u2019 is 2\u221a34 units.<\/p>\n\n\n\n

(iii) Co-ordinates of P\u2019 = (3,5)<\/strong><\/p>\n\n\n\n

Co-ordinates of P\u2019\u2019 = (-3,-5)<\/strong><\/p>\n\n\n\n

Here x1<\/sub> = 3, y1<\/sub> = 5 , x2<\/sub> = -3, y2<\/sub> = -5<\/p>\n\n\n\n

Let Q(x,y) be the midpoint of P\u2019P\u2019\u2019<\/p>\n\n\n\n

By midpoint formula,<\/p>\n\n\n\n

x = (x1<\/sub>+x2<\/sub>)\/2<\/p>\n\n\n\n

y = (y1<\/sub>+y2<\/sub>)\/2<\/p>\n\n\n\n

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

y = (5+-5)\/2 = 0\/2 = 0<\/p>\n\n\n\n

Hence the co-ordinate of midpoint of P\u2019P\u2019\u2019 is (0,0) .<\/p>\n\n\n\n

(iv) Co-ordinates of P = (3,-5)<\/strong><\/p>\n\n\n\n

Co-ordinates of P\u2019\u2019 = (-3,-5)<\/strong><\/p>\n\n\n\n

Here x1<\/sub> = 3, y1<\/sub> = -5 , x2<\/sub> = -3, y2<\/sub> = -5<\/p>\n\n\n\n

Let R(x,y) be the midpoint of PP\u2019\u2019<\/p>\n\n\n\n

By midpoint formula,<\/p>\n\n\n\n

x = (x1<\/sub>+x2<\/sub>)\/2<\/p>\n\n\n\n

y = (y1<\/sub>+y2<\/sub>)\/2<\/p>\n\n\n\n

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

y = (-5+-5)\/2 = -10\/2 = -5<\/p>\n\n\n\n

So the co-ordinate of midpoint of PP\u2019\u2019 is (0,-5) .<\/p>\n\n\n\n

Here x co-ordinate is zero.<\/p>\n\n\n\n

Hence the point lies on Y-axis.<\/p>\n\n\n\n

8. Use graph paper for this question. Take 1 cm = 1 unit on both axes. Plot the points A(3, 0) and B(0, 4).
(i) Write down the co-ordinates of A1, the reflection of A in the y-axis.<\/strong><\/p>\n\n\n\n

(ii) Write down the co-ordinates of B1, the reflection of B in the x-axis.<\/strong><\/p>\n\n\n\n

(iii) Assign the special name to the quadrilateral ABA1B1.
(iv) If C is the midpoint is AB. Write down the co-ordinates of the point C1, the reflection of C in the origin.
(v) Assign the special name to quadrilateral ABC1B1.<\/strong><\/p>\n\n\n\n

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

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

(i) Co-ordinates of point A are (3,0).<\/p>\n\n\n\n

When you reflect a point across the Y-axis, the y-coordinate remains the same,<\/p>\n\n\n\n

but the x-coordinate is transformed into its opposite (its sign is changed)<\/p>\n\n\n\n

Hence the reflection of A in the Y axis is (-3,0).<\/p>\n\n\n\n

(ii) Co-ordinates of point B are (0,4).<\/p>\n\n\n\n

When you reflect a point across the X-axis, the x-coordinate remains the same,<\/p>\n\n\n\n

but the y-coordinate is transformed into its opposite (its sign is changed).<\/p>\n\n\n\n

Hence the reflection of B in the X-axis is (0,-4)<\/p>\n\n\n\n

(iii) The quadrilateral ABA1B1 will be a rhombus.<\/p>\n\n\n\n

(iv) Let C be midpoint of AB.<\/p>\n\n\n\n

Co-ordinate of C = ((3+0)\/2 , (0+4)\/2) = (3\/2, 2) [Midpoint formula]<\/p>\n\n\n\n

In a point reflection in the origin, the image of the point (x,y) is the point (-x,-y).<\/p>\n\n\n\n

Hence the reflection of C in the origin is (-3\/2, -2)<\/p>\n\n\n\n

(v) In quadrilateral ABC1B1, ABB1C1<\/p>\n\n\n\n

Hence the quadrilateral ABC1B1 is a trapezium.<\/p>\n\n\n\n

9. The line segment joining A ( \u2013 3, 1) and B (5, \u2013 4) is a diameter of a circle whose centre is C. find the co-ordinates of the point C. (1990)<\/strong><\/p>\n\n\n\n

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

Given Co-ordinates of A = (-3,1)<\/strong><\/p>\n\n\n\n

Co-ordinates of B = (5,-4)<\/strong><\/p>\n\n\n\n

Here x1<\/sub> = -3, y1<\/sub> = 1 , x2<\/sub> = 5, y2<\/sub> = -4<\/p>\n\n\n\n

Let C(x,y) be the midpoint of AB<\/p>\n\n\n\n

By midpoint formula,<\/p>\n\n\n\n

x = (x1<\/sub>+x2<\/sub>)\/2<\/p>\n\n\n\n

y = (y1<\/sub>+y2<\/sub>)\/2<\/p>\n\n\n\n

x = (-3+5)\/2 = 2\/2 = 1<\/p>\n\n\n\n

y = (1+-4)\/2 = -3\/2<\/p>\n\n\n\n

Hence the co-ordinate of midpoint of AB is C(1,-3\/2) .<\/p>\n\n\n\n

10. The mid-point of the line segment joining the points (3m, 6) and ( \u2013 4, 3n) is (1, 2m \u2013 1). Find the values of m and n.<\/strong><\/p>\n\n\n\n

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

Let the midpoint of line joining the points A(3m,6) and B(-4,3n) be C(1,2m-1).<\/p>\n\n\n\n

Here x1<\/sub> = 3m, y1<\/sub> = 6 , x2<\/sub> = -4, y2<\/sub> = 3n<\/p>\n\n\n\n

x = 1 , y = 2m-1<\/p>\n\n\n\n

By Midpoint formula,<\/p>\n\n\n\n

x = (x1<\/sub>+x2<\/sub>)\/2<\/p>\n\n\n\n

1 = (3m+-4)\/2<\/p>\n\n\n\n

3m-4 = 2<\/p>\n\n\n\n

3m = 2+4<\/p>\n\n\n\n

3m = 6<\/p>\n\n\n\n

m = 6\/3 = 2<\/p>\n\n\n\n

By Midpoint formula,<\/p>\n\n\n\n

y = (y1<\/sub>+y2<\/sub>)\/2<\/p>\n\n\n\n

2m-1 = (6+3n)\/2<\/p>\n\n\n\n

4m-2 = 6+3n<\/p>\n\n\n\n

Put m = 2 in above equation<\/p>\n\n\n\n

4\u00d72-2 = 6+3n<\/p>\n\n\n\n

8-2-6 = 3n<\/p>\n\n\n\n

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

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

Hence the value of m and n are 2 and 0 respectively.<\/p>\n\n\n\n

11. The co-ordinates of the mid-point of the line segment PQ are (1, \u2013 2). The co-ordinates of P are ( \u2013 3, 2). Find the co-ordinates of Q.(1992)<\/strong><\/p>\n\n\n\n

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

Let the co-ordinates of Q be (x2<\/sub>, y2<\/sub>).<\/p>\n\n\n\n

Given co-ordinates of P = (-3,2)<\/p>\n\n\n\n

Co-ordinates of midpoint = (1,-2)<\/p>\n\n\n\n

Here x1<\/sub> = -3, y1<\/sub> = 2 , x = 1 , y = -2<\/p>\n\n\n\n

By Midpoint formula,<\/p>\n\n\n\n

x = (x1<\/sub>+x2<\/sub>)\/2<\/p>\n\n\n\n

1 = (-3+x2<\/sub>)\/2<\/p>\n\n\n\n

2 = -3+x2<\/sub><\/p>\n\n\n\n

x2<\/sub> = 2+3 = 5<\/p>\n\n\n\n

By Midpoint formula,<\/p>\n\n\n\n

y = (y1<\/sub>+y2<\/sub>)\/2<\/p>\n\n\n\n

-2 = (2+y2<\/sub>)\/2<\/p>\n\n\n\n

-4 = 2+y2<\/sub><\/p>\n\n\n\n

y2<\/sub> = -4-2<\/p>\n\n\n\n

y2<\/sub> = -6<\/p>\n\n\n\n

Hence the co-ordinates of Q are (5,-6).<\/p>\n\n\n\n

12. AB is a diameter of a circle with centre C ( \u2013 2, 5). If point A is (3, \u2013 7). Find:<\/strong>

(i) the length of radius AC.<\/strong>

(ii) the coordinates of B.<\/strong><\/p>\n\n\n\n

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

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

(i) Length of radius AC = d(A,C)<\/p>\n\n\n\n

Co-ordinates of A = (3,-7)<\/p>\n\n\n\n

Co-ordinates of C = (-2,5)<\/p>\n\n\n\n

Here x1<\/sub> = 3, y1<\/sub> = -7, x2<\/sub> = -2, y2<\/sub> = 5<\/p>\n\n\n\n

By distance formula, d(A,C) = \u221a[(x2<\/sub>-x1<\/sub>)2<\/sup>+(y2<\/sub>-y1<\/sub>)2<\/sup>]<\/p>\n\n\n\n

= \u221a[(-2-3)2<\/sup>+(5-(-7))2<\/sup>]<\/p>\n\n\n\n

= \u221a[(-5)2<\/sup>+(12)2<\/sup>]<\/p>\n\n\n\n

= \u221a[25+144]<\/p>\n\n\n\n

= \u221a169<\/p>\n\n\n\n

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

Hence the radius is 13 units.<\/p>\n\n\n\n

(ii)Given AB is the diameter and C is the centre of the circle.<\/p>\n\n\n\n

By midpoint formula, -2 = (x+3)\/2<\/p>\n\n\n\n

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

x = -4-3 = -7<\/p>\n\n\n\n

By midpoint formula, 5 = (-7+y)\/2<\/p>\n\n\n\n

10 = -7+y<\/p>\n\n\n\n

y = 10+7 = 17<\/p>\n\n\n\n

Hence the co-ordinates of B are (-7,17).<\/p>\n\n\n\n

13. Find the reflection (image) of the point (5, \u2013 3) in the point ( \u2013 1, 3).<\/strong><\/p>\n\n\n\n

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

Let the co-ordinates of the image of the point P(5,-3) be

P1(x, y) in the point (-1, 3) then the point (-1, 3) will be the midpoint of PP1.<\/p>\n\n\n\n

By midpoint formula, x = (x1<\/sub>+x2<\/sub>)\/2<\/p>\n\n\n\n

-1 = (5+x2<\/sub>)\/2 [x = -1, x1 <\/sub>= 5]<\/p>\n\n\n\n

-2 = 5+x2<\/sub><\/p>\n\n\n\n

x2<\/sub> = -2-5 = -7<\/p>\n\n\n\n

By midpoint formula, y = (y1<\/sub>+y2<\/sub>)\/2<\/p>\n\n\n\n

3 = (-3+y2<\/sub>)\/2 [y = 3, y1 <\/sub>= -3]<\/p>\n\n\n\n

6 = -3+y2<\/sub><\/p>\n\n\n\n

y2<\/sub> = 6+3 = 9<\/p>\n\n\n\n

Hence the co-ordinates of the image of P is (-7,9).<\/p>\n\n\n\n

14. The line segment joining A(-1,5\/3)  the points B (a, 5) is divided in the ratio 1 : 3 at P, the point where the line segment AB intersects y-axis. Calculate
(i) the value of a
(ii) the co-ordinates of P. (1994)<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> Let P(x,y) divides the line segment joining the points A(-1,5\/3), B(a,5) in the ratio 1:3,<\/p>\n\n\n\n

Here m:n = 1:3<\/p>\n\n\n\n

x1<\/sub> = -1 , y1<\/sub> = 5\/3 , x2<\/sub> = a, y2<\/sub> = 5<\/p>\n\n\n\n

By Section formula x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

x = (1\u00d7a+3\u00d7-1)\/(1+3)<\/p>\n\n\n\n

x = (a-3)\/4<\/p>\n\n\n\n

x = (a-3)\/4 \u2026..(i)<\/p>\n\n\n\n

By Section formula y = (my2<\/sub>+ny1<\/sub>)\/(m+n)<\/p>\n\n\n\n

y = (1\u00d75+3\u00d75\/3)\/(3+1)<\/p>\n\n\n\n

y = (5+5)\/4<\/p>\n\n\n\n

y = 10\/4<\/p>\n\n\n\n

y = 5\/2 \u2026.(ii)<\/p>\n\n\n\n

Given P meets Y axis. So its x co-ordinate will be zero.<\/p>\n\n\n\n

i.e, (a-3)\/4 = 0<\/p>\n\n\n\n

a-3 = 0<\/p>\n\n\n\n

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

(ii) x = (a-3)\/4 [From (i)]<\/p>\n\n\n\n

Substitute a = 3 in above equation.<\/p>\n\n\n\n

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

y = 5\/2 [From (ii)]<\/p>\n\n\n\n

Hence the co-ordinates of P are (0,5\/2).<\/p>\n\n\n\n

15.<\/strong> The point P ( \u2013 4, 1) divides the line segment joining the points A (2, \u2013 2) and B in the ratio of 3 : 5. Find the point B.<\/strong><\/p>\n\n\n\n

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

Let the co-ordinates of B be (x2<\/sub>,y2<\/sub>).<\/strong><\/p>\n\n\n\n

Given co-ordinates of A = (2,-2)<\/strong><\/p>\n\n\n\n

Co-ordinates of P = (-4,1)<\/strong><\/p>\n\n\n\n

Ratio m:n = 3:5<\/strong><\/p>\n\n\n\n

x1<\/sub> = 2, y1<\/sub> = -2, x = -4, y = 1<\/strong><\/p>\n\n\n\n

P divides AB in the ratio 3:5<\/strong><\/p>\n\n\n\n

By section formula, <\/strong>x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

-4 = (3\u00d7x2<\/sub>+5\u00d72)\/(3+5)<\/p>\n\n\n\n

-4 = (3x2<\/sub>+10)\/8<\/p>\n\n\n\n

-32 = 3x2<\/sub>+10<\/p>\n\n\n\n

3x2 <\/sub>= -32-10 = -42<\/p>\n\n\n\n

x2<\/sub> = -42\/3 <\/sub>= -14<\/p>\n\n\n\n

By Section formula y = (my2<\/sub>+ny1<\/sub>)\/(m+n)<\/p>\n\n\n\n

1= (3\u00d7y2<\/sub>+5\u00d7-2)\/(3+5)<\/p>\n\n\n\n

1 = (3y2<\/sub>-10)\/8<\/p>\n\n\n\n

8 = 3y2<\/sub>-10<\/p>\n\n\n\n

3y2<\/sub> = 8+10 = 18<\/p>\n\n\n\n

y = 18\/3 = 6<\/p>\n\n\n\n

Hence the co-ordinates of B are (-14,6).<\/p>\n\n\n\n

16.<\/strong> (i) In what ratio does the point (5, 4) divide the line segment joining the points (2, 1) and (7 ,6) ?
(ii) In what ratio does the point ( \u2013 4, b) divide the line segment joining the points P (2, \u2013 2), Q ( \u2013 14, 6) ? Hence find the value of b.<\/strong><\/p>\n\n\n\n

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

(i) Let the ratio that the point (5,4) divide the line segment joining the points (2,1) and (7,6) be m:n,<\/strong><\/p>\n\n\n\n

Here x1<\/sub> = 2 , y1<\/sub> = 1 , x2<\/sub> = 7, y2<\/sub> = 6, x = 5, y = 4<\/p>\n\n\n\n

By section formula, <\/strong>x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

5 = (m\u00d77+n\u00d72)\/(m+n)<\/p>\n\n\n\n

5 = (7m+2n)\/(m+n)<\/p>\n\n\n\n

5(m+n) = 7m+2n<\/p>\n\n\n\n

5m+5n = 7m+2n<\/p>\n\n\n\n

5m-7m = 2n-5n<\/p>\n\n\n\n

-2m = -3n<\/p>\n\n\n\n

m\/n = -3\/-2 = 3\/2<\/p>\n\n\n\n

Hence the ratio m:n is 3:2.<\/p>\n\n\n\n

(ii) Let the ratio that the point (-4,b) divide the line segment joining the points (2,-2) and (-14,6) be m:n,<\/strong><\/p>\n\n\n\n

Here x1<\/sub> = 2 , y1<\/sub> = -2 , x2<\/sub> = -14, y2<\/sub> = 6, x = -4, y = b<\/p>\n\n\n\n

By section formula, <\/strong>x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

-4 = (m\u00d7-14+n\u00d72)\/(m+n)<\/p>\n\n\n\n

-4= (-14m+2n)\/(m+n)<\/p>\n\n\n\n

-4(m+n) = -14m+2n<\/p>\n\n\n\n

-4m-4n = -14m+2n<\/p>\n\n\n\n

-4m+14m = 2n+4n<\/p>\n\n\n\n

10m = 6n<\/p>\n\n\n\n

m\/n = 6\/10 = 3\/5<\/p>\n\n\n\n

Hence the ratio m:n is 3:5.<\/p>\n\n\n\n

By Section formula y = (my2<\/sub>+ny1<\/sub>)\/(m+n)<\/p>\n\n\n\n

b= (3\u00d76+5\u00d7-2)\/(3+5)<\/p>\n\n\n\n

b = (18-10)\/8<\/p>\n\n\n\n

b = 8\/8<\/p>\n\n\n\n

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

Hence the value of b is 1 and the ratio m:n is 3:5.<\/p>\n\n\n\n

17. The line segment joining A (2, 3) and B (6, \u2013 5) is intercepted by the x-axis at the point K. Write the ordinate of the point k. Hence, find the ratio in which K divides AB. Also, find the coordinates of the point K.<\/strong><\/p>\n\n\n\n

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

Since the point K is on X axis, its y co-ordinate is zero.<\/strong><\/p>\n\n\n\n

Let the point K be (x,0).<\/strong><\/p>\n\n\n\n

Let the point K divides the line segment joining A(2,3) and B(6,-5) in the ratio m:n.<\/strong><\/p>\n\n\n\n

Here x1<\/sub> = 2 , y1<\/sub> = 3 , x2<\/sub> = 6, y2<\/sub> = -5, y = 0<\/p>\n\n\n\n

By Section formula y = (my2<\/sub>+ny1<\/sub>)\/(m+n)<\/p>\n\n\n\n

0 = (m\u00d7-5+n\u00d73)\/(m+n)<\/p>\n\n\n\n

0 = (-5m+3n)\/m+n<\/p>\n\n\n\n

-5m+3n = 0<\/p>\n\n\n\n

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

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

Hence the point K divides the line segment in the ratio 3:5.<\/p>\n\n\n\n

By section formula, <\/strong>x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

x = (3\u00d76+5\u00d72)\/(3+5)<\/p>\n\n\n\n

x = (18+10)\/8<\/p>\n\n\n\n

x = 28\/8 = 7\/2<\/p>\n\n\n\n

Hence the co-ordinates of K are (7\/2, 0).<\/p>\n\n\n\n

18. If A ( \u2013 4, 3) and B (8, \u2013 6), (i) find the length of AB.<\/strong>

(ii) in what ratio is the line joining AB, divided by the x-axis? (2008)<\/strong><\/p>\n\n\n\n

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

(i) Given points are A(-4,3) and B(8,-6).<\/p>\n\n\n\n

Here x1 <\/sub>= -4, y1<\/sub> = 3<\/p>\n\n\n\n

x2<\/sub> = 8, y2<\/sub> = -6<\/p>\n\n\n\n

By distance formula, d(AB) = \u221a[(x2<\/sub>-x1<\/sub>)2<\/sup>+(y2<\/sub>-y1<\/sub>)2<\/sup>]<\/p>\n\n\n\n

d(AB) = \u221a[(8-(-4))2<\/sup>+(-6-3)2<\/sup>]<\/p>\n\n\n\n

d(AB) = \u221a[(12)2<\/sup>+(-9)2<\/sup>]<\/p>\n\n\n\n

d(AB) = \u221a(144+81)<\/p>\n\n\n\n

d(AB) = \u221a225<\/p>\n\n\n\n

d(AB) = 15<\/p>\n\n\n\n

Hence the length of AB is 15 units.<\/p>\n\n\n\n

(ii)Let m:n be the ratio in which the line AB is divided by the X axis.<\/p>\n\n\n\n

Since the line meets X axis, its y co-ordinate is zero.<\/p>\n\n\n\n

By Section formula y = (my2<\/sub>+ny1<\/sub>)\/(m+n)<\/p>\n\n\n\n

0 = (m\u00d7-6+n\u00d73)\/(m+n)<\/p>\n\n\n\n

0 = (-6m+3n)\/m+n<\/p>\n\n\n\n

-6m+3n = 0<\/p>\n\n\n\n

-6m = -3n<\/p>\n\n\n\n

m\/n = -3\/-6 = 3\/6 = 1\/2<\/p>\n\n\n\n

Hence the ratio is 1:2.<\/p>\n\n\n\n

19. (i) Calculate the ratio in which the line segment joining (3, 4) and( \u2013 2, 1) is divided by the y-axis.<\/strong>

(ii) In what ratio does the line x \u2013 y \u2013 2 = 0 divide the line segment joining the points (3, \u2013 1) and (8, 9)? Also, find the coordinates of the point of division.<\/strong><\/p>\n\n\n\n

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

(i) Let m:n be the ratio in which the line segment joining (3,4) and (-2,1) is divided by the Y axis.<\/p>\n\n\n\n

Since the line meets Y axis, its x co-ordinate is zero.<\/p>\n\n\n\n

Here x1 <\/sub>= 3, y1<\/sub> = 4<\/p>\n\n\n\n

x2<\/sub> = -2, y2<\/sub> = 1<\/p>\n\n\n\n

By section formula, <\/strong>x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

0 = (m\u00d7-2+n\u00d73)\/(m+n)<\/p>\n\n\n\n

0 = (-2m+3n)\/( m+n)<\/p>\n\n\n\n

0 = -2m+3n<\/p>\n\n\n\n

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

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

Hence the ration m:n is 3:2.<\/p>\n\n\n\n

(ii)Let the line x-y-2 = 0 divide the line segment joining the points (3,-1) and (8,9) in the ratio m:n at the point P(x,y)<\/strong><\/p>\n\n\n\n

Here x1 <\/sub>= 3, y1<\/sub> = -1<\/p>\n\n\n\n

x2<\/sub> = 8 y2<\/sub> = 9<\/p>\n\n\n\n

By section formula, <\/strong>x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

x = (m\u00d78+n\u00d73)\/(m+n)<\/p>\n\n\n\n

x = (8m+3n)\/( m+n) \u2026. (i)<\/p>\n\n\n\n

By Section formula y = (my2<\/sub>+ny1<\/sub>)\/(m+n)<\/p>\n\n\n\n

y = (m\u00d79+n\u00d7-1)\/(m+n)<\/p>\n\n\n\n

y = (9m-n)\/(m+n) \u2026. (ii)<\/p>\n\n\n\n

Since the point P(x,y) lies on the line x-y-2 = 0,<\/p>\n\n\n\n

eqn (i) and (ii) will satify the equation x-y-2 = 0 \u2026(iii)<\/p>\n\n\n\n

Substitute (i) and (ii) in (iii)[(8m+3n)\/( m+n)] \u2013 [(9m-n)\/(m+n)]-2 = 0[(8m+3n)\/( m+n)] \u2013 [(9m-n)\/(m+n)]-[2(m+n)\/(m+n)] = 0<\/p>\n\n\n\n

8m+3n-(9m-n)-2(m+n) = 0<\/p>\n\n\n\n

8m+3n-9m+n-2m-2n = 0<\/p>\n\n\n\n

-3m+2n = 0<\/p>\n\n\n\n

-3m = -2n<\/p>\n\n\n\n

m\/n = -2\/-3 = 2\/3<\/p>\n\n\n\n

Hence the ratio m:n is 2:3.<\/p>\n\n\n\n

Substitute m and n in (i)<\/strong><\/p>\n\n\n\n

x = (8m+3n)\/( m+n)<\/p>\n\n\n\n

x = (8\u00d72+3\u00d73)\/(2+3)<\/p>\n\n\n\n

x = (16+9)\/5<\/p>\n\n\n\n

x = 25\/5 = 5<\/p>\n\n\n\n

Substitute m and n in (ii)<\/strong><\/p>\n\n\n\n

y = (9m-n)\/(m+n)<\/p>\n\n\n\n

y = (9\u00d72-3)\/(2+3)<\/p>\n\n\n\n

y = (18-3)\/5<\/p>\n\n\n\n

y = 15\/5 = 3<\/p>\n\n\n\n

Hence the co-ordinates of P are (5,3).<\/p>\n\n\n\n

20.<\/strong> Given a line segment AB joining the points A ( \u2013 4, 6) and B (8, \u2013 3). Find:<\/strong>

(i) the ratio in which AB is divided by the y-axis.<\/strong>

(ii) find the coordinates of the point of intersection.<\/strong>

(iii)the length of AB.<\/strong><\/p>\n\n\n\n

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

(i) Let m:n be the ratio in which the line segment joining A (-4,6) and B(8,-3) is divided by the Y axis.<\/p>\n\n\n\n

Since the line meets Y axis, its x co-ordinate is zero.<\/p>\n\n\n\n

Here x1 <\/sub>= -4, y1<\/sub> = 6<\/p>\n\n\n\n

x2<\/sub> = 8, y2<\/sub> = -3<\/p>\n\n\n\n

By section formula, <\/strong>x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

0 = (m\u00d78+n\u00d7-4)\/(m+n)<\/p>\n\n\n\n

0 = (8m+-4n)\/(m+n)<\/p>\n\n\n\n

0 = 8m+-4n<\/p>\n\n\n\n

8m = 4n<\/p>\n\n\n\n

m\/n = 4\/8 = 1\/2<\/p>\n\n\n\n

Hence the ration m:n is 1:2.<\/p>\n\n\n\n

(ii) By Section formula y = (my2<\/sub>+ny1<\/sub>)\/(m+n)<\/p>\n\n\n\n

Substitute m and n in above equation<\/p>\n\n\n\n

y = (1\u00d7-3+2\u00d76)\/(1+2)<\/p>\n\n\n\n

y = (-3+12)\/3<\/p>\n\n\n\n

y = 9\/3 = 3<\/p>\n\n\n\n

So the co-ordinates of the point of intersection are (0,3).<\/p>\n\n\n\n

(iii) By distance formula, d(AB) = \u221a[(x2<\/sub>-x1<\/sub>)2<\/sup>+(y2<\/sub>-y1<\/sub>)2<\/sup>]<\/p>\n\n\n\n

d(AB) = \u221a[(8-(-4))2<\/sup>+(-3-6)2<\/sup>]<\/p>\n\n\n\n

d(AB) = \u221a[(12)2<\/sup>+(-9)2<\/sup>]<\/p>\n\n\n\n

d(AB) = \u221a(144+81)<\/p>\n\n\n\n

d(AB) = \u221a225<\/p>\n\n\n\n

d(AB) = 15<\/p>\n\n\n\n

Hence the length of AB is 15 units.<\/p>\n\n\n\n

21.<\/strong> (i) Write down the co-ordinates of the point P that divides the line joining A ( \u2013 4, 1) and B (17,10) in the ratio 1 : 2.<\/strong>

(ii)Calculate the distance OP where O is the origin.<\/strong>

(iii)In what ratio does the y-axis divide the line AB ?<\/strong><\/p>\n\n\n\n

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

(i)Let P(x,y) divides the line segment joining the points A(-4,1), B(17,10) in the ratio 1:2,<\/p>\n\n\n\n

Here x1 <\/sub>= -4, y1<\/sub> = 1<\/p>\n\n\n\n

x2<\/sub> = 17, y2<\/sub> = 10<\/p>\n\n\n\n

m:n = 1:2<\/p>\n\n\n\n

By section formula, <\/strong>x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

x = (1\u00d717+2\u00d7-4)\/(1+2)<\/p>\n\n\n\n

x = (17+-8)\/3<\/p>\n\n\n\n

x = 9\/3<\/p>\n\n\n\n

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

By Section formula y = (my2<\/sub>+ny1<\/sub>)\/(m+n)<\/p>\n\n\n\n

y = (1\u00d710+2\u00d71)\/(1+2)<\/p>\n\n\n\n

y = (10+2)\/3<\/p>\n\n\n\n

y = 12\/3 = 4<\/p>\n\n\n\n

Hence the co-ordinates of the point P are (3,4).<\/p>\n\n\n\n

(ii)Since O is the origin, the co-ordinates of O are (0,0).<\/p>\n\n\n\n

By distance formula, d(OP) = \u221a[(x2<\/sub>-x1<\/sub>)2<\/sup>+(y2<\/sub>-y1<\/sub>)2<\/sup>]<\/p>\n\n\n\n

d(OP) = \u221a[(0-3)2<\/sup>+(0-4)2<\/sup>]<\/p>\n\n\n\n

d(OP) = \u221a[(3)2<\/sup>+(4)2<\/sup>]<\/p>\n\n\n\n

d(OP) = \u221a(9+16)<\/p>\n\n\n\n

d(OP) = \u221a25 = 5<\/p>\n\n\n\n

Hence the distance OP is 5 units.<\/p>\n\n\n\n

(iii)Let m:n be the ratio in which Y axis divide line AB.<\/p>\n\n\n\n

Since AB touches Y axis, its x co-ordinate will be zero.<\/p>\n\n\n\n

Here x1 <\/sub>= -4, y1<\/sub> = 1<\/p>\n\n\n\n

x2<\/sub> = 17, y2<\/sub> = 10<\/p>\n\n\n\n

By section formula, <\/strong>x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

0 = (m\u00d717+n\u00d7-4)\/(m+n)<\/p>\n\n\n\n

0 = (17m-4n)\/(m+n)<\/p>\n\n\n\n

17m-4n = 0<\/p>\n\n\n\n

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

m\/n = 4\/17<\/p>\n\n\n\n

m:n = 4:17<\/p>\n\n\n\n

Hence the ratio in which Y axis divide line AB is 4:17.<\/p>\n\n\n\n

22. Calculate the length of the median through the vertex A of the triangle ABC with vertices A (7, \u2013 3), B (5, 3) and C (3, \u2013 1).<\/strong><\/p>\n\n\n\n

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

Let M(x,y) be the median of \u0394ABC through A to BC.

M will be the midpoint of BC.<\/p>\n\n\n\n

x1 <\/sub>= 5, y1<\/sub> = 3<\/p>\n\n\n\n

x2 <\/sub>= 3, y2<\/sub> = -1<\/p>\n\n\n\n

By midpoint formula, x = (x1<\/sub>+x2<\/sub>)\/2<\/p>\n\n\n\n

x = (5+3)\/2 = 8\/2 = 4<\/p>\n\n\n\n

By midpoint formula, y = (y1<\/sub>+y2<\/sub>)\/2<\/p>\n\n\n\n

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

Hence the co-ordinates of M are (4,1).<\/p>\n\n\n\n

By distance formula, d(AM) = \u221a[(x2<\/sub>-x1<\/sub>)2<\/sup>+(y2<\/sub>-y1<\/sub>)2<\/sup>]<\/p>\n\n\n\n

x1 <\/sub>= 7, y1<\/sub> = -3<\/p>\n\n\n\n

x2 <\/sub>= 4, y2<\/sub> = 1<\/p>\n\n\n\n

d(AM) = \u221a[(4-7)2<\/sup>+(1-(-3))2<\/sup>]<\/p>\n\n\n\n

d(AM) = \u221a[(-3)2<\/sup>+(4)2<\/sup>]<\/p>\n\n\n\n

d(AM) = \u221a(9+16)<\/p>\n\n\n\n

d(AM) = \u221a25 = 5<\/p>\n\n\n\n

Hence the length of the median AM is 5 units.<\/p>\n\n\n\n

23. Three consecutive vertices of a parallelogram ABCD are A (1, 2), B (1, 0) and C (4, 0). Find the fourth vertex D.<\/strong><\/p>\n\n\n\n

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

Let M be the midpoint of the diagonals of the parallelogram ABCD.<\/p>\n\n\n\n

Co-ordinate of M will be the midpoint of diagonal AC.<\/p>\n\n\n\n

Given points are A(1,2), B(1,0) and C(4,0).<\/strong><\/p>\n\n\n\n

Consider line AC.<\/strong><\/p>\n\n\n\n

x1 <\/sub>= 1, y1<\/sub> = 2<\/p>\n\n\n\n

x2 <\/sub>= 4, y2<\/sub> = 0<\/p>\n\n\n\n

By midpoint formula, x = (x1<\/sub>+x2<\/sub>)\/2<\/p>\n\n\n\n

x = (1+4)\/2 = 5\/2<\/p>\n\n\n\n

By midpoint formula, y = (y1<\/sub>+y2<\/sub>)\/2<\/p>\n\n\n\n

y = (2+0)\/2 = 2\/2 = 1<\/p>\n\n\n\n

Hence the co-ordinates of M are (5\/2,1).<\/p>\n\n\n\n

M is also the midpoint of diagonal BD.<\/p>\n\n\n\n

Consider line BD and M as midpoint.<\/p>\n\n\n\n

x1 <\/sub>= 1, y1<\/sub> = 0<\/p>\n\n\n\n

x <\/sub>= 5\/2, y = 1<\/p>\n\n\n\n

By midpoint formula, x = (x1<\/sub>+x2<\/sub>)\/2<\/p>\n\n\n\n

5\/2 = (1+x2<\/sub>)\/2<\/p>\n\n\n\n

5 = 1+x2<\/sub><\/p>\n\n\n\n

x2<\/sub> = 5-1 = 4<\/p>\n\n\n\n

By midpoint formula, y = (y1<\/sub>+y2<\/sub>)\/2<\/p>\n\n\n\n

1 = (0+y2<\/sub>)\/2<\/p>\n\n\n\n

1 = y2<\/sub>\/2<\/p>\n\n\n\n

y2<\/sub> = 2<\/p>\n\n\n\n

Hence the co-ordinates of D are (4,2).<\/p>\n\n\n\n

24.<\/strong> If the points A ( \u2013 2, \u2013 1), B (1, 0), C (p, 3) and D (1, q) form a parallelogram ABCD, find the values of p and q.<\/strong><\/p>\n\n\n\n

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

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

Given vertices of the parallelogram are A(-2,-1), B(1, 0), C(p,3) and D(1,q).<\/strong><\/p>\n\n\n\n

Let M(x,y) be the midpoint of the diagonals of the parallelogram ABCD.<\/p>\n\n\n\n

Diagonals AC and BD bisect each other at M.<\/p>\n\n\n\n

When M is the midpoint of AC<\/p>\n\n\n\n

By midpoint formula,<\/p>\n\n\n\n

x = (-2+p)\/2 = (p-2)\/2 ..(i)<\/p>\n\n\n\n

y = (-1+3)\/2 = 2\/2 = 1 ..(ii)<\/p>\n\n\n\n

When M is the midpoint of BD<\/p>\n\n\n\n

By midpoint formula,<\/p>\n\n\n\n

x = (1+1)\/2 = 2\/2 = 1 ..(iii)<\/p>\n\n\n\n

y = (q+0)\/2 = q\/2 ..(iv)<\/p>\n\n\n\n

Equating (i) and (iii), we get<\/p>\n\n\n\n

(p-2)\/2 = 1<\/p>\n\n\n\n

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

p = 2+2 = 4<\/p>\n\n\n\n

Equating (i2) and (iv), we get<\/p>\n\n\n\n

q\/2 = 1<\/p>\n\n\n\n

q = 2<\/p>\n\n\n\n

Hence the value of p and q are 4 and 2 respectively.<\/p>\n\n\n\n

25. If two vertices of a parallelogram are (3, 2) ( \u2013 1, 0) and its diagonals meet at (2, \u2013 5), find the other two vertices of the parallelogram.<\/strong><\/p>\n\n\n\n

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

Let A(3,2) and B(-1,0) be the two vertices of the parallelogram ABCD.<\/p>\n\n\n\n

Let M(2,-5) be the point where diagonals meet.<\/p>\n\n\n\n

Since the diagonals of the parallelogram bisect each other, M is the midpoint of AC and BD.<\/p>\n\n\n\n

Consider A-M-C<\/p>\n\n\n\n

Let co-ordinate of C be (x2<\/sub>,y2<\/sub>)<\/p>\n\n\n\n

x1<\/sub> = 3, y1<\/sub> = 2<\/p>\n\n\n\n

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

By midpoint formula, x = (x1<\/sub>+x2<\/sub>)\/2<\/p>\n\n\n\n

2 = (3+x2<\/sub>)\/2<\/p>\n\n\n\n

3+x2<\/sub> = 4<\/p>\n\n\n\n

x2<\/sub> = 4-3 = 1<\/p>\n\n\n\n

By midpoint formula, y = (x1<\/sub>+x2<\/sub>)\/2<\/p>\n\n\n\n

-5 = (2+y2<\/sub>)\/2<\/p>\n\n\n\n

-10 = 2+y2<\/sub><\/p>\n\n\n\n

y2<\/sub> = -10-2 = -12<\/p>\n\n\n\n

Hence the co-ordinates of the point C are (1,-12).<\/p>\n\n\n\n

Consider B-M-D<\/p>\n\n\n\n

Let co-ordinate of D be (x2<\/sub>,y2<\/sub>)<\/p>\n\n\n\n

x1<\/sub> = -1, y1<\/sub> = 0<\/p>\n\n\n\n

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

By midpoint formula, x = (x1<\/sub>+x2<\/sub>)\/2<\/p>\n\n\n\n

2 = (-1+x2<\/sub>)\/2<\/p>\n\n\n\n

-1+x2<\/sub> = 4<\/p>\n\n\n\n

x2<\/sub> = 4+1 = 5<\/p>\n\n\n\n

By midpoint formula, y = (x1<\/sub>+x2<\/sub>)\/2<\/p>\n\n\n\n

-5 = (0+y2<\/sub>)\/2<\/p>\n\n\n\n

-10 = 0+y2<\/sub><\/p>\n\n\n\n

y2<\/sub> = -10<\/p>\n\n\n\n

Hence the co-ordinates of the point D are (5,-10).<\/p>\n\n\n\n

26. Prove that the points A ( \u2013 5, 4), B ( \u2013 1, \u2013 2) and C (5, 2) are the vertices of an isosceles right angled triangle. Find the co-ordinates of D so that ABCD is a square.<\/strong><\/p>\n\n\n\n

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

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

Given points are A(-5,4), B(-1,-2) and C(5,2) are given.

Since these are vertices of an isosceles triangle ABC then AB = BC.<\/p>\n\n\n\n

By distance formula, d(AB) = \u221a[(x2<\/sub>-x1<\/sub>)2<\/sup>+(y2<\/sub>-y1<\/sub>)2<\/sup>]<\/p>\n\n\n\n

Here x1<\/sub> = -5, y1<\/sub> = 4<\/p>\n\n\n\n

x2<\/sub> = -1, y2<\/sub> = -2<\/p>\n\n\n\n

d(AB) = \u221a[(-1-(-5))2<\/sup>+(-2-4)2<\/sup>]<\/p>\n\n\n\n

d(AB) = \u221a[(4)2<\/sup>+(6)2<\/sup>]<\/p>\n\n\n\n

d(AB) = \u221a(16+36)<\/p>\n\n\n\n

d(AB) = \u221a52 \u2026(i)<\/p>\n\n\n\n

By distance formula, d(BC) = \u221a[(x2<\/sub>-x1<\/sub>)2<\/sup>+(y2<\/sub>-y1<\/sub>)2<\/sup>]<\/p>\n\n\n\n

Here x1<\/sub> = -1, y1<\/sub> = -2<\/p>\n\n\n\n

x2<\/sub> = 5, y2<\/sub> = 2<\/p>\n\n\n\n

d(BC) = \u221a[(5-(-1))2<\/sup>+(2-(-2))2<\/sup>]<\/p>\n\n\n\n

d(BC) = \u221a[(6)2<\/sup>+(4)2<\/sup>]<\/p>\n\n\n\n

d(BC) = \u221a(36+16)<\/p>\n\n\n\n

d(BC) = \u221a52 \u2026(ii)<\/p>\n\n\n\n

From (i) and (ii) AB = BC<\/p>\n\n\n\n

So given points are the vertices of isosceles triangle.<\/p>\n\n\n\n

By distance formula, d(AC) = \u221a[(x2<\/sub>-x1<\/sub>)2<\/sup>+(y2<\/sub>-y1<\/sub>)2<\/sup>]<\/p>\n\n\n\n

Here x1<\/sub> = -5, y1<\/sub> = 4<\/p>\n\n\n\n

x2<\/sub> = 5, y2<\/sub> = 2<\/p>\n\n\n\n

d(AC) = \u221a[(5-(-5))2<\/sup>+(2-4)2<\/sup>]<\/p>\n\n\n\n

d(AC) = \u221a[(10)2<\/sup>+(-2)2<\/sup>]<\/p>\n\n\n\n

d(AC) = \u221a(100+4)<\/p>\n\n\n\n

d(AC) = \u221a104 \u2026.(iii)<\/p>\n\n\n\n

Apply Pythagoras theorem to triangle ABC<\/p>\n\n\n\n

AB2<\/sup>+BC2<\/sup> = (\u221a52)2<\/sup>+(\u221a52)2<\/sup><\/p>\n\n\n\n

= 52+52<\/p>\n\n\n\n

= 104 \u2026.(iv)<\/p>\n\n\n\n

AC2<\/sup> = (\u221a104)2<\/sup> = 104\u2026(v)<\/p>\n\n\n\n

From (iv) and (v) we got<\/p>\n\n\n\n

AB2<\/sup>+BC2<\/sup> = AC2<\/sup><\/p>\n\n\n\n

So Pythagoras theorem is satisfied.<\/p>\n\n\n\n

So the triangle is an isosceles right angled triangle.<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

If ABCD is a square, let the diagonals meet at O.<\/p>\n\n\n\n

Diagonals of a square bisect each other. So, O is the midpoint of AC and BD.<\/p>\n\n\n\n

Consider A-O-C<\/p>\n\n\n\n

x1 <\/sub>= -5, y1<\/sub> = 4<\/p>\n\n\n\n

x2 <\/sub>= 5, y2<\/sub> = 2<\/p>\n\n\n\n

By midpoint formula, x = (x1<\/sub>+x2<\/sub>)\/2<\/p>\n\n\n\n

x = (-5+5)\/2 = 0\/2 = 0<\/p>\n\n\n\n

By midpoint formula, y = (y1<\/sub>+y2<\/sub>)\/2<\/p>\n\n\n\n

y = (4+2)\/2 = 6\/2 = 3<\/p>\n\n\n\n

So co-ordinate of O is (0,3).<\/p>\n\n\n\n

Consider B-O-D<\/p>\n\n\n\n

Let co-ordinate of D be (x2<\/sub>,y2<\/sub>)<\/p>\n\n\n\n

x1<\/sub> = -1, y1<\/sub> = -2<\/p>\n\n\n\n

x = 0, y = 3<\/p>\n\n\n\n

By midpoint formula, x = (x1<\/sub>+x2<\/sub>)\/2<\/p>\n\n\n\n

0 = (-1+x2<\/sub>)\/2<\/p>\n\n\n\n

-1+x2<\/sub> = 0<\/p>\n\n\n\n

x2<\/sub> = 1<\/p>\n\n\n\n

By midpoint formula, y = (x1<\/sub>+x2<\/sub>)\/2<\/p>\n\n\n\n

3 = (-2+y2<\/sub>)\/2<\/p>\n\n\n\n

6 = -2+y2<\/sub><\/p>\n\n\n\n

y2<\/sub> = 6+2 = 8<\/p>\n\n\n\n

Hence the co-ordinates of the point D are (1,8).<\/p>\n\n\n\n

27.<\/strong> Find the third vertex of a triangle if its two vertices are ( \u2013 1, 4) and (5, 2) and midpoint of one sides is (0, 3).<\/strong><\/p>\n\n\n\n

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

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

Let A (-1,4) and B(5,2) are the vertices of the triangle and let D(0,3) is the midpoint of side AC.<\/strong><\/p>\n\n\n\n

Let co-ordinate of C be (x,y).<\/strong><\/p>\n\n\n\n

Consider D(0,3) as midpoint of AC<\/strong><\/p>\n\n\n\n

By midpoint formula,<\/strong><\/p>\n\n\n\n

(-1+x)\/2 = 0<\/strong><\/p>\n\n\n\n

-1 +x = 0<\/strong><\/p>\n\n\n\n

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

By midpoint formula,<\/strong><\/p>\n\n\n\n

(4+y)\/2 = 3<\/strong><\/p>\n\n\n\n

4+y = 6<\/strong><\/p>\n\n\n\n

y = 6-4 = 2<\/strong><\/p>\n\n\n\n

So the co-ordinates of C are (1,2).<\/strong><\/p>\n\n\n\n

Consider D(0,3) as midpoint of BC<\/strong><\/p>\n\n\n\n

By midpoint formula,<\/strong><\/p>\n\n\n\n

(5+x)\/2 = 0<\/strong><\/p>\n\n\n\n

5 +x = 0<\/strong><\/p>\n\n\n\n

x = -5<\/strong><\/p>\n\n\n\n

By midpoint formula,<\/strong><\/p>\n\n\n\n

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

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

y = 6-2 = 4<\/strong><\/p>\n\n\n\n

So the co-ordinates of C are (-5,4).<\/strong><\/p>\n\n\n\n

Hence the co-ordinates of the point C will be (1,2) or (-5,4).<\/strong><\/p>\n\n\n\n

28.<\/strong> Find the coordinates of the vertices of the triangle the middle points of whose sides are (0, \u00bd ) , ( \u00bd , \u00bd) and ( \u00bd , 0).<\/strong><\/p>\n\n\n\n

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

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

Let A(x1<\/sub>,y1<\/sub>), B(x2<\/sub>,y2<\/sub>) and C(x3<\/sub>,y3<\/sub>) be the vertices of the triangle ABC.<\/strong><\/p>\n\n\n\n

Consider AB<\/strong><\/p>\n\n\n\n

By midpoint formula, (x1<\/sub>+x2<\/sub>)\/2 = 0<\/strong><\/p>\n\n\n\n

x1<\/sub>+x2<\/sub> = 0<\/strong><\/p>\n\n\n\n

x1<\/sub> = -x2<\/sub> ..(i)<\/strong><\/p>\n\n\n\n

By midpoint formula, (y1<\/sub>+y2<\/sub>)\/2 = \u00bd<\/strong><\/p>\n\n\n\n

y1<\/sub>+y2<\/sub> = 1 \u2026(ii)<\/strong><\/p>\n\n\n\n

Consider AC<\/strong><\/p>\n\n\n\n

By midpoint formula, (x1<\/sub>+x3<\/sub>)\/2 = \u00bd<\/strong><\/p>\n\n\n\n

x1<\/sub>+x3<\/sub> = 1 \u2026(iii)<\/strong><\/p>\n\n\n\n

By midpoint formula, (y1<\/sub>+y3<\/sub>)\/2 = 0<\/strong><\/p>\n\n\n\n

y1<\/sub>+y3<\/sub> = 0<\/strong><\/p>\n\n\n\n

y1<\/sub> = -y3<\/sub> \u2026(iv)<\/strong><\/p>\n\n\n\n

Consider BC<\/strong><\/p>\n\n\n\n

By midpoint formula, (x2<\/sub>+x3<\/sub>)\/2 = \u00bd<\/strong><\/p>\n\n\n\n

x2<\/sub>+x3<\/sub> = 1 \u2026(v)<\/strong><\/p>\n\n\n\n

By midpoint formula, (y2<\/sub>+y3<\/sub>)\/2 = \u00bd<\/strong><\/p>\n\n\n\n

y2<\/sub>+y3<\/sub> = 1 \u2026(vi)<\/strong><\/p>\n\n\n\n

Substitute (i) in (iii)<\/strong><\/p>\n\n\n\n

Then (iii) becomes -x2<\/sub>+x3<\/sub> = 1<\/strong><\/p>\n\n\n\n

Equation (v) x2<\/sub>+x3<\/sub> = 1<\/strong><\/p>\n\n\n\n

Adding above two equations, we get<\/strong><\/p>\n\n\n\n

2x3 <\/sub>= 2<\/strong><\/p>\n\n\n\n

x3 <\/sub>= 2\/2 = 1<\/strong><\/p>\n\n\n\n

Substitute x3<\/sub> = 1 in (iii), we get x1<\/sub> = 0<\/strong><\/p>\n\n\n\n

x2<\/sub> = 0 [From (i)]<\/strong><\/p>\n\n\n\n

So x1<\/sub> = 0, x2<\/sub> = 0, x3<\/sub> = 1<\/strong><\/p>\n\n\n\n

Substitute (iv) in (ii)<\/strong><\/p>\n\n\n\n

Then (ii) becomes -y3<\/sub>+y2<\/sub> = 1<\/strong><\/p>\n\n\n\n

Equation (vi) y2<\/sub>+y3<\/sub> = 1<\/strong><\/p>\n\n\n\n

Adding above two equations, we get<\/strong><\/p>\n\n\n\n

2y2<\/sub> = 2<\/strong><\/p>\n\n\n\n

y2<\/sub> = 2\/2 = 1<\/strong><\/p>\n\n\n\n

Substitute y2<\/sub> = 1 in (i), we get y1<\/sub> = 0<\/strong><\/p>\n\n\n\n

y3<\/sub> = 0<\/strong><\/p>\n\n\n\n

So y1<\/sub> = 0, y2<\/sub> = 1, y3<\/sub> = 0<\/strong><\/p>\n\n\n\n

Hence the Co-ordinates of vertices are A(0,0), B(0,1) and C(1,0).<\/strong><\/p>\n\n\n\n

29. Show by section formula that the points (3, \u2013 2), (5, 2) and (8, 8) are collinear.<\/strong><\/p>\n\n\n\n

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

Let the point B(5,2) divides the line joining A(3,-2) and C(8,8) in the ratio m:n.<\/strong><\/p>\n\n\n\n

Then by section formula, <\/strong>x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

5 = (m\u00d78+n\u00d73)\/(m+n)<\/strong><\/p>\n\n\n\n

5 = (8m+3n)\/(m+n)<\/strong><\/p>\n\n\n\n

5m+5n = 8m+3n<\/strong><\/p>\n\n\n\n

2n = 3m<\/strong><\/p>\n\n\n\n

m\/n = 2\/3 \u2026(i)<\/strong><\/p>\n\n\n\n

By section formula, <\/strong>y = (my2<\/sub>+ny1<\/sub>)\/(m+n)<\/p>\n\n\n\n

2 = (m\u00d78+n\u00d7-2)\/(m+n)<\/strong><\/p>\n\n\n\n

2 = (8m-2n)\/(m+n)<\/strong><\/p>\n\n\n\n

2m+2n = 8m-2n<\/strong><\/p>\n\n\n\n

6m = 4n<\/strong><\/p>\n\n\n\n

m\/n = 4\/6 = 2\/3 \u2026(ii)<\/strong><\/p>\n\n\n\n

Here ratios are same.<\/strong><\/p>\n\n\n\n

So the points are collinear.<\/strong><\/p>\n\n\n\n

30.<\/strong> Find the value of p for which the points ( \u2013 5, 1), (1, p) and (4, \u2013 2) are collinear.<\/strong><\/p>\n\n\n\n

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

Let A(-5,1) divides the line joining (1,p) and (4,-2) in the ratio m:n<\/strong><\/p>\n\n\n\n

Then by section formula, <\/strong>x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

-5 = (m\u00d74+n\u00d71)\/(m+n)<\/strong><\/p>\n\n\n\n

-5 = (4m+n)\/(m+n)<\/strong><\/p>\n\n\n\n

-5m-5n = 4m+n<\/strong><\/p>\n\n\n\n

-9m = 6n<\/strong><\/p>\n\n\n\n

m\/n = -9\/6 = -2\/3 \u2026(i)<\/strong><\/p>\n\n\n\n

By section formula, <\/strong>y = (my2<\/sub>+ny1<\/sub>)\/(m+n)<\/p>\n\n\n\n

1 = (m\u00d7-2+n\u00d7p)\/(m+n)<\/strong><\/p>\n\n\n\n

1 = (-2m+pn)\/(m+n)<\/strong><\/p>\n\n\n\n

m+n = -2m+pn<\/strong><\/p>\n\n\n\n

3m = (p-1)n<\/strong><\/p>\n\n\n\n

m\/n = (p-1)\/3 \u2026.(ii)<\/strong><\/p>\n\n\n\n

Equating (i) and (ii)<\/strong><\/p>\n\n\n\n

(p-1)\/3 = -2\/3<\/strong><\/p>\n\n\n\n

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

p = -2+1 = -1<\/strong><\/p>\n\n\n\n

Hence the value of p is -1.<\/strong><\/p>\n\n\n\n

31.<\/strong> A (10, 5), B (6, \u2013 3) and C (2, 1) are the vertices of triangle ABC. L is the mid point of AB, M is the mid-point of AC. Write down the co-ordinates of L and M. Show that LM = \u00bd BC.<\/strong><\/p>\n\n\n\n

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

Given points are A(10,5), B(6,-3) and C(2,1).<\/strong><\/p>\n\n\n\n

Let L(x,y) be the midpoint of AB.<\/strong><\/p>\n\n\n\n

Here x1<\/sub>= 10, y1<\/sub> = 5<\/strong><\/p>\n\n\n\n

x2 <\/sub>= 6, y2<\/sub> = -3<\/strong><\/p>\n\n\n\n

By midpoint formula, x = (x1<\/sub>+x2<\/sub>)\/2<\/p>\n\n\n\n

x = (10+6)\/2 = 16\/2 = 8<\/strong><\/p>\n\n\n\n

By midpoint formula, y = (y1<\/sub>+y2<\/sub>)\/2<\/p>\n\n\n\n

y = (5-3)\/2 = 2\/2 = 1<\/strong><\/p>\n\n\n\n

So co-ordinates of L are (8,1).<\/strong><\/p>\n\n\n\n

Let M(x,y) be the midpoint of AC.<\/strong><\/p>\n\n\n\n

Here x1<\/sub>= 10, y1<\/sub> = 5<\/strong><\/p>\n\n\n\n

x2 <\/sub>= 2, y2<\/sub> = 1<\/strong><\/p>\n\n\n\n

By midpoint formula, x = (x1<\/sub>+x2<\/sub>)\/2<\/p>\n\n\n\n

x = (10+2)\/2 = 12\/2 = 6<\/strong><\/p>\n\n\n\n

By midpoint formula, y = (y1<\/sub>+y2<\/sub>)\/2<\/p>\n\n\n\n

y = (5+1)\/2 = 6\/2 = 3<\/strong><\/p>\n\n\n\n

So co-ordinates of M are (6,3).<\/strong><\/p>\n\n\n\n

By distance formula, d(LM) = \u221a[(x2<\/sub>-x1<\/sub>)2<\/sup>+(y2<\/sub>-y1<\/sub>)2<\/sup>]<\/p>\n\n\n\n

The points are L(8,1) and M(6,3)<\/strong><\/p>\n\n\n\n

So x1<\/sub>= 8, y1<\/sub> = 1<\/strong><\/p>\n\n\n\n

x2 <\/sub>= 6, y2<\/sub> = 3<\/strong><\/p>\n\n\n\n

d(LM) = \u221a[(x2<\/sub>-x1<\/sub>)2<\/sup>+(y2<\/sub>-y1<\/sub>)2<\/sup>]<\/p>\n\n\n\n

d(LM) = \u221a[(6-8)2<\/sup>+(3-1)2<\/sup>]<\/p>\n\n\n\n

d(LM) = \u221a[(-2)2<\/sup>+(2)2<\/sup>]<\/p>\n\n\n\n

d(LM) = \u221a(4+4)<\/p>\n\n\n\n

d(LM) = \u221a8 = 2\u221a2 \u2026(i)<\/p>\n\n\n\n

By distance formula, d(BC) = \u221a[(x2<\/sub>-x1<\/sub>)2<\/sup>+(y2<\/sub>-y1<\/sub>)2<\/sup>]<\/p>\n\n\n\n

The points are B(6,-3) and C(2,1).<\/strong><\/p>\n\n\n\n

So x1<\/sub>= 6, y1<\/sub> = -3<\/strong><\/p>\n\n\n\n

x2 <\/sub>= 2, y2<\/sub> = 1<\/strong><\/p>\n\n\n\n

d(BC) = \u221a[(x2<\/sub>-x1<\/sub>)2<\/sup>+(y2<\/sub>-y1<\/sub>)2<\/sup>]<\/p>\n\n\n\n

d(BC) = \u221a[(2-6)2<\/sup>+(1-(-3))2<\/sup>]<\/p>\n\n\n\n

d(BC) = \u221a[(-4)2<\/sup>+(4)2<\/sup>]<\/p>\n\n\n\n

d(BC) = \u221a(16+16)<\/p>\n\n\n\n

d(BC) = \u221a32 = 4\u221a2 \u2026(ii)<\/p>\n\n\n\n

From (i) and (ii), LM = \u00bd BC<\/p>\n\n\n\n

32.<\/strong> A (2, 5), B ( \u2013 1, 2) and C (5, 8) are the vertices of a triangle ABC. P and Q are points on AB and AC respectively such that AP : PB = AQ : QC = 1 : 2.<\/strong>

(i) Find the co-ordinates of P and Q.<\/strong>

(ii) Show that PQ = 1\/3 BC<\/strong><\/p>\n\n\n\n

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

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

(i) Given vertices of the ABC are A(2,5), B(-1,2) and C(5,8).<\/strong><\/p>\n\n\n\n

P and Q are points on AB and AC respectively such that AP:PB = AQ :QC = 1:2.<\/strong><\/p>\n\n\n\n

P(x,y) divides AB in the ratio 1:2.<\/strong><\/p>\n\n\n\n

x1<\/sub>= 2, y1<\/sub> = 5<\/strong><\/p>\n\n\n\n

x2 <\/sub>= -1, y2<\/sub> = 2<\/strong><\/p>\n\n\n\n

m:n = 1:2<\/strong><\/p>\n\n\n\n

By section formula, <\/strong>x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

x = (1\u00d7-1+2\u00d72)\/(1+2)<\/strong><\/p>\n\n\n\n

x = (-1+4)\/(3)<\/strong><\/p>\n\n\n\n

x = 3\/3 = 1<\/strong><\/p>\n\n\n\n

By section formula, <\/strong>y = (my2<\/sub>+ny1<\/sub>)\/(m+n)<\/p>\n\n\n\n

y = (1\u00d72+2\u00d75)\/(1+2)<\/strong><\/p>\n\n\n\n

y = (2+10)\/(3)<\/strong><\/p>\n\n\n\n

y = 12\/3 = 4<\/strong><\/p>\n\n\n\n

Co-ordinates of P are (1,4).<\/strong><\/p>\n\n\n\n

Q(x,y) divides AC in the ratio 1:2.<\/strong><\/p>\n\n\n\n

x1<\/sub>= 2, y1<\/sub> = 5<\/strong><\/p>\n\n\n\n

x2 <\/sub>= 5, y2<\/sub> = 8<\/strong><\/p>\n\n\n\n

m:n = 1:2<\/strong><\/p>\n\n\n\n

By section formula, <\/strong>x = (mx2<\/sub>+nx1<\/sub>)\/(m+n)<\/p>\n\n\n\n

x = (1\u00d75+2\u00d72)\/(1+2)<\/strong><\/p>\n\n\n\n

x = (5+4)\/(3)<\/strong><\/p>\n\n\n\n

x = 9\/3 = 3<\/strong><\/p>\n\n\n\n

By section formula, <\/strong>y = (my2<\/sub>+ny1<\/sub>)\/(m+n)<\/p>\n\n\n\n

y = (1\u00d78+2\u00d75)\/(1+2)<\/strong><\/p>\n\n\n\n

y = (8+10)\/(3)<\/strong><\/p>\n\n\n\n

y = 18\/3 = 6<\/strong><\/p>\n\n\n\n

Co-ordinates of Q are (3,6).<\/strong><\/p>\n\n\n\n

(ii)<\/strong> By distance formula, d(PQ) = \u221a[(x2<\/sub>-x1<\/sub>)2<\/sup>+(y2<\/sub>-y1<\/sub>)2<\/sup>]<\/p>\n\n\n\n

Points are P(1,4) and Q(3,6).<\/strong><\/p>\n\n\n\n

So x1<\/sub>= 1, y1<\/sub> = 4<\/strong><\/p>\n\n\n\n

x2 <\/sub>= 3, y2<\/sub> = 6<\/strong><\/p>\n\n\n\n

d(PQ) = \u221a[(x2<\/sub>-x1<\/sub>)2<\/sup>+(y2<\/sub>-y1<\/sub>)2<\/sup>]<\/p>\n\n\n\n

d(PQ) = \u221a[(3-1)2<\/sup>+(6-4)2<\/sup>]<\/p>\n\n\n\n

d(PQ) = \u221a[(2)2<\/sup>+(2)2<\/sup>]<\/p>\n\n\n\n

d(PQ) = \u221a(4+4)<\/p>\n\n\n\n

d(PQ) = \u221a8 = 2\u221a2 ..(i)<\/p>\n\n\n\n

By distance formula, d(BC) = \u221a[(x2<\/sub>-x1<\/sub>)2<\/sup>+(y2<\/sub>-y1<\/sub>)2<\/sup>]<\/p>\n\n\n\n

Points are B(-1,2) and C(5,8).<\/strong><\/p>\n\n\n\n

So x1<\/sub>= -1, y1<\/sub> = 2<\/strong><\/p>\n\n\n\n

x2 <\/sub>= 5, y2<\/sub> = 8<\/strong><\/p>\n\n\n\n

d(BC) = \u221a[(x2<\/sub>-x1<\/sub>)2<\/sup>+(y2<\/sub>-y1<\/sub>)2<\/sup>]<\/p>\n\n\n\n

d(BC) = \u221a[(5-(-1))2<\/sup>+(8-2)2<\/sup>]<\/p>\n\n\n\n

d(BC) = \u221a[(6)2<\/sup>+(6)2<\/sup>]<\/p>\n\n\n\n

d(BC) = \u221a(36+36)<\/p>\n\n\n\n

d(BC) = \u221a72 = \u221a(36\u00d72) = 6\u221a2 ..(ii)<\/p>\n\n\n\n

BC\/3 = 6\u221a2\/3 = 2\u221a2 = PQ<\/p>\n\n\n\n

PQ = 1\/3 BC.<\/strong><\/p>\n\n\n\n

Hence proved.<\/strong><\/p>\n\n\n\n

33.<\/strong> The mid-point of the line segment AB shown in the adjoining diagram is (4, \u2013 3). Write down the co-ordinates of A and B.<\/strong><\/p>\n\n\n\n

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

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

Let P(4,-3) be the midpoint of line joining the points A and B.<\/strong><\/p>\n\n\n\n

Since A lies on X axis, its co-ordinates are (x2<\/sub>,0)<\/strong><\/p>\n\n\n\n

Since B lies on Y axis, its co-ordinates are (0,y1<\/sub>)<\/strong><\/p>\n\n\n\n

By midpoint formula, x = (x1<\/sub>+x2<\/sub>)\/2<\/p>\n\n\n\n

4 = (0+x2<\/sub>)\/2<\/strong><\/p>\n\n\n\n

x2 <\/sub>= 4\u00d72 = 8<\/strong><\/p>\n\n\n\n

By midpoint formula, y = (y1<\/sub>+y2<\/sub>)\/2<\/p>\n\n\n\n

-3 = (y1<\/sub>+0)\/2<\/strong><\/p>\n\n\n\n

y1 <\/sub>= -3\u00d72 = -6<\/strong><\/p>\n\n\n\n

Hence the co-ordinates of A and B are (8,0)and (0,-6) respectively.<\/strong><\/p>\n\n\n\n

34. Find the co-ordinates of the centroid of a triangle whose vertices are A ( \u2013 1, 3), B(1, \u2013 1) and C (5, 1) (2006)<\/strong><\/p>\n\n\n\n

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

Given vertices of the triangle are A(-1,3), B(1,-1) and C(5,1)<\/strong><\/p>\n\n\n\n

Co-ordinates of the centroid of a triangle, whose vertices are (x1<\/sub>,y1<\/sub>), (x2<\/sub>,y2<\/sub>) and (x3<\/sub>,y3<\/sub>) are<\/p>\n\n\n\n

[(<\/strong>x1 <\/sub>+ x2<\/sub>+ x3<\/sub>)\/3, (y1 <\/sub>+ y2<\/sub>+ y3<\/sub>)\/3]<\/p>\n\n\n\n

(x1<\/sub>,y1<\/sub>) = (-1,3)<\/strong><\/p>\n\n\n\n

(x2<\/sub>,y2<\/sub>) = (1,-1)<\/strong><\/p>\n\n\n\n

(x3<\/sub>,y3<\/sub>) = (5,1)<\/strong><\/p>\n\n\n\n

(<\/strong>x1 <\/sub>+ x2<\/sub>+ x3<\/sub>)\/3 = (-1+1+5)\/3 = 5\/3<\/p>\n\n\n\n

(y1 <\/sub>+ y2<\/sub>+ y3<\/sub>)\/3 = (3-1+1)\/3 = 3\/3 = 1<\/p>\n\n\n\n

Hence the co-ordinates of centroid are (5\/3, 1).<\/strong><\/p>\n\n\n\n

35. Two vertices of a triangle are (3, \u2013 5) and ( \u2013 7, 4). Find the third vertex given that the centroid is (2, \u2013 1).<\/strong><\/p>\n\n\n\n

Solution:
Let third vertex be C(x3<\/sub>,y3<\/sub>).<\/strong><\/p>\n\n\n\n

Given (x1<\/sub>,y1<\/sub>) = (3,-5)<\/strong><\/p>\n\n\n\n

(x2<\/sub>,y2<\/sub>) = (-7,4)<\/strong><\/p>\n\n\n\n

Co-ordinates of centroid are (2,-1)<\/strong><\/p>\n\n\n\n

Co-ordinates of the centroid of a triangle, whose vertices are (x1<\/sub>,y1<\/sub>), (x2<\/sub>,y2<\/sub>) and (x3<\/sub>,y3<\/sub>) are<\/p>\n\n\n\n

[(<\/strong>x1 <\/sub>+ x2<\/sub>+ x3<\/sub>)\/3, (y1 <\/sub>+ y2<\/sub>+ y3<\/sub>)\/3]<\/p>\n\n\n\n

(<\/strong>x1 <\/sub>+ x2<\/sub>+ x3<\/sub>)\/3 = (3+-7+x3<\/sub>)\/3 = 2 [x co-ordinate of centroid]<\/p>\n\n\n\n

-4+x3<\/sub> = 2\u00d73<\/p>\n\n\n\n

-4+x3<\/sub> = 6<\/p>\n\n\n\n

x3<\/sub> = 6+4<\/p>\n\n\n\n

x3<\/sub> = 10<\/p>\n\n\n\n

(y1 <\/sub>+ y2<\/sub>+ y3<\/sub>)\/3 = -1 [y co-ordinate of centroid]<\/p>\n\n\n\n

-5+4+y3<\/sub> = -1\u00d73<\/p>\n\n\n\n

-1+y3<\/sub> = -3<\/p>\n\n\n\n

y3<\/sub> = -3+1<\/p>\n\n\n\n

y3<\/sub> = -2<\/p>\n\n\n\n

Hence the third vertex is (10,-2).<\/p>\n\n\n\n

36. The vertices of a triangle are A ( \u2013 5, 3), B (p, \u2013 1) and C (6, q). Find the values of p and q if the centroid of the triangle ABC is the point (1, \u2013 1).<\/strong><\/p>\n\n\n\n

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

Given vertices of the triangle are A(-5,3), B(p,-1) and C(6,q).<\/strong><\/p>\n\n\n\n

Co-ordinates of centroid are (1,-1).<\/strong><\/p>\n\n\n\n

Co-ordinates of the centroid of a triangle, whose vertices are (x1<\/sub>,y1<\/sub>), (x2<\/sub>,y2<\/sub>) and (x3<\/sub>,y3<\/sub>) are<\/p>\n\n\n\n

[(<\/strong>x1 <\/sub>+ x2<\/sub>+ x3<\/sub>)\/3, (y1 <\/sub>+ y2<\/sub>+ y3<\/sub>)\/3]<\/p>\n\n\n\n

(x1<\/sub>,y1<\/sub>) = (-5,3)<\/strong><\/p>\n\n\n\n

(x2<\/sub>,y2<\/sub>) = (p,-1)<\/strong><\/p>\n\n\n\n

(x3<\/sub>,y3<\/sub>) = (6,q)<\/strong><\/p>\n\n\n\n

x co-ordinate of centroid, (<\/strong>x1 <\/sub>+ x2<\/sub>+ x3<\/sub>)\/3 = (-5+p+6)\/3 = 1<\/p>\n\n\n\n

p+1 = 3<\/p>\n\n\n\n

p = 3-1<\/p>\n\n\n\n

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

y co-ordinate of centroid, <\/strong>(y1 <\/sub>+ y2<\/sub>+ y3<\/sub>)\/3 = (3-1+q)\/3 = -1<\/p>\n\n\n\n

2+q = 3\u00d7-1<\/p>\n\n\n\n

2+q = -3<\/p>\n\n\n\n

q = -3-2<\/p>\n\n\n\n

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

Hence the value of p and q are 2 and -5 respectively.<\/strong><\/p>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

ML Aggarwal Solutions for Class 10 Maths Chapter 11- Section Formula<\/p>\n\n\n\n

Download PDF<\/strong>: ML Aggarwal Solutions for Class 10 Maths Chapter 11- Section Formula PDF<\/a><\/p>\n\n\n\n

Chapterwise ML Aggarwal Solutions for Class 10 Maths :<\/strong><\/h2>\n\n\n\n