Class 10: Maths Chapter 6 solutions. Complete Class 10 Maths Chapter 6 Notes.
Contents
- 1 RS Aggarwal Solutions for Class 10 Maths Chapter 6–Coordinate Geometry
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RS Aggarwal Solutions for Class 10 Maths Chapter 6–Coordinate Geometry
RS Aggarwal 10th Maths Chapter 6, Class 10 Maths Chapter 6 solutions
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Question 1:
Find the distance between the points:
(i) A(9, 3) and B(15, 11)
(ii) A(7, −4) and B(−5, 1)
(iii) A(−6, −4) and B(9, −12)
(iv) A(1, −3) and B(4, −6)
(v) P(a + b, a − b) and Q(a −b, a + b)
(vi) P(a sin α, a cos α) and Q(a cos α, −a sin α)
ANSWER:
(i) A(9, 3) and B(15, 11)
The given points are A(9, 3) and B(15, 11).
Then (x1 = 9, y1 = 3) and (x2 = 15, y2 = 11)
(ii) A(7, −4) and B(−5, 1)
The given points are A(7, −4) and B(−5, 1).
Then (x1= 7, y1 = −4) and (x2 = −5, y2 = 1)
(iii) A(−6, −4) and B(9, −12)
The given points are A(−6, −4) and B(9, −12).
Then (x1 = −6, y1 = −4) and (x2 = 9, y2 = −12)
(iv) A(1, −3) and B(4, −6)
The given points are A(1, −3) and B(4, −6).
Then (x1 = 1, y1 = −3) and (x2 = 4, y2 = −6)
(v) P(a + b, a − b) and Q(a −b, a + b)
The given points are P(a + b, a − b) and Q(a − b, a + b).
Then (x1 = a + b, y1 = a − b) and (x2 = a − b, y2 = a + b)
(vi) P(a sin α, a cos α) and Q(a cos α, −a sin α)
The given points are P(a sin α, a cos α) and Q(a cos α, −a sin α).
Then (x1 = a sin α, y1 = a cos α) and (x2 = a cos α, y2 = −a sin α)
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Question 2:
Find the distance of each of the following points from the origin:
(i) A(5, −12)
(ii) B(−5, 5)
(iii) C(−, −6)4
ANSWER:
(i) A(5, −12)
Let O(0, 0) be the origin.
(ii) B(−5, 5)
Let O(0, 0) be the origin.
(iii) C(−4, −6)
Let O(0,0) be the origin.
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Question 3:
Find all possible values of x for which the distance between the points A(x, −1) and B(5, 3) is 5 units.
ANSWER:
Given AB = 5 units
Therefore, (AB)2 = 25 units
Therefore, x = 2 or 8.
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Question 4:
Find all possible values of y for which the distance between the points A(2,−3) and B(10,y)A2,-3 and B10,y is 10 units.
ANSWER:
The given points are A(2,−3) and B(10,y)A2,-3 and B10,y.
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Question 5:
Find the values of x for which the distance between the points P(x, 4) and Q(9, 10) is 10 units.
ANSWER:
The given points are P(x, 4) and Q(9, 10).
Hence, the values of x are 1 and 17.
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Question 6:
If the point A(x, 2) is equidistant from the points B(8, − 2) and C(2, − 2), find the value of x. Also, find the length of AB.
ANSWER:
As per the question
Squaring both sides, we get
Now,
Hence, x = 5 and AB = 5 units.
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Question 7:
If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find the value of p. Also, find the length of AB.
ANSWER:
As per the question
Squaring both sides, we get
Now,
Hence, p = 1 and AB = 10−−√10 units.
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Question 8:
Find the coordinates of the point on x-axis which is equidistant from the points (–2, 5) and (2, –3).
ANSWER:
Let the point on the x – axis be (x, 0).
Point on the x – axis is (−-2, 0).
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Question 9:
Find points on the x-axis, each of which is at a distance of 10 units from the point A(11, −8).
ANSWER:
Let P (x, 0) be the point on the x-axis. Then as per the question, we have
Hence, the points on the x-axis are (5, 0) and (17, 0).
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Question 10:
Find the point on the y-axis which is equidistant from the points A(6, 5) and B(− 4, 3).
ANSWER:
Let P (0, y) be a point on the y-axis. Then as per the question, we have
Hence, the point on the y-axis is (0, 9).
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Question 11:
If the point P(x, y) is equidistant from the points A(5, 1) and B(− 1, 5), prove that 3x = 2y.
ANSWER:
As per the question, we have
Hence, 3x = 2y.
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Question 12:
If P(x, y) is a point equidistant from the points A(6, −1) and B(2, 3), show that x − y = 3.
ANSWER:
The given points are A(6, −1) and B(2, 3). The point P(x, y) is equidistant from the points A and B. So, PA = PB.
Also, (PA)2 = (PB)2
Hence proved.
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Question 13:
Find the coordinates of the point equidistant from the three points A(5, 3), B(5, −5) and C(1, −5).
ANSWER:
Let the required point be P(x, y). Then AP = BP = CP
That is, (AP)2=(BP)2=(CP)2
This means (AP)2=(BP)2
Hence, the required point is (3, −1).
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Question 14:
If the points A(4, 3) and B(x, 5) lie on a circle with centre O(2, 3), find the value of x.
ANSWER:
Given, the points A(4, 3) and B(x, 5) lie on a circle with centre O(2, 3).
Then OA = OB
Also (OA)2 = (OB)2
Therefore, x = 2.
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Question 15:
If the point C(− 2, 3) is equidistant from the points A(3, − 1) and B(x, 8), find the value of x. Also, find the distance BC.
ANSWER:
As per the question, we have
AC=BC
Hence, x = 2 or −6 and BC=41−−√ unitsBC=41 units.
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Question 16:
If the point P(2, 2) is equidistant from the points A(−2, k) and B(−2k, −3), find k. Also find the length of AP.
ANSWER:
As per the question, we have
Now for k=−1
For k=−3
Hence, k=−1,−3 ; AP=5 units for k=−1 and AP=41−−√ units for k=−3k=-1,-3 ; AP=5 units for k=-1 and AP=41 units for k=-3.
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Question 17:
(i) If the point (x, y) is equidistant from the points (a + b, b − a) and (a − b, a + b), prove that bx = ay.
(ii) If the distances of P(x, y) from A(5, 1) and B(–1, 5) are equal then prove that 3x = 2y.
ANSWER:
(i) As per the question, we have
Hence, bx = ay.
(ii)
As per the question, we have
Hence, 3x = 2y.
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Question 18:
Using the distance formula, show that the given points are collinear:
(i) (1, −1), (5, 2) and (9, 5) (ii) (6, 9), (0, 1) and (−6, −7)
(iii) (−1, −1), (2, 3) and (8, 11) (iv) (−2, 5), (0, 1) and (2, −3)
ANSWER:
(i)
Let A(1, −1), B(5, 2) and C(9, 5) be the given points. Then
∴AB+BC=(5+5) units=10 units=AC∴AB+BC=5+5 units=10 units=AC
Hence, the given points are collinear.
(ii)
Let A(6, 9), B(0, 1) and C(−6, −7) be the given points. Then
∴AB+BC=(10+10) units=20 units=AC
Hence, the given points are collinear.
(iii)
Let A(−1, −1), B(2, 3) and C(8, 11) be the given points. Then
∴AB+BC=(5+10) units=15 units=AC
∴AB+BC=(25–√+25–√) units=45–√ units=AC∴AB+BC=25+25 units=45 units=AC
Hence, the given points are collinear.
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Question 19:
Prove that the points A(7, 10), B(−2, 5) and C(3, −4) are the vertices of an isosceles right triangle.
ANSWER:
The given points are A(7, 10), B(−2, 5) and C(3, −4).
Since, AB and BC are equal, they form the vertices of an isosceles triangle.
Also, (AB)2+(BC)2 = (106−−−√)2+ (106−−−√)2=2121062+ 1062=212
and (AC)2 = (212−−−√)22122 = 212
Thus, (AB)2+(BC)2 = (AC)2
This show that ΔABC∆ABC is right- angled at B.
Therefore, the points A(7, 10), B(−2, 5) and C(3, −4) are the vertices of an isosceles right-angled triangle.
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Question 20:
Show that the points A(3, 0), B(6, 4) and C(− 1, 3) are the vertices of an isosceles right triangle.
ANSWER:
The given points are A(3, 0), B(6, 4) and C(− 1, 3). Now
Therefore, A(3, 0), B(6, 4) and C(− 1, 3) are the vertices of an isosceles right triangle.
Page No 312:
Question 21:
Show that the points A(5, 2), B(2, − 2) and C(− 2, t) are the vertices of a right triangle with ∠B=90∘∠B=90∘, then find the value of t.
ANSWER:
Hence, t = 1.
Page No 312:
Question 22:
Prove that the points A(2, 4), B(2, 6) and C(2+3–√, 5)C2+3, 5 are the vertices of an equilateral triangle.
ANSWER:
The given points are A(2, 4), B(2, 6) and C(2+3–√, 5)C2+3, 5. Now
Hence, the points A(2, 4), B(2, 6) and C(2+3–√, 5)C2+3, 5 are the vertices of an equilateral triangle.
Page No 312:
Question 23:
Show that the points (− 3, − 3), (3, 3) and (−33–√, 33–√)-33, 33 are the vertices of an equilateral triangle.
ANSWER:
Let the given points be A(− 3, − 3), B(3, 3) and C(−33–√, 33–√)-33, 33. Now
Hence, the given points are the vertices of an equilateral triangle.
Page No 312:
Question 24:
Show that the points A(−5, 6) B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle. Calculate its area.
ANSWER:
Let the given points be A(−5, 6) B(3, 0) and C(9, 8).
Also, (AB)2+(BC)2 = (10)2+ (10)2=200102+ 102=200
and (AC)2 = (102–√)2=2001022=200
Thus, (AB)2+(BC)2 = (AC)2
This show that ΔABC∆ABC is right- angled at B.
Therefore, the points A(−5, 6) B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle.
Also, area of a triangle = 12×base×heightarea of a triangle = 12×base×height
If AB is the height and BC is the base,Area = 12×10×10 = 50 square units
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Question 25:
Show that the points O(0, 0) A(3, 3–√3) and B(3, −3–√3) are the vertices of an equilateral triangle. Find the area of this triangle.
ANSWER:
The given points are O(0, 0) A(3, 3–√3) and B(3, –√3).
Thus, the points O(0, 0) A(3, 3–√3)and B(3, − 3–√3) are the vertices of an equilateral triangle.
Also, the area of the triangle OAB = 3√4×(side)2
Page No 313:
Question 26:
Show that the following points are the vertices of a square.
(i) A(3, 2), B(0, 5), C(−3, 2) and D(0, −1)
(ii) A(6, 2), B(2, 1), C(1, 5) and D(5, 6)
(iii) P(0, −2), Q(3, 1), R(0, 4) and S(−3, 1)
ANSWER:
(i) The given points are A(3, 2), B(0, 5), C(−3, 2) and D(0, −1).
Therefore, the given points form a square.
(ii) The given points are A(6, 2), B(2, 1), C(1, 5) and D(5, 6).
Therefore, the given points form a square.
(iii) The given points are P(0, −2), Q(3, 1), R(0, 4) and S(−3, 1).
Therefore, the given points form a square.
Page No 313:
Question 27:
Show that the points A(−3, 2), B(−5, −5), C(2, −3) and D(4, 4) are the vertices of a rhombus. Find the area of this rhombus.
ANSWER:
The given points are A(−3, 2), B(−5, −5), C(2, −3) and D(4, 4).
Therefore, ABCD is a quadrilateral with equal sides and unequal diagonals.
Hence, ABCD is a rhombus.
Page No 313:
Question 28:
Show that the points A(3, 0), B(4, 5), C(−1, 4) and D(−2, −1) are the vertices of a rhombus. Find its area.
ANSWER:
The given points are A(3, 0), B(4, 5), C(− 1, 4) and D(− 2, − 1).
∵AB=BC=CD=AD=62–√ and AC≠BD∵AB=BC=CD=AD=62 and AC≠BD
Therefore, the given points are the vertices of a rhombus.
Hence, the area of the rhombus is 24 sq. units.
Page No 313:
Question 29:
Show that the points A(6, 1), B(8, 2), C(9, 4) and D(7, 3) are the vertices of a rhombus. Find its area.
ANSWER:
The given points are A(6, 1), B(8, 2), C(9, 4) and D(7, 3).
∵AB=BC=CD=AD=5–√ and AC≠BD
Therefore, the given points are the vertices of a rhombus. Now
Hence, the area of the rhombus is 3 sq. units.
Page No 313:
Question 30:
Show that the points A(2, 1), B(5, 2), C(6, 4) and D(3, 3) are the angular points of a parallelogram. Is this figure a rectangle?
ANSWER:
The given points are A(2, 1), B(5, 2), C(6, 4) and D(3, 3).
But diagonal AC is not equal to diagonal BD.
Hence, the given points do not form a rectangle.
Page No 313:
Question 31:
Show that A(1, 2), B(4, 3), C(6, 6) and D(3, 5) are the vertices of a parallelogram. Show that ABCD is not a rectangle.
ANSWER:
The given vertices are A(1, 2), B(4, 3), C(6, 6) and D(3, 5).
∵AB=CD=10−−√ units and BC=AD=13−−√ units∵AB=CD=10 units and BC=AD=13 units
Therefore, ABCD is a parallelogram. Now
Thus, the diagonals AC and BD are not equal and hence ABCD is not a rectangle.
Page No 313:
Question 32:
Show that the following points are the vertices of a rectangle.
(i) A(−4, −1), B(−2, −4) C(4, 0) and D(2, 3)
(ii) A(2, −2), B(14, 10) C(11, 13) and D(−1, 1)
(iii) A(0, −4), B(6, 2) C(3, 5) and D(−3, −1)
ANSWER:
(i)
(ii) The given points are A(2, −2), B(14, 10) C(11, 13) and D(−1, 1).
Also, diagonal AC = diagonal BD
Hence, the given points form a rectangle.
(iii) The given points are A(0, −4), B(6, 2) C(3, 5) and D(−3, −1).
Also, diagonal AC = diagonal BD.
Hence, the given points form a rectangle.
Page No 313:
Question 33:
Show that ΔABC with vertices A(–2, 0), B(0, 2) and C(2, 0) is similar to ΔDEF with vertices D(–4, 0), F(4, 0) and E(0, 4). [CBSE 2017]
ANSWER:
In ΔABC, the coordinates of the vertices are A(–2, 0), B(0,2), C(2,0).
In ΔDEF, the coordinates of the vertices are D(–4, 0), E(4, 0), F(0, 4).
Now, for ΔABC and ΔDEF to be similar, the corresponding sides should be proportional.
Page No 324:
Question 1:
(i) Find the coordinates of the point that divides the join of A(−1, 7) and B(4, −3) in the ratio 2 : 3.
(ii) Find the coordinates of the point which divides the join of A(−5, 11) and B(4, −7) in the ratio 7 : 2.
ANSWER:
(i) The end points of AB are A(−1, 7) and B(4, −3).
Therefore, (x1 = −1, y1 = 7) and (x2 = 4, y2 = −3)
Also, m = 2 and n = 3
Let the required point be P(x, y).
By section formula, we get:
Hence, the coordinates of the required point are (1, 3).
(ii) The end points of AB are A(−5, 11) and B(4, −7).
Therefore, (x1 = −5, y1 = 11) and (x2 = 4, y2 = −7).
Also, m = 7 and n = 2
Let the required point be P(x, y).
By section formula, we have:
Hence, the required point is P(2, −3).
Page No 324:
Question 2:
Find the coordinates of the points of trisection of the line segment joining the points A(7, –2) and B(1, –5). [CBSE 2017]
ANSWER:

Consider the figure.
Here points P and Q trisect AB.
Therefore, P divides AB into 1 : 2 and Q divides AB into 2 : 1.
Using section formula, coordinates of P are;
Similarly, coordinates of Q are;
Therefore, coordinates of points P and Q are (5, −-3) and (3, −-4) respectively.
Page No 324:
Question 3:
If the coordinates of points A and B are (−2, −2) and (2, −4) respectively, find the coordinates of the point P
such that AP=37ABAP=37AB, where P lies on the line segment AB. [CBSE 2015]
ANSWER:
The coordinates of the points A and B are (−2, −2) and (2, −4) respectively, where AP=37ABAP=37AB and P lies on the line segment AB. So
Let (x, y) be the coordinates of P which divides AB in the ratio 3 : 4 internally. Then
Hence, the coordinates of point P are
Page No 325:
Question 4:
Point A lies on the line segment PQ joining P(6, −6) and Q(−4, −1) in such a way that PAPQ=25PAPQ=25.
If the point A also lies on the line 3x + k (y + 1) = 0, find the value of k.
ANSWER:
Let the coordinates of A be (x, y). Here, PAPQ=25PAPQ=25. So,
Let (x, y) be the coordinates of A, which divides PQ in the ratio 2 : 3 internally. Then using section formula, we get
Now, the point (2, −4) lies on the line 3x + k (y + 1) = 0, therefore
Hence, k = 2.
Page No 325:
Question 5:
Point P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) in five equal parts.
Find the coordinates of the points P, Q and R.
ANSWER:
Since, the points P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) in five equal parts, so
AP = PQ = QR = RS = SB
Here, point P divides AB in the ratio of 1 : 4 internally. So using section formula, we get
The point Q divides AB in the ratio of 2 : 3 internally. So using section formula, we get
The point R divides AB in the ratio of 3 : 2 internally. So using section formula, we get
Hence, the coordinates of the points P, Q and R are (2, 3), (3, 4) and (4, 5) respectively.
Page No 325:
Question 6:
Points P, Q and R, in that order, divide a line segment joining A(1, 6) and B(5, −2) in four equal parts. Find the coordinates of P, Q and R.
ANSWER:
The given points are A(1, 6) and B(5, −2).
Then, P(x, y) is a point that divides the line AB in the ratio 1:3.
By the section formula:
Therefore, the coordinates of point P are (2, 4).
Let Q be the mid point of AB.
Then, Q(x, y):
Therefore, the coordinates of Q are (3, 2).
Let R (x, y) be a point that divides AB in the ratio 3:1.
Then, by the section formula:
Therefore, the coordinates of R are (4, 0).
Hence, the coordinates of point P, Q and R are (2, 4), (3, 2) and (4, 0) respectively.
Page No 325:
Question 7:
The line segment joining the points A(3, −4) and B(1, 2) is trisected at the points P(p, −2) and Q(53,q)Q53,q. Find the values of p and q.
ANSWER:
Let P and Q be the points of trisection of AB.
Then, P divides AB in the ratio 1:2.
So, the coordinates of P are
Hence, the coordinates of P are (7373, −2).
But (p, −2) are the coordinates of P.
So, p=73p=73
Also, Q divides the line AB in the ratio 2:1.
So, the coordinates of Q are
But the given coordinates of Q are (53, q).Q are 53, q.
So, q = 0
Thus, p=73p=73 and q=0q=053
Page No 325:
Question 8:
Find the coordinates of the midpoints of the line segment joining:
(i) A(3, 0) and B(−5, 4)
(ii) P(−11, −8) and Q(8, −2)
ANSWER:
(i) The given points are A(3, 0) and B(−5, 4).
Let (x, y) be the mid point of AB. Then:
Therefore, (−1, 2) are the coordinates of mid point of AB.
(ii) The given points are P(−11, −8) and Q(8, −2).
Let (x, y) be the mid point of PQ. Then:
Therefore, (−32, −5)-32, -5 are the coordinates of midpoint of PQ.
Page No 325:
Question 9:
If (2, p) is the midpoint of the line segment joining the points A(6, −5) and B(−2, 11), find the value of p.
ANSWER:
The given points are A(6, −5) and B(−2, 11).
Let (x, y) be the mid point of AB. Then:
So, the midpoint of AB is (2, 3).
But it is given that the midpoint of AB is (2, p).
Therefore, the value of p = 3.
Page No 325:
Question 10:
The midpoint of the line segment joining A(2a, 4) and B(−2, 3b) is C(1, 2a + 1). Find the values of a and b.
ANSWER:
The points are A(2a, 4) and B(−2, 3b).
Let C(1, 2a + 1) be the mid point of AB. Then:
Page No 325:
Question 11:
The line segment joining A(−2, 9) and B(6, 3) is a diameter of a circle with centre C. Find the coordinates of C.
ANSWER:
The given points are A(−2, 9) and B(6, 3).
Then, C(x, y) is the midpoint of AB.
Therefore, the coordinates of point C are (2, 6).
Page No 325:
Question 12:
Find the coordinates of a point A, where AB is the diameter of a circle with centre C(2, −3) and the other end of the diameter is B(1, 4).
ANSWER:
C(2, −3) is the centre of the given circle. Let A(a, b) and B(1, 4) be the two end-points of the given diameter AB. Then, the coordinates of C are
Therefore, the coordinates of point A are (3, -10).
Page No 325:
Question 13:
In what ratio does the point P(2, 5) divide the join of A(8, 2) and B(−6, 9)?
ANSWER:
Let the point P(2, 5) divide AB in the ratio k : 1.
Then, by section formula, the coordinates of P are
Therefore, the point P(2, 5) divides AB in the ratio 3 : 4.
Page No 325:
Question 14:
Find the ratio in which the point P(34, 512)P34, 512 divides the line segment joining the points
A(12, 32)A12, 32 and (2,−5)2,-5. [CBSE 2015]
ANSWER:
Let k : 1 be the ratio in which the point P(34, 512)P34, 512 divides the line segment joining the points A(12, 32)A12, 32 and (2,−5)2,-5. Then
Hence, the required ratio is 1 : 5.
Page No 325:
Question 15:
Find the ratio in which the point P(m, 6) divides the join of A(−4, 3) and B(2, 8). Also, find the value of m.
ANSWER:
Let the point P(m, 6) divide the line AB in the ratio k : 1.
Then, by the section formula:
Page No 325:
Question 16:
Find the ratio in which the point (−3, k) divides the join of A(−5, −4) and B(−2, 3). Also, find the value of k.
ANSWER:
Let the point P(−3, k) divide the line AB in the ratio s : 1.
Then, by the section formula:
Page No 325:
Question 17:
In what ratio is the line segment joining A(2, −3) and B(5, 6) divided by the x-axis? Also, find the coordinates of the point of division.
ANSWER:
Let AB be divided by the x-axis in the ratio k : 1 at the point P.
Then, by section formula the coordinates of P are
P=(5k+2k+1, 6k−3k+1)P=5k+2k+1, 6k-3k+1
But P lies on the x-axis; so, its ordinate is 0.
Therefore, the required ratio is 1212 : 1, which is same as 1 : 2.
Thus, the x-axis divides the line AB in the ratio 1 : 2 at the point P.
Applying k = 1212, we get the coordinates of point :
Hence, the point of intersection of AB and the x-axis is P(3, 0).
Page No 325:
Question 18:
In what ratio is the line segment joining the points A(−2, −3) and B(3, 7) divided by the y-axis? Also, find the coordinates of the points of division.
ANSWER:
Let AB be divided by the x-axis in the ratio k : 1 at the point P.
Then, by section formula the coordinates of P are
But P lies on the y-axis; so, its abscissa is 0.
Therefore, the required ratio is 2323 : 1, which is same as 2 : 3.
Thus, the x-axis divides the line AB in the ratio 2:3 at the point P.
Applying k=2323, we get the coordinates of point P:
Hence, the point of intersection of AB and the x-axis is P(0, 1).
Page No 326:
Question 19:
In what ratio does the line x − y − 2 = 0 divide the line segment joining the points A(3, −1) and B(8, 9)?
ANSWER:
Let the line x − y − 2 = 0 divide the line segment joining the points A(3, −1) and B(8, 9) in the ratio k : 1 at P.
Then, the coordinates of P are
Since, P lies on the line x − y − 2 = 0, we have:
So, the required ratio is 2323 : 1, which is equal to 2 : 3.
Page No 326:
Question 20:
Find the lengths of the median of ∆ABC whose vertices are A(0, −1), B(2, 1) and C(0, 3).
ANSWER:
The vertices of ∆ABC are A(0, −1), B(2, 1) and C(0, 3).
Let AD, BE and CF be the medians of ∆ABC.

Let D be the midpoint of BC. So, the coordinates of D are
Let E be the midpoint of AC. So, the coordinates of E are
Let F be the midpoint of AB. So, the coordinates of F are
Therefore, the lengths of the medians: AD = 10−−√10 units, BE = 2 units and CF = 10−−√10 units
Page No 326:
Question 21:
Find the centroid of ∆ABC whose vertices are A(−1, 0), B(5, −2) and C(8, 2).
ANSWER:
Here, (x1 = −1, y1 = 0), (x2 = 5, y2 = −2) and (x3 = 8, y3 = 2).
Let G(x, y) be the centroid of the ∆ABC. Then,
Hence, the centroid of ∆ABC is G(4, 0).
Page No 326:
Question 22:
If G(−2, 1) is the centroid of a ∆ABC and two of its vertices are A(1, −6) and B(−5, 2), find the third vertex of the triangle.
ANSWER:
Two vertices of ∆ABC are A(1, −6) and B(−5, 2). Let the third vertex be C(a, b).
Then the coordinates of its centroid are
But it is given that G(−2, 1) is the centroid. Therefore,
Therefore, the third vertex of ∆ABC is C(−2, 7).
Page No 326:
Question 23:
Find the third vertex of ∆ABC if two of its vertices are B(−3, 1) and C(0, −2) and its centroid is at the origin.
Answer:
Two vertices of ∆ABC are B(−3,1) and C(0, −2). Let the third vertex be A(a, b).
Then, the coordinates of its centroid are
But it is given that the centroid is at the origin, that is G(0, 0). Therefore,
Therefore, the third vertex of ∆ABC is A(3, 1).
Page No 326:
Question 24:
Show that the points A(3, 1), B(0, −2), C(1, 1) and D(4, 4) are the vertices of a parallelogram ABCD.
Answer:
The points are A(3, 1), B(0, −2), C(1, 1) and D(4, 4).
Join AC and BD, intersecting at O.
We know that the diagonals of a parallelogram bisect each other.
Thus, the diagonals AC and BD have the same midpoint.
Therefore, ABCD is a parallelogram.
Page No 326:
Question 25:
If the points P(a, −11), Q(5, b), R(2, 15) and S(1, 1) are the vertices of a parallelogram PQRS, find the values of a and b.
Answer:
The points are P(a, −11), Q(5, b), R(2, 15) and S(1, 1).
Join PR and QS, intersecting at O.
We know that the diagonals of a parallelogram bisect each other.
Therefore, O is the midpoint of PR as well as QS.
Page No 326:
Question 26:
If three consecutive vertices of a parallelogram ABCD are A(1, −2), B(3, 6) and C(5, 10), find its fourth vertex D.
Answer:
Let A(1, −2), B(3, 6) and C(5, 10) be the three vertices of a parallelogram ABCD and the fourth vertex be D(a, b).
Join AC and BD intersecting at O.
We know that the diagonals of a parallelogram bisect each other.
Therefore, O is the midpoint of AC as well as BD.
Therefore, the fourth vertex is D(3, 2).
Page No 326:
Question 27:
In what ratio does y-axis divide the line segment joining the points (−4, 7) and (3, −7)? [CBSE 2012]
Answer:
Let y-axis divides the line segment joining the points (−4, 7) and (3, −7) in the ratio k : 1. Then
Hence, the required ratio is 4 : 3.
Page No 326:
Question 28:
If the point lies on the line segment joining the points A(3, −5) and B(−7, 9) then find the
ratio in which P divides AB. Also, find the value of y.
Answer:
Let the point divides the line segment joining the points A(3, −5) and B(−7, 9) in the ratio k : 1. Then
Now, substituting in , we get
Hence, required ratio is 1 : 3 and .
Page No 326:
Question 29:
Find the ratio in which the line segment joining the points A(3, −3) and B(−2, 7) is divided by x-axis.
Also, find the point of division. [CBSE 2015]
Answer:
The line segment joining the points A(3, − 3) and B(− 2, 7) is divided by x-axis. Let the required ratio be k : 1. So,
Now
Hence, the required ratio is 3 : 7 and the point of division is .
Page No 326:
Question 30:
The base QR of an equilateral triangle PQR lies on x-axis. The coordinates of the point Q are (−4, 0)
and origin is the midpoint of the base. Find the coordinates of the points P and R.
Answer:
Let (x, 0) be the coordinates of R. Then
Thus, the coordinates of R are (4, 0).
Here, PQ = QR = PR and the coordinates of P lies on y-axis. Let the coordinates of P be (0, y). Then
Hence, the required coordinates are .
Page No 326:
Question 31:
The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point C are (0, −3).
The origin is the midpoint of the base. Find the coordinates of the points A and B. Also, find
the coordinates of another point D such that ABCD is a rhombus.
Answer:
Let (0, y) be the coordinates of B. Then
Thus, the coordinates of B are (0, 3).
Here, AB = BC = AC and by symmetry the coordinates of A lies on x-axis. Let the coordinates of A be (x, 0). Then
If the coordinates of point A are , then the coordinates of D are .
If the coordinates of point A are , then the coordinates of D are .
Hence, the required coordinates are or .
Page No 326:
Question 32:
Find the ratio in which the point P( −1, y), lying on the line segment joining points A(−3, 10) and B(6, −8) divides it. Also, find the value of y. Also, find the value of y. [CBSE 2013]
Answer:
Let k be the ratio in which P( −1, y) divides the line segment joining the points A(−3, 10) and B(6, −8). Then
Substituting in , we get
Hence, the required ratio is 2 : 7 and y = 6.
Page No 326:
Question 33:
ABCD is a rectangle formed by the points . If P, Q, R and S be the mid points of
AB, BC, CD and DA respectively, show that PQRS is a rhombus.
Answer:
Here, the points P, Q, R and S are the mid points of AB, BC, CD and DA respectively. Then
Now
Thus, PQ = QR = RS = SP and therefore PQRS is a rhombus.
Page No 327:
Question 34:
The midpoint P of the line segment joining the points A(−10, 4) and B(−2, 0) lies on the line segment joining the points C(−9, −4) and D(−4, y). Find the ratio in which P divides CD. Also find the value of y.
Answer:
The midpoint of AB is .
Let k be the ratio in which P divides CD. So
Now, substituting in , we get
Hence, the required ratio is 3 : 2 and y = 6.
Page No 327:
Question 35:
A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, –5) is the midpoint of PQ then find the coordinates of P and Q. [CBSE 2017]
Answer:
Using mid-point formula, we have
Thus, the coordinates of P and Q are (0, −10) and (4, 0), respectively.
Page No 327:
Question 36:
In what ratio does the point divide the line segment joining the points P(2, –2) and Q(3, 7)? Also, find the value of y. [CBSE 2017]
Answer:
Let the point P divides the line PQ in the ratio k : 1.
Then, by the section formula:
Page No 327:
Question 37:
The midpoints of the sides BC, CA and AB of a ΔABC are D(3, 4), E(8, 9) and F(6, 7) respectively. Find the coordinates of the vertices of the triangle. [CBSE 2017]
Answer:
Let the coordinates of A, B, C be
Because D is the mid-point of BC, using mid-point formula, we have
Similarly, E is the mid point of AC. Using mid-point formula, we have;
Again, F is the mid point of AB. Using mid point formula, we have
Adding (i), (ii) and (iii), we get
On solving equation (iv), using equations (i), (ii) and (iii), we get
Similarly,
Hence, the points are: A(11,12), B(1,2) and C(5,6).
Page No 327:
Question 38:
If two adjacent vertices of a parallelogram are (3, 2) and (–1, 0) and the diagonals intersect at (2, –5) then find the coordinates of the other two vertices. [CBSE 2017]
Answer:
Let ABCD be the parallelogram with two adjacent vertices A(3, 2) and B(−1, 0). Suppose O(2, −5) be the point of intersection of the diagonals AC and BD.
Let C(x1, y1) and D(x2, y2) be the coordinates of the other vertices of the parallelogram.
We know that the diagonals of the parallelogram bisect each other. Therefore, O is the mid-point of AC and BD.
Using the mid-point formula, we have
So, the coordinates of C are (1, −12).
Also,
So, the coordinates of D are (5, −10).
Page No 327:
Question 39:
Point A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD. Find the values of a and b.
Answer:
Given: Point A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD.
Diagonals of a parallelogram bisect each other.
∴ Mid point of AC = Mid point of BD
Hence, the values of a and b is 6 and 3, respectively.
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Question 40:
The line segment joining the point A(2, 1) and B(5, –8) is trisected at the points P and Q such that P is nearer to A. If P also lies on the line given by 2x – y + k = 0, find the value of k.
Answer:
Let the points A(2, 1) and B(5, –8) is trisected at the points P(x, y) and Q(a, b).
Thus, AP = PQ = QB
Therefore, P divides AB internally in the ratio 1 : 2.
Section formula: if the point (x, y) divides the line segment joining the points (x1, y1) and (x2, y2) internally in the ratio m : n, then the coordinates (x, y) =
Therefore, using section formula, the coordinates of P are:
Hence, the coordinates of P are (3, –2).
Since, P also lies on the line given by 2x – y + k = 0,
Therefore, (3, –2) satisfies the equation 2x – y + k = 0
Hence, the values of k is –8.
Page No 327:
Question 41:
Find the ratio in which y-axis divides the line segment joining the points A(5, –6) and B(–1, 4) Also, find the coordinates of the point of division.
Answer:
Section formula: if the point (x, y) divides the line segment joining the points (x1, y1) and (x2, y2) internally in the ratio k : 1, then the coordinates (x, y) =
Let the point P(0, y) divides the line segment joining the points A(5, –6) and B(–1, 4) in the ratio k : 1.
Therefore, using section formula, the coordinates of P are:
Hence, the y-axis divides the line segment joining the points A(5, –6) and B(–1, 4) in the ratio 5 : 1.
and the coordinates of the point of division are .
Page No 340:
Question 1:
Find the area of ∆ABC whose vertices are:
(i) A(1, 2), B(−2, 3) and C(−3, −4)
(ii) A(−5, 7), B(−4, −5) and C(4, 5)
(iii) A(3, 8), B(−4, 2) and C(5, −1)
(iv) A(10, −6), B(2, 5) and C(−1, 3)
Answer:
(i) A(1, 2), B(−2, 3) and C(−3, −4) are the vertices of ∆ABC. Then,
(x1 = 1, y1 = 2), (x2 = −2, y2 = 3) and (x3 = −3, y3 = -4)
(ii) A(−5, 7), B(−4, −5) and C(4, 5) are the vertices of ∆ABC. Then,
(x1 = −5, y1 = 7), (x2 = −4, y2 = −5) and (x3 = 4, y3 = 5)
(iii) A(3, 8), B(−4, 2) and C(5, −1) are the vertices of ∆ABC. Then,
(x1 = 3, y1 = 8), (x2 = −4, y2 = 2) and (x3 = 5, y3 = −1)
(iv) A(10, −6), B(2, 5) and C(−1, −3) are the vertices of ∆ABC. Then,
(x1 = 10, y1 = −6), (x2 = 2, y2 = 5) and (x3 = −1, y3 = 3)
Page No 341:
Question 2:
Find the area of quadrilateral ABCD whose vertices are A(3, −1), B(9, −5), C(14, 0) and D(9, 19). [CBSE 2012]
Answer:
By joining A and C, we get two triangles ABC and ACD.
Let . Then
So, the area of the quadrilateral is 25 + 107 = 132 sq. units.
Page No 341:
Question 3:
Find the area of quadrilateral PQRS whose vertices are P(−5, −3), Q(−4, −6), R(2, −3) and S(1, 2).
[CBSE 2015]
Answer:
By joining P and R, we get two triangles PQR and PRS.
Let . Then
So, the area of the quadrilateral PQRS is sq. units.
Page No 341:
Question 4:
Find the area of quadrilateral ABCD whose vertices are A(−3, −1), B(−2, −4), C(4, −1) and D(3, 4).
[CBSE 2013C]
Answer:
By joining A and C, we get two triangles ABC and ACD.
Let . Then
So, the area of the quadrilateral ABCD is sq. units.
Page No 341:
Question 5:
If A(–7, 5), B(–6, –7), C(–3, –8) and D(2, 3) are the vertices of a quadrilateral ABCD then find the area of the quadrilateral.
Answer:
Consider the figure.
Construction: Produce AC by joining points A to C to form two triangles, .
In
We know that,
Similarly, in
Now, ar(quad. ABCD) =
ar(quad. ABCD) =
Therefore, area of quadrilateral ABCD is 72 sq. units
Disclaimer: The answer thus calculated does not match with the answer given in the book.
Page No 341:
Question 6:
Find area of the triangle formed by joining the midpoints of the sides of the triangle
whose vertices are A(2, 1), B(4, 3) and C(2, 5).
Answer:
The vertices of the triangle are A(2, 1), B(4, 3) and C(2, 5).
Now
Hence, the area of the required triangle is 1 sq. unit.
Page No 341:
Question 7:
A(7, −3), B(5, 3), C(3, −1) are the vertices of a and AD is its median. Prove that the median
AD divides into two triangles of equal areas.
Answer:
The vertices of the triangle are A(7, −3), B(5, 3), C(3, −1).
For the area of the triangle ADC, let . Then
Now, for the area of triangle ABD, let . Then
Thus, .
Hence, AD divides into two triangles of equal areas.
Page No 341:
Question 8:
Find the area of with A(1, −4) and midpoints of sides through A being (2, −1) and (0, −1). [CBSE 2015]
Answer:
Let be the coordinates of B and C respectively. Since, the coordinates of A are (1, −4), therefore
Let . Now
Hence, the area of the triangle is 12 sq. units.
Page No 341:
Question 9:
A(6, 1), B(8, 2) and C(9, 4) are the vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of . [CBSE 2015]
Answer:
Let (x, y) be the coordinates of D and be the coordinates of E. Since, the diagonals of a parallelogram bisect
each other at the same point, therefore
Thus, the coordinates of D are (7, 3).
E is the midpoint of DC, therefore
Thus, the coordinates of E are .
Let . Now
Hence, the area of the triangle is .
Page No 341:
Question 10:
(i) If the vertices of be A(1, −3), B(4, p) and C(−9, 7) and its area is 15 square units, find the values of p. [CBSE 2012]
(ii) The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, –2). If the third vertex is , find the value of y. [CBSE 2017]
Answer:
(i) Let . Now
Therefore
Hence, .
(ii)
Let
Now
Hence, y =
Page No 341:
Question 11:
Find the value of k so that the area of the triangle with vertices A(k + 1, 1), B(4, −3) and C(7, −k) is 6 square units.
Answer:
Let . Now
Hence, k = 3.
Page No 341:
Question 12:
For what value of k (k > 0) is the area of the triangle with vertices (−2, 5), (k, −4) and
(2k + 1, 10) equal to 53 square units?
Answer:
Let be the vertices of
the triangle. So
Hence, k = 3.
Page No 341:
Question 13:
Show that the following points are collinear:
(i) A(2, −2), B(−3, 8) and C(−1, 4)
(ii) A(−5, 1), B(5, 5) and C(10, 7)
(iii) A(5, 1), B(1, −1) and C(11, 4)
(iv) A(8, 1), B(3, −4) and C(2, −5)
Answer:
(i)
Let A(x1 = 2, y1 = −2), B(x2 = −3, y2 = 8) and C(x3 = −1, y3 = 4) be the given points. Now
Hence, the given points are collinear.
(ii)
Let A(x1 = −5, y1 = 1), B(x2 = 5, y2 = 5) and C(x3 = 10, y3 = 7) be the given points. Now
Hence, the given points are collinear.
(iii)
Let A(x1 = 5, y1 = 1), B(x2 = 1, y2 = −1) and C(x3 = 11, y3 = 4) be the given points. Now
Hence, the given points are collinear.
(iv)
Let A(x1 = 8, y1 = 1), B(x2 = 3, y2 = −4) and C(x3 = 2, y3 = −5) be the given points. Now
Hence, the given points are collinear.
Page No 342:
Question 14:
Find the value of x for which the points are collinear. [CBSE 2015]
Answer:
Let . So, the condition for three collinear points is
Hence, x = − 63.
Page No 342:
Question 15:
For what value of x are the points A(−3, 12), B(7, 6) and C(x, 9) collinear?
Answer:
A(−3, 12), B(7, 6) and C(x, 9) are the given points. Then:
(x1 = −3, y1 = 12), (x2 = 7, y2 = 6) and (x3 = x, y3 = 9)
It is given that the points A, B and C are collinear. Therefore,
Therefore, when x= 2, the given points are collinear.
Page No 342:
Question 16:
Find the value of p for which the points A(–5, 1), B(1, p) and C(4, –2) are collinear.
Answer:
If the area of the triangle formed by three points is equal to zero, then the points are collinear.
Area of the triangle formed by the vertices .
Now, the given points A(–5, 1), B(1, p) and C(4, –2) are collinear.
Therefore, Area of triangle formed by them is equal to zero.
Hence, the value of p is –1.
Page No 342:
Question 17:
Find the value of y for which the points A(−3, 9), B(2, y) and C(4, −5) are collinear.
Answer:
Let A(x1 = −3, y1 = 9), B(x2 = 2, y2 = y) and C(x3 = 4, y3 = −5) be the given points.
The given points are collinear if
Page No 342:
Question 18:
For what values of k are the points A(8, 1), B(3, −2k) and C(k, −5) collinear? [CBSE 2015]
Answer:
Let A(x1 = 8, y1 = 1), B(x2 = 3, y2 = −2k) and C(x3 = k, y3 = −5) be the given points.
The given points are collinear if
Hence, .
Page No 342:
Question 19:
Find a relation between x and y, if the points A(2, 1), B(x, y) and C(7, 5) are collinear. [CBSE 2009C]
Answer:
Let A(x1 = 2, y1 = 1), B(x2 = x, y2 = y) and C(x3 = 7, y3 = 5) be the given points.
The given points are collinear if
Hence, the required relation is 4x − 5y − 3 = 0.
Page No 342:
Question 20:
Find a relation between x and y, if the points A(x, y), B(−5, 7) and C(−4, 5) are collinear. [CBSE 2015]
Answer:
Let A(x1 = x, y1 = y), B(x2 = −5, y2 = 7) and C(x3 = −4, y3 = 5) be the given points.
The given points are collinear if
Hence, the required relation is 2x + y + 3 = 0.
Page No 342:
Question 21:
Prove that the points A(a, 0), B(0, b) and C(1, 1) are collinear if
Answer:
Consider the points A(a, 0), B(0, b) and C(1, 1).
Here, (x1 = a, y1 = 0), (x2 = 0, y2 = b) and (x3 = 1, y3 = 1).
It is given that the points are collinear. So,
Therefore, the given points are collinear if = 1.
Page No 342:
Question 22:
If the points P(−3, 9), Q(a, b) and R(4, −5) are collinear and a + b = 1, find the values of a and b. [CBSE 2014]
Answer:
Let A(x1 = −3, y1 = 9), B(x2 = a, y2 = b) and C(x3 = 4, y3 = −5) be the given points.
The given points are collinear if
Now, solving a + b = 1 and 2a + b = 3, we get a = 2 and b = −1.
Hence, a = 2 and b = −1.
Page No 342:
Question 23:
Find the area of with vertices A(0, −1), B(2, 1) and C(0, 3). Also, find the area of the triangle formed
by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is 4 : 1. [CBSE 2014]
Answer:
Let A(x1 = 0, y1 = −1), B(x2 = 2, y2 = 1) and C(x3 = 0, y3 = 3) be the given points. Then
So, the area of the triangle is 4 sq. units.
Let D(a1, b1), E(a2, b2) and F(a3, b3) be the midpoints of AB, BC and AC respectively. Then
Thus, the coordinates of D, E and F are D(a1 = 1, b1 = 0), E(a2 = 1, b2 = 2) and F(a3 = 0, b3 = 1). Now
So, the area of the triangle is 1 sq. unit.
Hence, .
Page No 342:
Question 24:
If a ≠ b ≠ c, prove that (a, a2), (b, b2), (0, 0) will not be collinear. [CBSE 2017]
Answer:
Let A(a, a2), B(b, b2) and C(0, 0) be the coordinates of the given points.
We know that the area of triangle having vertices is square units.
So,
Area of ∆ABC
Since the area of the triangle formed by the points (a, a2), (b, b2) and (0, 0) is not zero, so the given points are not collinear.
Page No 342:
Question 25:
If the area of the triangle with vertices (x, 3), (4, 4) and (3, 5) is 4 square units, find x.
Answer:
Area of the triangle formed by the vertices .
Now, the given vertices are (x, 3), (4, 4) and (3, 5)
and the given area is 4 square units.
Therefore,
Hence, the value of x is 13 and −3.
Page No 344:
Question 1:
Points A(−1, y) and B(5, 7) lie on a circle with centre O(2, −3y). Find the values of y. [CBSE 2014]
Answer:
The given points are A(−1, y), B(5, 7) and O(2, −3y).
Here, AO and BO are the radii of the circle. So
Hence, y = 7 or y = −1.
Page No 344:
Question 2:
If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find p. [CBSE 2014]
Answer:
The given points are A(0, 2), B(3, p) and C(p, 5).
Hence, p = 1.
Page No 345:
Question 3:
ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). Find the length of one of its diagonal.
[CBSE 2014]
Answer:
The given vertices are B(4, 0), C(4, 3) and D(0, 3). Here, BD is one of the diagonals. So
Hence, the length of the diagonal is 5 units..
Page No 345:
Question 4:
If the point P(k − 1, 2) is equidistant from the points A(3, k) and B(k, 5), find the values of k. [CBSE 2014]
Answer:
The given points are P(k − 1, 2), A(3, k) and B(k, 5).
Hence, k = 1 or k = 5.
Page No 345:
Question 5:
Find the ratio in which the point P(x, 2) divides the join of A(12, 5) and B(4, −3). [CBSE 2014]
Answer:
Let k be the ratio in which the point P(x, 2) divides the line joining the points A(x1 = 12, y1 = 5) and B(x2 = 4, y2 = −3). Then
Now
Hence, the required ratio is 3 : 5.
Page No 345:
Question 6:
Prove that the diagonals of a rectangle ABCD with vertices A(2, −1), B(5, −1), C(5, 6) and D(2, 6) are
equal and bisect each other. [CBSE 2014]
Answer:
The vertices of the rectangle ABCD are A(2, −1), B(5, −1), C(5, 6) and D(2, 6). Now
Since, the midpoints of AC and BD coincide, therefore the diagonals of rectangle ABCD bisect each other.
Page No 345:
Question 7:
Find the lengths of the medians AD and BE of whose vertices are A(7, −3), B(5, 3) and C(3, −1). [CBSE 2014]
Answer:
The given vertices are A(7, −3), B(5, 3) and C(3, −1).
Since D and E are the midpoints of BC and AC respectively, therefore
Now
Hence, AD = BE = 5 units.
Page No 345:
Question 8:
If the point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2 : 3 then find the value of k. [CBSE 2013C]
Answer:
Here, the point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2 : 3. So
Hence, .
Page No 345:
Question 9:
Find the point on x-axis which is equidistant from points A(−1, 0) and B(5, 0). [CBSE 2013C]
Answer:
Let P(x, 0) be the point on x-axis. Then
Hence, x = 2.
Page No 345:
Question 10:
Find the distance between the points
Answer:
The given points are .
Then,
Therefore,
Page No 345:
Question 11:
Find the value of a, so that the point (3, a) lies on the line 2x − 3y = 5.
Answer:
.
Page No 345:
Question 12:
If the points A(4, 3) and B(x, 5) lie on the circle with centre O(2, 3), find the value of x.
Answer:
.
Page No 345:
Question 13:
If P(x, y) is equidistant from the points A(7, 1) and B(3, 5), find the relation between x and y.
Answer:
Let the point be equidistant from the points A(7, 1) and B(3, 5).
Then,
Page No 345:
Question 14:
If the centroid of ∆ABC, which has vertices A(a, b), B(b, c) and C(c, a), is the origin, find the value of (a + b + c).
Answer:
The given points are A(a, b), B(b, c) and C(c, a).
Here,
Let the centroid be (x, y).
Then,
But it is given that the centroid of the triangle is the origin.
Then, we have:
Page No 345:
Question 15:
Find the centroid of ∆ABC whose vertices are A(2, 2), B(−4, −4) and C(5, −8).
Answer:
The given points are A(2, 2), B(−4, −4) and C(5, −8).
Here,
Let G(x, y) be the centroid of . Then,
Hence, the centroid of .
Page No 345:
Question 16:
In what ratio does the point C(4, 5) divide the join of A(2, 3) and B(7, 8)?
Answer:
Let the required ratio be .
Then, by section formula, the coordinates of C are
Therefore,
Page No 345:
Question 17:
If the points A(2, 3) B(4, k) and C(6, −3) are collinear, find the value of k.
Answer:
The given points are .
Here,
It is given that the points A, B and C are collinear. Then,
Page No 348:
Question 1:
The distance of the point P(−6, 8) from the origin is [CBSE 2013C]
(a) 8 (b) (c) 6 (d) 10
Answer:
The distance of a point (x, y) from the origin O(0, 0) is .
Let P(x = −6, y = 8) be the given point. Then
Hence, the correct answer is option (d).
Page No 348:
Question 2:
The distance of the point (−3, 4) from x-axis is [CBSE 2012]
(a) 3 (b) −3 (c) 4 (d) 5
Answer:
The distance of a point (x, y) from x-axis is .
Here, the point is (−3, 4). So, its distance from x-axis is .
Hence, the correct answer is option (c).
Page No 348:
Question 3:
The point on x-axis which is equidistant from points A(−1, 0) and B(5, 0) is [CBSE 2013]
(a) (0, 2) (b) (2, 0) (c) (3, 0) (d) (0, 3)
Answer:
Let P(x, 0) the point on x-axis, then
Thus, the required point is (2, 0).
Hence, the correct answer is option (b).
Page No 348:
Question 4:
If R(5, 6) is the midpoint of the line segment AB joining the points A(6, 5) and B(4, y) then y equals
(a) 5 (b) 7 (c) 12 (d) 6
Answer:
Since R(5, 6) is the midpoint of the line segment AB joining the points A(6, 5) and B(4, y), therefore
Hence, the correct option is (b).
Page No 348:
Question 5:
If the point C(k, 4) divides the join of the points A(2, 6) and B(5, 1) in the ratio 2 : 3 then the value of k is [CBSE 2013C]
(a) 16 (b) (c) (d)
Answer:
The point C(k, 4) divides the join of the points A(2, 6) and B(5, 1) in the ratio 2 : 3. So
Hence, the correct answer is option (c).
Page No 348:
Question 6:
The perimeter of the triangle with vertices (0, 4), (0, 0) and (3, 0) is [CBSE 2014]
(a) (b) 5 (c) 10 (d) 12
Answer:
Let A(0, 4), B(0, 0) and C(3, 0) be the given vertices. So
Therefore
AB + BC + AC = 4 + 3 + 5 = 12
Hence, the correct answer is option (d).
Page No 348:
Question 7:
If A(1, 3), B(−1, 2) and C(2, 5) and D(x, 4) are the vertices of a ||gm ABCD then the value of x is [CBSE 2012]
(a) 3 (b) 4 (c) 0 (d)
Answer:
The diagonals of a parallelogram bisect each other. The vertices of the ||gm ABCD are A(1, 3), B(−1, 2) and C(2, 5) and D(x, 4).
Here, AC and BD are the diagonals. So
Hence, the correct answer is option (b).
Page No 348:
Question 8:
If the points A(x, 2), B(−3, −4) and C(7, −5) are collinear then the value of x is [CBSE 2014]
(a) −63 (b) 63 (c) 60 (d) −60
Answer:
Let A(x1 = x, y1 = 2), B(x2 = −3, y2 = −4) and C(x3 = 7, y3 = −5) be collinear points. Then
Hence, the correct answer is option (a).
Page No 348:
Question 9:
The area of a triangle with vertices A(5, 0), B(8, 0) and C(8, 4) in square units is [CBSE 2012]
(a) 20 (b) 12 (c) 6 (d) 16
Answer:
Let A(x1 = 5, y1 = 0), B(x2 = 8, y2 = 0) and C(x3 = 8, y3 = 4) be the vertices of the triangle. Then
Hence, the correct answer is option (c).
Page No 348:
Question 10:
The area of with vertices A(a, 0), O(0, 0) and B(0, b) in square units is
(a) ab (b) (c) (d)
Answer:
Let A(x1 = a, y1 = 0), O(x2 = 0, y2 = 0) and B(x3 = 0, y3 = b) be the given vertices. So
Hence, the correct answer is (b).
Page No 348:
Question 11:
If is the midpoint of the line segment joining the points A(−6, 5) and B(−2, 3) then the value of a is [CBSE 2011]
(a) −8 (b) 3 (c) −4 (d) 4
Answer:
The point is the midpoint of the line segment joining the points A(−6, 5) and B(−2, 3). So
Hence, the correct answer is option (a).
Page No 349:
Question 12:
ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). The length of one of its diagonals is [CBSE 2014]
(a) 5 (b) 4 (c) 3 (d) 25
Answer:
Here, AC and BD are two diagonals of the rectangle ABCD. So
Hence, the correct answer is option (a).
Page No 349:
Question 13:
The coordinates of the point P dividing the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2 : 1 is [CBSE 2012]
(a) (2, 4) (b) (3, 5) (c) (4, 2) (d) (5, 3)
Answer:
Here, the point P divides the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2 : 1. Then
Hence, the correct answer is option (b).
Page No 349:
Question 14:
If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (−2, 5), then the coordinates of the other end of the diameter are [CBSE 2012]
(a) (−6, 7) (b) (6, −7) (c) (4, 2) (d) (5, 3)
Answer:
Let (x, y) be the coordinates of the other end of the diameter. Then
Hence, the correct answer is option (a).
Page No 349:
Question 15:
In the given figure P(5, −3) and Q(3, y) are the points of trisection of the line segment
joining A(7, −2) and B(1, −5). Then, y equals [CBSE 2012]
(a) 2 (b) 4 (c) −4 (d)
Answer:
Here, AQ : BQ = 2 : 1. Then
Hence, the correct answer is option (c).
Page No 349:
Question 16:
The midpoint of segment AB is P(0, 4). If the coordinates of B are (−2, 3), then the coordinates of A are [CBSE 2011]
(a) (2, 5) (b) (−2, −5) (c) (2, 9) (d) (−2, 11)
Answer:
Let (x, y) be the coordinates of A. Then
Thus, the coordinates of A are (2, 5).
Hence, the correct answer is option (a).
Page No 349:
Question 17:
The point P which divides the line segment joining the points A(2, −5) and B(5, 2) in the ratio 2 : 3 lies in the quadrant [CBSE 2011]
(a) I (b) II (c) III (d) IV
Answer:
Let (x, y) be the coordinates of P. Then
Thus, the coordinates of point P are and so it lies in the fourth quadrant.
Hence, the correct answer is option (d).
Page No 349:
Question 18:
If A(−6, 7) and B(−1, −5) are two given points then the distance 2AB is [CBSE 2011]
(a) 13 (b) 26 (c) 169 (d) 238
Answer:
The given points are A(−6, 7) and B(−1, −5). So
Thus, 2AB = 26.
Hence, the correct answer is option (b).
Page No 349:
Question 19:
Which point on x-axis is equidistant from the points A(7, 6) and B(−3, 4)?
(a) (0, 4) (b) (−4, 0) (c) (3, 0) (d) (0, 3)
Answer:
Let P(x, 0) be the point on x-axis. Then as per the question
Thus, the required point is (3, 0).
Hence, the correct answer is option (c).
Page No 349:
Question 20:
The distance of the point P(3, 4) from the x-axis is
(a) 3 units
(b) 4 units
(c) 7 units
(d) 1 unit
Answer:
(b) 4 units
The y-coordinate is the distance of the point from the x-axis.
Here, the y-coordinate is 4.
Page No 349:
Question 21:
In what ratio does the x-axis divide the join of A(2, −3) and B(5, 6)?
(a) 2 : 3
(b) 3 : 5
(c) 1 : 2
(d) 2 : 1
Answer:
(c) 1 : 2
Let AB be divided by the x axis in the ratio at the point P.
Then, by section formula, the coordinates of P are
But P lies on the x axis: so, its ordinate is 0.
Hence, the required ratio is , which is same as 1 : 2.
Page No 349:
Question 22:
In what ratio does the y-axis divide the join of P(−4, 2) and Q(8, 3)?
(a) 3 : 1
(b) 1 : 3
(c) 2 : 1
(d) 1 : 2
Answer:
(d) 1 : 2
Let AB be divided by the y axis in the ratio at the point P.
Then, by section formula, the coordinates of P are
But, P lies on the y axis; so, its abscissa is 0.
Hence, the required ratio is , which is same as 1 : 2.
Page No 349:
Question 23:
If P(−1, 1) is the midpoint of the line segment joining A(−3, b) and B(1, b + 4), then b = ?
(a) 1
(b) −1
(c) 2
(d) 0
Answer:
(b) −1
The given points are A(−3, b) and B(1, b+4).
Then,
Therefore,
and
But the midpoint is .
Therefore,
Page No 350:
Question 24:
The line 2x + y −4 = 0 divides the line segment joining A(2, −2) and B(3, 7) in the ratio
(a) 2 : 5
(b) 2 : 9
(c) 2 : 7
(d) 2 : 3
Answer:
(b) 2 : 9
Let the line divide the line segment in the ratio k : 1 at the point P.
Then, by section formula, the coordinates of P are
Since P lies on the line , we have:
Hence, the required ratio is , which is same as 2 : 9.
Page No 350:
Question 25:
If A(4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ABC and AD is median, then the coordinates of D are
(a)
(b)
(c)
(d) None of these
Answer:
(c)
D is the midpoint of BC.
So, the coordinates of D are
Page No 350:
Question 26:
If A(−1, 0), B(5, −2) and C(8, 2) are the vertices of a ∆ABC, then its centroid is
(a) (12, 0)
(b) (6, 0)
(c) (0, 6)
(d) (4, 0)
Answer:
(d) (4, 0)
The given points are .
Here,
Let G(x, y) be the centroid of . Then,
and
Hence, the centroid of is G(4, 0).
Page No 350:
Question 27:
Two vertices of ∆ABC are A(−1, 4) and B(5, 2) and its centroid is G(0, −3). Then, the coordinates of C are
(a) (4, 3)
(b) (4, 15)
(c) (−4, −15)
(d) (−15, −4)
Answer:
(c) (−4, −15)
Two vertices of .
Let the third vertex be C(a, b).
Then, the coordinates of its centroid are
But it is given that the centroid is .
Therefore,
Hence, the third vertex of .
Page No 350:
Question 28:
The points A(−4, 0), B(4, 0) and C(0, 3) are the vertices of a triangle, which is
(a) isosceles
(b) equilateral
(c) scalene
(d) right-angled
Answer:
(a) isosceles
Let A(−4, 0), B(4, 0) and C(0, 3) be the given points. Then,
BC = AC = 5 units
Therefore, is isosceles.
Page No 350:
Question 29:
The points P(0, 6), Q(−5, 3) and R(3, 1) are the vertices of a triangle, which is
(a) equilateral
(b) isosceles
(c) scalene
(d) right-angled
Answer:
(d) right-angled
Let P(0, 6), Q(−5, 3) and R(3, 1) be the given points. Then,
Therefore, ∆PQR is right-angled.
Page No 350:
Question 30:
If the points A(2, 3), B(5, k) and C(6, 7) are collinear, then
(a) k = 4
(b) k = 6
(c)
(d)
Answer:
(b) k = 6
The given points are A(2, 3), B(5, k) and C(6, 7).
Here,
Points A,B and C are collinear. Then,
Page No 350:
Question 31:
If the points A(1, 2), O(0, 0) and C(a, b) are collinear, then
(a) a = b
(b) a = 2b
(c) 2a = b
(d) a + b = 0
Answer:
(c) 2a = b
The given points are A(1, 2), O(0, 0) and C(a, b).
Here, .
Points A, O and C are collinear.
Page No 350:
Question 32:
The area of ∆ABC with vertices A(3, 0), B(7, 0) and C(8, 4) is
(a) 14 sq units
(b) 28 sq units
(c) 8 sq units
(d) 6 sq units
Answer:
(c) 8 sq units
The given points are .
Here,
Therefore,
Page No 350:
Question 33:
AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 0). Length of each of its diagonal is
(a) 5 units
(b) 3 units
(c) units
(d) 4 units
Answer:
(c) units
are the three vertices of a rectangle; let C be the fourth vertex.
Then, the length of the diagonal,
Since, the diagonals of rectangle are equal .
Hence, the length of its diagonals is .
Page No 350:
Question 34:
If the distance between the points A(4, p) and B(1, 0) is 5, then
(a) p = 4 only
(b) p = −4 only
(c) p = ±4 only
(d) p = 0
Answer:
(c) p = ±4 only
The given points are A(4, p) and B(1, 0) and AB = 5.
Then,
Therefore,
RS Aggarwal Solutions for Class 10 Maths Chapter 6: Download PDF
RS Aggarwal Solutions for Class 10 Maths Chapter 6–Coordinate Geometry
Download PDF: RS Aggarwal Solutions for Class 10 Maths Chapter 6–Coordinate Geometry PDF
Chapterwise RS Aggarwal Solutions for Class 10 Maths :
- Chapter 1–Real Numbers
- Chapter 2–Polynomials
- Chapter 3–Linear Equations In Two Variables
- Chapter 4–Quadratic Equations
- Chapter 5–Arithmetic Progression
- Chapter 6–Coordinate Geometry
- Chapter 7–Triangles
- Chapter 8–Circles
- Chapter 9–Constructions
- Chapter 10–Trigonometric Ratios
- Chapter 11–T Ratios Of Some Particular Angles
- Chapter 12–Trigonometric Ratios Of Some Complementary Angles
- Chapter 13–Trigonometric Identities
- Chapter 14–Height and Distance
- Chapter 15–Perimeter and Areas of Plane Figures
- Chapter 16–Areas of Circle, Sector and Segment
- Chapter 17–Volume and Surface Areas of Solids
- Chapter 18–Mean, Median, Mode of Grouped Data
- Chapter 19–Probability
About RS Aggarwal Class 10 Book
Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal’s book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.
He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.
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