# RD SHARMA Solutions for Class 10 Maths Chapter 6 - Co-ordinate Geometry

## Chapter 6 - Co-ordinate Geometry Exercise Ex. 6.1

## Chapter 6 - Co-ordinate Geometry Exercise Ex. 6.2

Show that the points A(1,-2), B(3,6), C(5,10) and D(3,2) are the vertices of a parallelogram.

Prove that the points (1, 7), (4, 2), (-1, -1) and (-4, 4) are the vertices of a square.

The points A(2, 9), B(a, 5) and C(5, 5) are the vertices of triangle ABC right angled at B. Find the values of a and hence the area of ⧍ABC.

Find a point which is equidistant from the points

A (-5, 4) and B (-1, 6). How many such points are there?

The centre of a circle is (2a, a - 7). Find the values of a if the circle passes through the point (11, -9) and has diameter units.

Ayush starts walking from his house to office, Instead of going to the office directly, he goes to bank first, from there to his daughter's school and then reaches the office. What is the extra distance travelled by Ayush in reaching the office? (Assume that all distance covered are in straight lines). If the house is situated at (2, 4), bank at (5, 8) school at (13, 14) and office at (13, 26) and coordinates are in kilometer.

If (-4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the (i) interior (ii) exterior of the triangle.

Show that the points A(5,6), B(1,5), C(2,1) and D(6,2) are the vertices of a square.

Prove that the points A(2, 3), B(-2, 2), C(-1, -2), and D (3, -1) are the vertices of a square ABCD.

If the point P(x,y) is equidistant from the points A(5,1) and B(1,5), prove that x = y.

If the point P(k-1, 2) is equidistant from the points A(3, k) and B(k, 5), find the values of k.

If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find p. Also, find the length of AB.

Find the equation of the perpendicular bisector of the line segment joining points (7,1) and (3,5).

Prove that the points (3,0), (4,5), (-1,4) and (-2,-1), taken in order, form a rhombus. Also, find its area.

If a point A(0,2) is equidistant from the points B(3,p) and C(p,5), then find the value of p.

Prove that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right triangle.

If the point P(x, 3) is equidistant from the points A(7, -1) and B(6, 8), find the value of x and the find the distance AP.

If A(3, y) is equidistant from the points P(8, - 3) and Q(7, 6), find the value of y and find the distance AQ.

If (0, -3) and (0, 3) are the two vertices of an equilateral triangle, find the coordinates of its third vertex.

For an equilateral triangle, the perpendicular bisector of any side passes through the opposite vertex.

Both the points, (0, -3) and (0, 3), lie on the y-axis equidistant from the origin. Hence, the perpendicular bisector joining these two points is the x-axis.

Any point on the x-axis has the coordinates (a, 0).

The distance between (0, -3) and (0, 3) is 6.

Hence, the distance between (a, 0) and (0, 3) should also be 6.

6^{2 }= (a - 0)^{2 }+ (0 - 3)^{2}

36 = a^{2 }+ 9

a^{2 }= 27

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.

Show that ∆ABC, where A (-2, 0), B (2, 0) C (0, 2) and ∆PQR, where P (-4, 0), Q (4, 0), R (0, 4) are similar.

Using the distance formula,

AB=

AC=

BC=

PQ=

PR=

QR=

Now,

ΔABC ~ ΔPQR

by the SSS test.

Find the circumcentre of the triangle whose vertices are (-2,-3), (-1,0), (7,-6).

Find the centre of the circle passing through (5, -8), (2, -9) and (2, 1).

## Chapter 6 - Co-ordinate Geometry Exercise Ex. 6.3

Find the points of trisection of the line segment joining the points:

(i) (5, -6) and (-7, 5), (ii) (3, -2) and (-3,-4), (iii) (2,-2) and (-7,4).

(i)

(ii)

(iii)

If P(9a - 2, -b) divides the line segment joining A(3a + 1, -3) and B(8a, 5) in the ratio 3 : 1, find the values of a and b.

If (a, b) is the mid-point of the line segment joining the points A (10, -6), B(k, 4) and a - 2b = 18, find the value of k and the distance AB.

If the points P,Q(x, 7), R, S(6, y) in this order divide the line segment joining A(2, p) and B (7, 10) in 5 equal parts, find x, y and p.

Prove that (4,3), (6,4), (5,6) and (3,5) are the angular points of a square.

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 coordinates of the point of division.

Let the point on the x-axis be (a, 0).

Let this point divide the line segment AB in the ratio of r : 1.

Using the section formula for the y-coordinate, we get

Find the ratio in which the point P(x, 2) divides the line segment joining the points A (12, 5) and B (4, -3). Also, find the value of x.

Find the ratio in which the point P(-1, y) lying on the line segment joining A(-3, 10) and B(6, -8) divides it. Also find the value of y.

In what ratio does the point (-4,6) divide the line segment joining the points A(-6,10) and B(3,-8)?

Find the ratio in which P(4, m) divides the line segment joining the points A(2, 3) and B(6, -3). Hence, find m.

Let P divides the line segment AB is the ratio k: 1.

So, the ratio is 1:1.

Also,

Show that the points A(1,0) B(5,3), C(2,7) and D(-2,4) are the vertices of a parallelogram.

Points P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) in 5 equal parts. Find the coordinates of the points P, Q, and R.

The difference between the x-coordinates of A and B is 6 - 1 = 5

Similarly, the difference between the y-coordinates of A and B is 7 - 2 = 5

Hence, if the line segment joining A(1, 2) and B(6, 7) is divided into 5 equal parts by the points P, Q, R and S, then the coordinates of P, Q, R and S can be found out by increasing the x and the y coordinates of A by 1 successively.

Hence, the coordinates of P are (1 + 1, 2 + 1) = (2, 3)

The coordinates of Q are (2 + 1, 3 + 1) = (3, 4)

The coordinates of R are (3 + 1, 4 + 1) = (4, 5)

If two vertices of a parallelogram are (3,2), (-1,0) and the diagonals cut at (2,-5), find the other vertices of the parallelogram.

If the coordinates of the mid-points of the sides of a triangle are (3,4), (4,6), and (5,7), find its vertices.

Points 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.

Given: ABCD is a parallelogram

We know that, diagonals of a parallelogram bisect each other.

Therefore, midpoints of diagonals coincide

The midpoints of AC and BD coincide.

Points P and Q trisect the line segment joining the points A(-2, 0) and B(0, 8) such that P is near to A. Find the coordinates of P and Q.

As P and Q trisect AB and P is near to A.

Therefore, P divides AB in the ratio 1:2.

Also, Q divides AB in the ration 2:1.

A point P divides the line segment joining the points A(3,-5) and B(-4,8), such that . If P lies on the line x + y = 0, then find the value of k.

Given points are A(3,-5) and B(-4,8).

P divides AB in the ratio k : 1.

Using the section formula, we have:

Coordinate of point P are {(-4k+3/k+1)(8k-5/k+1)}

Now it is given, that P lies on the line x+y = 0

Therefore,

-4k+3/k+1 + 8k-5/k+1 =0

=> -4k+3+8k-5 =0

=> 4k -2 =0

=> k=2/4

=> k=1/2

Thus, the value of k is 1/2.

The mid - point 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.

If the point C(-1, 2) divides internally the line segment joining the points A(2, 5) and B(x, y) in the ratio 3 : 4, find the value of x^{2} + y^{2}.

ABCD is a parallelogram with vertices A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C (x_{3}, y_{3}). Find the coordinates of the fourth vertex D in terms of x_{1}, x_{2}, x_{3}, y_{1}, y_{2} and y_{3}.

The points A (x_{1}, y_{1}), B(x_{2}, y_{2}) and C (x_{3}, y_{3}) are the vertices of ⧍ABC.

i. The median from A meets BC at D. Find the coordinates of the point D.

ii. Find the coordinates of the point P on AD such that AP : PD = 2 : 1.

iii. Find the points of coordinates Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.

iv. What are the coordinates of the centroid of the triangle ABC

## Chapter 6 - Co-ordinate Geometry Exercise Ex. 6.4

Prove analytically that the line segment joining the middle points of two sides of a triangle is equal to half of the third side.

## Chapter 6 - Co-ordinate Geometry Exercise Ex. 6.5

Find the area of a triangle whose vertices are

(1, -1), (-4, 6) and (-3, -5)

Area of triangle
with vertices (x_{1}, y_{1}), (x_{2}, y_{2})
and (x_{3}, y_{3}) is

Therefore, area of triangle with given vertices is

Hence, the area of triangle will be 24 sq. units.

Find the area of a triangle whose vertices are

(-5, 7), (-4, -5) and (4, 5)

Area of triangle
with vertices (x_{1}, y_{1}), (x_{2}, y_{2})
and (x_{3}, y_{3}) is

Therefore, area of triangle with given vertices is

Hence, the area of triangle will be 53 sq. units.

Find the area of the quadrilaterals, the coordinates of whose vertices are

(-4,-2), (-3,-5), (3,-2), (2,3)

Find the area of a quadrilateral ABCD, the coordinates of whose vertices are A(-3, 2), B(5, 4), C(7, -6) and D(-5, -4).

In ⧍ABC, the coordinates of vertex A are (0, -1) and D(1, 0) and E(0, 1) respectively the mid-points of the sides AB and AC. If F is the mid-point of side BC, find the area of ⧍DEF.

Find the area of the triangle PQR with Q (3, 2) and the mid-points of the sides through Q being (2, -1) and (1, 2).

If P(-5, -3), Q(-4, -6), R(2, -3) and S(1, 2) are the vertices of a quadrilateral PQRS, find its area.

If A (-3, 5), B(-2, -7), C(1, -8) and D(6, 3) are the vertices of a quadrilateral ABCD, find its area.

If the vertices of a triangle are (1,-3), (4,p) and (-9, 7) and its area is 15 sq. units, find the value (s) of p.

Find the value of k so that the area of triangle ABC with A(k + 1, 1), B(4, -3) and C(7, -k) is 6 square units.

Area of triangle
with vertices (x_{1}, y_{1}), (x_{2}, y_{2})
and (x_{3}, y_{3}) is

Hence, the value of k is 3.

Find the value of a for which the area of the triangle formed by the points A(a, 2a), B(-2, 6) and C (3, 1) is 10 sequare units.

If a ≠ b ≠ 0, prove that the points (a, a^{2}), (b, b^{2}), (0, 0) are never collinear.

Let the points (a, a^{2}), (b, b^{2}), (0, 0) represent a triangle. If we can prove that the area of the triangle so formed is not equal to zero, then we can prove that the points (a, a^{2}), (b, b^{2}), (0, 0) are never collinear.

Area of a triangle is given by

Now b≠a≠0.

So,

b - a≠0,

Δ≠0

Thus, points (a, a^{2}), (b, b^{2}), (0, 0) are never collinear.

The area of a triangle is 5 sq. units. Two of its vertices are at (2, 1) and (3, -2). If the third vertex is (7/2, y), find y.

Area of a triangle is given by

Find the value (s) of k for which the points (3k - 1, k - 2), (k, k - 7) and (k - 1, -k - 2) are collinear.

Let the points be A, B and C respectively.

If A, B and C are collinear, then the area of ∆ABC is zero.

If the points A(-1, -4), B(b, c) and C(5, -1) are collinear and 2b + c = 4, find the values of b and c.

If the points A(-2, 1), B(a, b) and C(4, -1) are collinear and a - b = 1, find the values of a and b.

If the points A(1, -2), B(2, 3), C(a, 2) and D(-4, -3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base.

A(6, 1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ABCD. If E is the mid-point of DC, find the area of ⧍ADE.

If D(-1/5, 5/2), E(7, 3) and F(7/2, 7/2) are the mid-points of sides of ⧍ABC, find the area of ⧍ABC

## Chapter 6 - Co-ordinate Geometry Exercise 6.63

A line Segement is of length 10 units. If the coordinates of its one end are (2,3) and the abscissa of the other end is 10, then its ordinate is

(a) 9,6

(b) 3,-9

(c) -3,9

(d) 9,-6

So, the correct option is (d).

## Chapter 6 - Co-ordinate Geometry Exercise 6.64

If (-1,2),(2,-1) and (3,1) are any three vertices of a parallelogram, then

(a) a = 2, b = 0

(b) a = -2, b = 0

(c) a = -2, b = 6

(d) a = 6, b = 2

Note: The answer does not match the options.

If A (5,3), B(11,-5) and P(12,y) are the vertices of a right triangle right angled at P, then y =

(a) - 2,4

(b) -2 ,4

(c) 2, -4

(d) 2,4

The area of the triangle formed by (a,b+c), (b,c+a) and (c,a+b) is

(a) a+b+c

(b) abc

(c) (a+b+c)^{2}

(d) 0

If (x,2),(-3,-4) and (7,-5) are collinear, then x =

(a) 60

(b) 63

(c) -63

(d) -60

The line segment joining points (-3,4), and (1,-2) is divided by y - axis in the ratio

(a) 1 : 3

(b) 2 : 3

(c) 3 : 1

(d) 2: 3

The ratio in which (4,5) divides the join of (2,3) and (7,8) is

(a) -2 : 3

(b) -3 : 2

(c) 3 : 2

(d) 2 : 3

The ratio in which the x-axis divides the segment joining (3,6) and (12,-3) is

(a) 2:1

(b) 1 :2

(c) -2 : 1

(d) 1 : -2

If the centrroid of the triangle formed by the points (a,b),(b,c) and (c,a) is at the origin, then a^{3} + b^{3} + c^{3} =

(a) abc

(b) 0

(c) a+b+c

(d) 3abc

If the centroid of the triangle formed by (7, x), (y, -6) and (9, 10) is at (6, 3), then (x, y) =

(a) (4, 5)

(b) (5, 4)

(c) (-5, -2)

(d) (5, 2)

If points A(3, 1), B(5, p) and C(7, -5) are collinear, then p =

(a) -2

(b) 2

(c) -1

(d) 1

As the points A, B and C are collinear, then the area formed by these three points is 0.

Area of triangle with vertices (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) is

Hence, the value of p is -2.

## Chapter 6 - Co-ordinate Geometry Exercise 6.65

The distance of the point (4, 7) from the x-axis is

(a) 4

(b) 7

(c) 11

(d)

We know, distance of point (x, y) from x-axis is y.

Hence, distance of point (4, 7) from x-axis is 7.

Hence, correct option is (b).

The distance of the point (4, 7) from the y-axis is

(a) 4

(b) 7

(c) 11

(d)

We know distance of point (x, y) from y-axis is x.

Hence distance of point (4, 7) from y-axis is 4.

Hence, correct option is (a).

If P is a point on x-axis such that its distance from the origin is 3 units, then the coordinates of a point Q on OY such that OP = OQ, are

(a) (0, 3)

(b) (3, 0)

(c) (0, 0)

(d) (0, -3)

Given P is a point on x-axis

Hence P = (x, 0)

Distance from the origin is 3

Hence P = (3, 0)

Given Q is a point on y-axis

So Q is (0, y)

Given that OP = OQ

implies OQ = 3

Distance of Q from the origin is 3

Hence y = 3

implies Q = (0, 3)

Hence, correct option is (a).

If the point (x, 4) lies on a circle whose centre is at the origin and radius is 5, then x =

(a) ±5

(b) ±3

(c) 0

(d) ±4

If the point P (x, y) is equidistant from A(5, 1) and B (-1, 5), then

(a) 5x = y

(b) x = 5y

(c) 3x = 2y

(d) 2x = 3y

If points A (5, p), B (1, 5), C (2, 1) and D (6, 2) form a square ABCD, then p =

(a) 7

(b) 3

(c) 6

(d) 8

The coordinates of a point on x-axis which lies on the perpendicular bisector of the line segment joining the points (7, 6) and (-3, 4) are

(a) (0, 2)

(b) (3, 0)

(c) (0, 3)

(d) (2, 0)

If the centroid of the triangle formed by the points (3, -5), (-7, 4), (10, -k) is at the point (k, - 1), then k =

(a) 3

(b) 1

(c) 2

(d) 4

If (-2, 1) is the Centroid of the triangle having its vertices at (x, 2), (10, -2), (-8, y), then x, y satisfy the relation

a. 3x + 8y = 0

b. 3x - 8y = 0

c. 8x + 3y = 0

d. 8x = 3y

The Centroid of the triangle is given by

x=-8 and y=3 satisfy(a) 3x + 8y = 0.

The coordinates of the fourth vertex of the rectangle formed by the points (0, 0), (2, 0), (0, 3) are

(a) (3, 0)

(b) (0, 2)

(c) (2, 3)

(d) (3, 2)

The length of a line segment joining A(2, -3) and B is 10 units. If the abscissa of B is 10 units, then its ordinates can be

(a) 3 or -9

(b) -3 or 9

(c) 6 or 27

(d) -6 or -27

The ratio in which the line segment joining P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) is divided by x-axis is

(a) y_{1} : y_{2}

(b) -y_{1} : y_{2}

(c) x_{1} : x_{2}

(d) -x_{1} : x_{2}

The ratio in which the line segment joining points A(a_{1}, b_{1}) and B(a_{2}, b_{2}) is divided by y-axis is

(a) -a_{1} : a_{2}

(b) a_{1} : a_{2}

(c) b_{1} : b_{2}

(d) -b_{1} : b_{2}

## Chapter 6 - Co-ordinate Geometry Exercise 6.66

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

(a) (-6, 7)

(b) (6, -7)

(c) (6, 7)

(d) -6, -7)

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 are

(a) (2, 4)

(b) (3, 5)

(c) (4, 2)

(d) (5, 3)

In the figure, the area of ΔABC (in square units) is

(a) 15

(b) 10

(c) 7.5

(d) 2.5

The point on the x-axis which is equidistant from points (-1, 0) and (5, 0) is

(a) (0, 2)

(b) (2, 0)

(c) (3, 0)

(d) (0, 3)

If P(2, 4), Q(0,3), R (3, 6) and S (5, y) are the vertices of a parallelogram PQRS, then the value of y is

(a) 7

(b) 5

(c) -7

(d) -8

If A(x, 2), B (-3, -4) and C (7, -5) are collinear, then the value of x is

(a) -63

(b) 63

(c) 60

(d) -60

## Chapter 6 - Co-ordinate Geometry Exercise 6.67

If the point P(2, 1) lies on the line segment joining points A (4, 2) and B (8, 4), then

A line intersects the y-axis and x-axis at P and Q, respectively, If (2, -5) is the mid-point of PQ, then the coordinates of P and Q are respectively

a. (0, -5) and (2, 0)

b. (0, 10) and (-4, 0)

c. (0, 4) and (-10, 0)

d. (0, -10) and (4, 0)

If the point (k, 0) divides the line segment joining the points A(2, -2) and B(-7, 4) in the ratio 1 : 2, then the value of k is

(a) 1

(b) 2

(c) -2

(d) -1

As the point (k, 0) divides the line segment AB in the 1:2

Hence, option (d) is correct.

### Other Chapters for CBSE Class 10 Mathematics

Chapter 1- Real Numbers Chapter 2- Polynomials Chapter 3- Pairs of Linear Equations in Two Variables Chapter 4- Quadratic Equations Chapter 5- Arithmetic Progressions Chapter 7- Triangles Chapter 8- Circles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- Trigonometric Identities Chapter 12- Heights and Distances Chapter 13- Areas Related to Circles Chapter 14- Surface Areas and Volumes Chapter 15- Statistics Chapter 16- Probability### RD SHARMA Solutions for CBSE Class 10 Subjects

### Kindly Sign up for a personalised experience

- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions

#### Sign Up

#### Verify mobile number

Enter the OTP sent to your number

Change