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

## Co-ordinate Geometry Exercise Ex. 6.1

### Solution 1

### Solution 2

### Solution 3

## Co-ordinate Geometry Exercise Ex. 6.2

### Solution 1 (i)

### Solution 1 (ii)

### Solution 1 (iii)

### Solution 1 (iv)

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

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### Solution 13

### Solution 14

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### Solution 18

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### Solution 20

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### Solution 22

### Solution 23

### Solution 24

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### Solution 26

### Solution 27

### Solution 28

### Solution 29(i)

### Solution 29(ii)

### Solution 29(iii)

### Solution 30

### Solution 31

### Solution 32

### Solution 33

### Solution 34

### Solution 35

### Solution 36

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### Solution 39

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### Solution 48

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

### Solution 49

### Solution 50

Using the distance formula,

AB=

AC=

BC=

PQ=

PR=

QR=

Now,

ΔABC ~ ΔPQR

by the SSS test.

### Solution 51

### Solution 52

### Solution 53

### Solution 54

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### Solution 56

### Solution 57

## Co-ordinate Geometry Exercise Ex. 6.3

### Solution 1

### Solution 2

(i)

(ii)

(iii)

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11(i)

### Solution 11(ii)

### Solution 12

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### Solution 15

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### Solution 19

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

### Solution 20

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### Solution 29

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

So, the ratio is 1:1.

Also,

### Solution 30

### Solution 31

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### Solution 33

### Solution 34

### Solution 35

### Solution 36

### Solution 37

### Solution 38

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)

### Solution 39

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### Solution 48

### Solution 49

### Solution 50 (i)

### Solution 50 (ii)

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.

### Solution 51

### Solution 52 (i)

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.

### Solution 52 (ii)

### Solution 53

### Solution 54

### Solution 55

### Solution 56

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.

### Solution 57

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### Solution 60

## Co-ordinate Geometry Exercise Ex. 6.4

### Solution 1(i)

### Solution 1(ii)

### Solution 2

### Solution 3

### Solution 4

### Solution 5

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### Solution 10

## Co-ordinate Geometry Exercise Ex. 6.5

### Solution 1

### Solution 1 (iv)

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.

### Solution 1 (v)

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.

### Solution 2(i)

### Solution 2(ii)

### Solution 2(iii)

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

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### Solution 11

### Solution 12

### Solution 13 (i)

### Solution 13 (ii)

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.

### Solution 14

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### Solution 21

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.

### Solution 22

Area of a triangle is given by

### Solution 23

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### Solution 30

Let the points be A, B and C respectively.

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

### Solution 31

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## Co-ordinate Geometry Exercise 6.63

### Solution 1

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So, the correct option is (d).

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## Co-ordinate Geometry Exercise 6.64

### Solution 11

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Note: The answer does not match the options.

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### Solution 24

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.

## Co-ordinate Geometry Exercise 6.65

### Solution 26

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).

### Solution 27

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).

### Solution 28

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).

### Solution 29

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### Solution 31

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### Solution 35

The Centroid of the triangle is given by

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

### Solution 36

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## Co-ordinate Geometry Exercise 6.66

### Solution 40