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

## Co-ordinate Geometry Exercise Ex. 6.2

### 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.

62 = (a - 0)2 + (0 - 3)2

36 = a2 + 9

a2 = 27

### Solution 50

Using the distance formula,

AB=

AC=

BC=

PQ=

PR=

QR=

Now,

ΔABC ~ ΔPQR

by the SSS test.

## Co-ordinate Geometry Exercise Ex. 6.3

(i)

(ii)

(iii)

### Solution 14

*Note: (i) Answer given in the book is incorrect.

### 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 29

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

So, the ratio is 1:1.

Also,

### 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 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 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 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.

## Co-ordinate Geometry Exercise Ex. 6.5

### Solution 1 (iv)

Area of triangle with vertices (x1, y1), (x2, y2) and (x3, y3) 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 (x1, y1), (x2, y2) and (x3, y3) is

Therefore, area of triangle with given vertices is

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

### Solution 13 (ii)

Area of triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is

Hence, the value of k is 3.

### Solution 21

Let the points (a, a2), (b, b2), (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, a2), (b, b2), (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, a2), (b, b2), (0, 0) are never collinear.

### Solution 22

Area of a triangle is given by

### 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.

## Co-ordinate Geometry Exercise 6.63

### Solution 7

So, the correct option is (d).

## Co-ordinate Geometry Exercise 6.64

### Solution 12

Note: The answer does not match the options.

### 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 (x1, y1), (x2, y2) and (x3, y3) 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 35

The Centroid of the triangle is given by

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