Question

x-3y-5=0 is the perpendicular bisector of the line segment joining the points A,B.if A=(-1,-3),find the co-ordinates of B

Answer 1

Solution :

▪ We have to find the line segment AB which is perpendicularly bisected by the line x - 3y - 5 = 0, i.e., AB is perpendicular to x - 3y - 5 = 0.

Let, the equation of AB is

3x + y = k ..... (1)

Given that, the coordinates of A is (- 1, -3), i.e., (- 1, - 3) lies on (1) no. line and this gives

3 (- 1) + (- 3) = k

or, - 3 - 3 = k

or, k = - 6

From (1), we get

3x + y = - 6,

which is the line AB.

▪ We have to find the point of bisection of x - 3y - 5 = 0 and the line AB : 3x + y = - 6.

The lines are

x - 3y - 5 = 0

3x + y = - 6

Multiplying the first equation by 3 and the second equation by 1, we get

3x - 9y - 15 = 0

3x + y = - 6

On subtraction, we get

- 9y - y - 15 = 6

or, - 10y = 6 + 15 = 21

or, y = - 9/10

Putting y = - 9/10 in x - 3y - 5 = 0, we get

x - 3 (- 9/10) - 5 = 0

or, x = 5 - 27/10 = (50 - 27)/10

or, x = 23/10

Thus, the coordinates of the point of bisection is (23/10, - 9/10).

▪ We have to find the coordinates of the point B.

The point A is (- 1, - 3) and the point of bisection is (23/10, - 9/10).

Let, the point B is (p, q).

Then, (23/10, - 9/10) is the middle point on the line segment joining the points (- 1, - 3) and (p, q).

So,

(- 1 + p)/2 = 23/10

or, 10 (- 1 + p) = 2 * 23

or, - 10 + 10p = 46

or, 10p = 10 + 46 = 56

or, p = 56/10

or, p = 28/5

and (- 3 + q)/2 = - 9/10

or, 10 (- 3 + q) = - 9 * 2

or, - 30 + 10q = - 18

or, 10q = 30 - 18 = 12

or, q = 12/10

or, q = 6/5

∴ the coordinates of B is (28/5, 6/5).

Answer 2

Answer:

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