# The equations of perpendicular bisectors of the sides AB and AC of a triangle ABC are x – y + 5 = 0 and x + 2y = 0 respectively. If the point A is (1, -2), find the equation of the line BC.

Given:

Sides AB and AC of a triangle ABC are x – y + 5 = 0 and x + 2y = 0.

To find:

The equation of the line BC.

Explanation:

Diagram:

Let the perpendicular bisectors x y + 5 = 0 and x + 2y = 0 of the sides AB and AC intersect at D and E, respectively.

Let (x1,y1) and (x2,y2) be the coordinates of points B and C.

Coordinates of D

And coordinates of E

Point D lies on the line x y + 5 = 0

x1 – y1 + 13 = 0 … (1)

Point E lies on the line x + 2y = 0

x2 + 2y2 – 3 = 0 … (2)

Side AB is perpendicular to the line x y + 5 = 0

x1 + y1 + 1 … (3)

Similarly, side AC is perpendicular to the line x + 2y = 0

2x2 - y2 - 4 = 0… (4)

Now, solving eq (1) and eq (3) by cross multiplication, we get:

x1 = -7, y1 = 6

Thus, the coordinates of B are (-7, 6)

Similarly, solving (2) and (4) by cross multiplication, we get:

Thus, coordinates of C are

Therefore, equation of line BC is

14x + 23 y – 40 = 0

Hence, the equation of line BC is 14x + 23 y – 40 = 0

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