Question

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.

Answer 1

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. 

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

∴ \(-\frac{1}{2}\times\frac{y_2+2}{x_2-1}\) = -1

⇒ 2x₂ - y₂ - 4 = 0… (4)

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

\(\frac{x_1}{-1-13}=\frac{y_1}{13-1}=\frac{1}{1+1}\)

⇒ x₁ = -7, y₁ = 6 

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

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

\(\frac{x_2}{-8-3}=\frac{y_2}{-6+4}=\frac{1}{-1-4}\) 

⇒ x₂ = \(\frac{11}{5}\) , y₂ = \(\frac{2}{5}\)

Thus, coordinates of C are \(\Big(\frac{11}{5},\frac{2}{5}\Big)\) 

Therefore, equation of line BC is

⇒ 14x + 23 y – 40 = 0 

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