Answer :
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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