Question: The equation 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) $ , then the equation of line BC is[IIT 1986]
Options:
A) $ 23x+14y-40=0 $
B) $ 14x-23y+40=0 $
C) $ {{\tan }^{-1}}(2) $
D) $ 14x+23y-40=0 $
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Answer:
Correct Answer: D
Solution:
- Let the equation of perpendicular bisector FN of AB is $ x-y+5=0 $……(i) The middle point F of AB is $ ( \frac{x_1+1}{2},\frac{y_1-2}{2} ) $ lies on line (i). Therefore $ x_1-y_1=-13 $ ?..(ii) Also AB is perpendicular to FN. So the product of their slopes is ?1. i.e. $ \frac{y_1+2}{x_1-1}\times 1=-1 $ or $ x_1+y_1=-1 $??(iii) On solving (ii) and (iii), we get $ B(-7,6) $ . Similarly $ C( \frac{11}{5},\frac{2}{5} ) $ . Hence the equation of BC is $ 14x+23y-40=0 $ .
BETA
