Question: Given vertices $ A(1,1),B(4,-2) $ and $ C(5,5) $ of a triangle, then the equation of the perpendicular dropped from C to the interior bisector of the angle A is [Roorkee 1994]
Options:
A) $ y-5=0 $
B) $ x-5=0 $
C) $ y+5=0 $
D) $ x+5=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
- The internal bisector of the angle A will divide the opposite side $ BC $ at $ D $ in the ratio of arms of the angle i.e. $ AB=3\sqrt{2} $ and $ AC=4\sqrt{2} $ . Hence by ratio formula the point D is $ ( \frac{31}{7},1 ) $ . Slope of $ AD $ by $ \frac{y_2-y_1}{x_2-x_1}=0 $ . \ Slope of a line perpendicular to $ AD $ is $ \infty $ . Any line through C perpendicular to this bisector is $ \frac{y-5}{x-5}=m=\infty $ ; \ $ x-5=0 $ .
BETA
