Question: The line $ x+3y-2=0 $ bisects the angle between a pair of straight lines of which one has equation $ x-7y+5=0 $ . The equation of the other line is
Options:
A) $ 3x+3y-1=0 $
B) $ x-3y+2=0 $
C) $ 5x+5y-3=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] The family of line through the given lines is $ L\equiv x-7y+5+\lambda (x+3y-2)=0 $ ?. (i) For line L =0 in the diagram, the distance of any point say (2, 0) on the line $ x+3y-2=0 $ from the line $ x-7y+5=0 $ and the line $ L=0 $ must be the same. Therefore, $ | \frac{2+5}{\sqrt{50}} |=| \frac{2+2\lambda +5-2\lambda }{\sqrt{{{(1+\lambda )}^{2}}+{{(3\lambda -7)}^{2}}}} | $ or $ 10{{\lambda }^{2}}-40\lambda =0 $ i.e., $ \lambda =4or0 $ Hence, $ L=0,\lambda =4. $ Therefore, the required line is $ 5x+5y-3=0. $
BETA
