Pair Of Straight Lines Question 14
Question: One bisector of the angle between the lines given by $ a{{(x-1)}^{2}}+2h(x-1)y+by^{2}=0 $ is $ 2x+y-2=0 $ . The other bisector is
Options:
A) $ x-2y+1=0 $
B) $ 2x+y-1=0 $
C) $ x+2y-1=0 $
D) $ x-2y-1=0 $
Show Answer
Answer:
Correct Answer: D
Solution:
We have $ a{{(x-1)}^{2}}+2h(x-1)y+by^{2}=0 $ or $ a{{(x-1)}^{2}}+2h(x-1)(y-0)+b{{(y-0)}^{2}}=0 $
This equation represents a pair of straight lines intersecting at (1, 0).
Therefore shifting the origin at (1, 0), we have $ x=X+1,y=Y+0 $ and the equation reduces to $ aX^{2}+2hXY+bY^{2}=0 $ …..(i)
One of the bisectors of the angles between the lines given by (i) is $ 2x+y-2=0 $ or $ 2(X+1)+Y-2=0 $ i.e. $ 2X+Y=0 $ .
Since the bisector are always at right angle, therefore the other bisector is $ X-2Y=0 $ i.e., $ x-1-2y=0 $ or $ x-2y-1=0 $ .
BETA
