Question: A line is such that its segment between the straight lines $ 5x-y-4=0 $ and $ 3x+4y-4=0 $ is bisected at the point (1, 5), then its equation is [Roorkee 1988]
Options:
A) $ 83x-35y+92=0 $
B) $ 35x-83y+92=0 $
C) $ 35x+35y+92=0 $
D)None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- Any line through the middle point M(1, 5) of the intercept AB may be taken as $ \frac{x-1}{\cos \theta }=\frac{y-5}{\sin \theta }=r $ ?..(i) where ?r? is the distance of any point (x, y) on the line (i) from the point M(1, 5). Since the points A and B are equidistant from M and on the opposite sides of it, therefore if the coordinates of A are obtained by putting r=d in (i), then the co-ordinates of B are given by putting $ r=-d $ . Now the point $ A(1+d\cos \theta ,5+d\sin \theta ) $ lies on the line $ 5x-y-4=0 $ and point $ B(1-d\cos \theta ,5-d\sin \theta ) $ lies on the line $ 3x+4y-4=0 $ . Therefore, $ 5(1+d\cos \theta )-(5+d\sin \theta )-4=0 $and $ 3(1-d\cos \theta )+ $ $ 4(5-d\sin \theta )-4=0 $ Eliminating ?d? from the two, we get $ \frac{\cos \theta }{35}=\frac{\sin \theta }{83} $ . Hence the required line is $ \frac{x-1}{35}=\frac{y-5}{83} $ or $ 83x-35y+92=0 $ .
BETA
