Question

A line is such that its segment between the straight lines $5 x - y - 4 = 0$ and $3 x + 4 y - 4 = 0$ is bisected at the point (1, 5), then its equation is.

• A. $83 x - 35 y + 92 = 0$
• B. $35 x - 83 y + 92 = 0$
• C. $35 x + 35 y + 92 = 0$
• D. None of these

Answer 1

Answer: (A)

$83 x - 35 y + 92 = 0$

Answer 2

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 $5 x - y - 4 = 0$ and point $B ( 1 - d \cos \theta , 5 - d \sin \theta )$ lies on the line $3 x + 4 y - 4 = 0$.

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

$83 x - 35 y + 92 = 0$.