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 $(1,5)$. Obtain its equation.

A line is such that its segment between the straight lines and <im

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

Option: 1
$\mathrm{83 x+35 y-92=0}$

Option: 2
$\mathrm{83 x-35 y-92=0}$

Option: 3
$\mathrm{73 x+49 y-92=0}$

Option: 4
$\mathrm{83 x-35 y+92=0}$

Answers (1)

Let the required line intersect the lines $\mathrm{3 x+4 y-4 =0}$ and $\mathrm{5 x+ y-4 =0}$ at $\mathrm{A}$ and $\mathrm{B}$ respectively.
Suppose line $\mathrm{AB}$ makes an angle $\mathrm{\Theta }$ with $\mathrm{x }$ -axis. The segment $\mathrm{ AB }$ is bisected at $\mathrm{ C(1,5) \text {. Let } A C=C B=r \text {. } }$
Equation of the line $\mathrm{ AB }$ is $\mathrm{ \frac{x-1}{\cos \theta}=\frac{y-5}{\sin \theta}\ \ \ .........\left ( i \right ) }$
The coordinates of $\mathrm{A}$ and $\mathrm{B}$ are given by

$\mathrm{\frac{x-1}{\cos \theta}=\frac{y-5}{\sin \theta}=-r \text { and } \frac{x-1}{\cos \theta}=\frac{y-5}{\sin \theta}=r}$ respectively.
Thus the coordinates of $\mathrm{A}$ and $\mathrm{B}$ are $\mathrm{A}$ $\mathrm{\left ( 1-r\ cos\ \Theta,5-r\ sin\ \Theta \right )}$ , $\mathrm{\mathrm{B}(1+\mathrm{r}\ \cos \theta, 5+\mathrm{r}\ \sin \theta)}$
Point $\mathrm{A}$ lies on $\mathrm{3 x+4 y-4=0}$ . Therefore, $\mathrm{3\left ( 1-r\ cos\ \Theta \right )+4(5-r \sin \theta)-4=0}$
$\mathrm{\Rightarrow r=\frac{19}{3 \cos \theta+4 \sin \theta}\ \ \ ..........\left ( ii \right )}$
Point $\mathrm{B}$ lies on $\mathrm{5 x-y-4=0.}$ Therefore, $\mathrm{5\left ( 1+r\ cos\ \Theta \right )-(5+r \sin \theta)-4=0}$
$\mathrm{\Rightarrow r=\frac{4}{5 \cos \theta-\sin \theta}\ \ \ \ ..........\left ( iii \right )}$
from $\mathrm{\left ( ii \right )}$ and $\mathrm{\left ( iii \right )}$ , we get $\mathrm{\frac{19}{3 \cos \theta+4 \sin \theta}=}$ $\mathrm{\frac{4}{5 \cos \theta- \sin \theta}}$
$\mathrm{\begin{aligned} & \Rightarrow 19(5 \cos \theta-\sin \theta)=4(3 \cos \theta+4 \sin \theta) \\ & 83 \cos \theta=35 \sin \theta \quad \Rightarrow \tan \theta=83 / 35 \end{aligned}}$
putting the value of $\mathrm{\tan \theta \text { in (i), }}$ we obtain the equation of the required line as $\mathrm{y-5=\frac{83}{35}\left ( x-1 \right )}$
$\mathrm{\text { or, } 83 x-35 y+92=0}$