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

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

Answer 1

The correct option is A $83x-35y+92=0$

$P_1{P}_2$ be the intercept between the lines $5x-y-4=0$ and $3x+4y-4=0.$

$P_1{P}_2$ makes an angle $\theta$ with the positive $x-axis.$

$(x_1,y_1)=A(1,5)$

So, the equation of the line passing through $A(1,5)$ is

$\Rightarrow$ $\frac{x-x_1}{\cos \theta }=\frac{y-y_1}{\sin \theta }$

$\Rightarrow$ $\frac{x-1}{\cos \theta }=\frac{y-5}{\sin \theta }$

$\Rightarrow$ $\frac{y-5}{x-1}=\tan \theta$

Let $A{P}_1=A{P}_2=r$

Then, the co-ordinates of $P_1$ and $P_2$ are given by

$\Rightarrow$ $\frac{x-1}{\cos \theta }=\frac{y-5}{\sin \theta }=r$ and $\frac{x-1}{\cos \theta }=\frac{y-5}{\sin \theta }=-r$

So, the co-ordinates of $P_1$ and $P_2$ are $1+r\cos \theta ,$ $5+r\sin \theta$ and $1-r\cos \theta ,5-r\sin \theta$ respectively.

Clearly, $P_1$ and $P_2$ lie on $5x-y-4=0$ and $3x+4y-4=0$ respectively.

$\therefore$ $5(1+r\cos \theta )-5-r\sin \theta -4=0$ and

$3(1-r\cos \theta )+4(5-r\sin \theta )-4=0$

$\Rightarrow$ $r=\frac{4}{5\cos \theta -\sin \theta }$ and $r=\frac{19}{3\cos \theta +4\sin \theta }$

$\Rightarrow$ $\frac{4}{5\cos \theta -\sin \theta }=\frac{19}{3\cos \theta +4\sin \theta }$

$\Rightarrow$ $95\cos \theta -19\sin \theta =12\cos \theta +16\sin \theta$

$\Rightarrow$ $83\cos \theta =35\sin \theta$

$\Rightarrow$ $\frac{\sin \theta }{\cos \theta }=\frac{83}{35}$

Thus, the equation of the required line is

$\frac{y-5}{x-1}=\tan \theta$

$\Rightarrow$ $\frac{y-5}{x-1}=\frac{83}{35}$

$\Rightarrow$ $35y-175=83-83$

$\Rightarrow$ $83x-35y+92=0$

$\therefore$ The equation of line is $83x-35y+92=0$