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

Hint: In this question, we are to find the equation of the line which is segmented between the given lines. To do this, the symmetric form of the line when it is segmented between two lines is used.

Formula used: The symmetric form of the equation of the line passing through the point $({{x}_{1}},{{y}_{1}})$ and having inclination of $\theta $ is
$\dfrac{x-{{x}_{1}}}{\cos \theta }=\dfrac{y-{{y}_{1}}}{\sin \theta }$ here $\theta \in \left( 0,\dfrac{\pi }{2} \right)\cup \left( \dfrac{\pi }{2},\pi \right)$
If a point $P(x,y)$ on the line which is passing through the point $A({{x}_{1}},{{y}_{1}})$ and having an inclination $\theta $, then $x={{x}_{1}}+r\cos \theta $, $y={{y}_{1}}+r\sin \theta $ where $\left| r \right|$ is the distance from the points $A$ to $P$. These equations are called parametric equations.

Complete step by step solution: Given that, the required line is bisected at the point $(1,5)$ and segmented between the two lines.
So, the line is
$\begin{align}
  & \dfrac{x-{{x}_{1}}}{\cos \theta }=\dfrac{y-{{y}_{1}}}{\sin \theta } \\
 & \Rightarrow \dfrac{x-1}{\cos \theta }=\dfrac{y-5}{\sin \theta }=r\text{ }...(1) \\
\end{align}$
Consider the given point as $M$. i.e., $M(1,5)$
Here $r$ is the distance of any point $(x,y)$ on the given lines from the point $M(1,5)$.
So, for the two given lines $5x-y-4=0$ and $3x+4y-4=0$, consider two points $A$ and $B$ which lie on them.
So, we can write,
$\begin{align}
  & A(1+r\cos \theta ,5+r\sin \theta ) \\
 & B(1-r\cos \theta ,5-r\sin \theta ) \\
\end{align}$
Thus, the point $A(1+r\cos \theta ,5+r\sin \theta )$ lies on the line $5x-y-4=0$.
On substituting, we get
$\begin{align}
  & 5x-y-4=0 \\
 & \Rightarrow 5(1+r\cos \theta )-(5+r\sin \theta )-4=0 \\
 & \Rightarrow 5+5r\cos \theta -5-r\sin \theta -4=0 \\
 & \Rightarrow 5r\cos \theta -r\sin \theta -4=0 \\
 & \Rightarrow r(5\cos \theta -\sin \theta )=4\text{ }...(2) \\
\end{align}$
The point $B(1-r\cos \theta ,5-r\sin \theta )$ lies on the line $3x+4y-4=0$.
On substituting, we get
$\begin{align}
  & 3x+4y-4=0 \\
 & \Rightarrow 3(1-r\cos \theta )+4(5-r\sin \theta )-4=0 \\
 & \Rightarrow 3-3r\cos \theta \text{+}20-4r\sin \theta -4=0 \\
 & \Rightarrow 3r\cos \theta -4r\sin \theta =19 \\
 & \Rightarrow r(3\cos \theta -4\sin \theta )=19\text{ }...(3) \\
\end{align}$
From (2) and (3), we get
\[\begin{align}
  & \dfrac{4}{5\cos \theta -\sin \theta }=\dfrac{19}{3\cos \theta -4\sin \theta } \\
 & \Rightarrow 12\cos \theta -16\sin \theta =95\cos \theta -19\sin \theta \\
 & \Rightarrow 83\cos \theta =35\sin \theta \\
 & \Rightarrow \dfrac{83}{\sin \theta }=\dfrac{35}{\cos \theta }\text{ }...(4) \\
\end{align}\]
Thus, from (1) and (4), we get
$\begin{align}
  & \dfrac{x-1}{35}=\dfrac{y-5}{83} \\
 & \Rightarrow 83x-83=35y-175 \\
 & \Rightarrow 83x-35y+92=0 \\
\end{align}$
Therefore, the line is $83x-35y+92=0$.

Thus, Option (A) is correct.

Note: Here we need to remember that, the parameter $r$ is the distance from the point $M(1,5)$ to any points on the given lines. So, this will help us to find the required equation of the line.