Question

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

• A. $107x-3y+92=0$
• B. $3x+107y+92=0$
• C. $107x-3y-92=0$
• D. $3x-107y-92=0$

Answer 1

Answer: (C)

Given equations of lines are
$5x-y+4=0...(i)$ and
$3x+4y-4=0...(ii)$
Let the required line intersect the lines $(i)$ and
$(ii)$ at the points $(x_1,y_1)$ and $(x_2,y_2)$ respectively. Therefore
$5x_1-y_1+4=0$ and
$3x_2+4y_2-4=0$
or $y_1=5x_1+4$ and $y_2=\frac{4-3x_2}{4}$.
We are given that the mid point of the segment of the required line between $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $\left(1,5\right)$.
Therefore $\frac{x_1+x_2}{2}=1$ and $\frac{y_1+y_2}{2}=5$
or $x_1+x_2=2$ and $\frac{5x_1+4+\frac{4-3x_2}{4}}{2}=5$,
or $x_1+x_2=2\dots \left(iii\right)$
and $20x_1-3x_2=20\dots \left(iv\right)$
Solving $\left(iii\right)$ and $\left(iv\right)$, we get
$x_1=\frac{26}{23}$ and $x_2=\frac{20}{23}$
$\therefore y_1=5\cdot \frac{26}{23}+4=\frac{222}{23}$.
Equation of the required line passing through $\left(1,5\right)$ and
$\left(x_1,y_1\right)$ is $y-5=\frac{\frac{222}{23}-5}{\frac{26}{23}-1}\left(x-1\right)$
$\Rightarrow 107x-3y-92=0$