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

The Correct Option is C

Given equations of lines are $5x - y + 4 = 0\quad ...(i)$ and $3x + 4y - 4 = 0\quad ...(ii)$ 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 $x_{1}+x_{2}=2\quad\ldots\left(iii\right)$ and $20x_{1}-3x_{2}=20\quad\ldots\left(iv\right)$ 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 $y-5=\frac{\frac{222}{23}-5}{\frac{26}{23}-1}\left(x-1\right)$ $\Rightarrow 107x-3y-92=0$