Question

If the segment intercepted between the lines $x+6y-13=0$ and $x-y+3=0$ is bisected at $\left(6,8\right)$ , then the square of the length of segment is

• A. $268$
• B. $244$
• C. $212$
• D. $252$

Answer 1

Answer: (B)

Let the segment is $AB$ as shown in the figure

Let $A=\left(13 - 6 y_{1} , y_{1}\right)$
$B=\left(y_{2} - 3 , y_{2}\right)$
$\therefore $ mid-point $AB=\left(6,8\right)$
$\therefore \frac{13 - 6 y_{1} + y_{2} - 3}{2}=6.............\left(i\right)$
And $\frac{y_{1} + y_{2}}{2}=8.............\left(i i\right)$
on solving $\left(i\right)$ and $\left(i i\right)$ we get
$A=\left(1,2\right)$ $and$ $B=\left(11,14\right)$
$\Rightarrow \left(\right. A B \left.\right)^{2}=\left(10\right)^{2}+\left(12\right)^{2}=100+144=244$