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 $A=\left(13-6y_1,y_1\right)$
$B=\left(y_2-3,y_2\right)$
$\therefore$ mid-point $AB=\left(6,8\right)$
$\therefore \frac{13-6y_1+y_2-3}{2}=\mathrm{6.............}\left(i\right)$
And $\frac{y_1+y_2}{2}=\mathrm{8.............}\left(ii\right)$
on solving $\left(i\right)$ and $\left(ii\right)$ we get
$A=\left(1,2\right)$ $and$ $B=\left(11,14\right)$
$\Rightarrow (AB{)}^2={\left(10\right)}^2+{\left(12\right)}^2=100+144=244$