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

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 $AB=\left(\mathrm{6,8}\right)$
$\therefore \frac{13-6y_1+y_2-3}{2}=6.............\left(\mathrm{i}\right)$
And $\frac{y_1+y_2}{2}=8.............\left(\mathrm{i}\mathrm{i}\right)$
on solving $\left(\mathrm{i}\right)$ and $\left(\mathrm{i}\mathrm{i}\right)$ we get

$A=\left(\mathrm{1,2}\right)$ $\mathrm{a}\mathrm{n}\mathrm{d}$ $B=\left(\mathrm{11,14}\right)$
$\Rightarrow {\left(AB\right)}^2={10}^2+{12}^2=100+144=244$