Question

If a line parallel to $2x+6y-7=0$ makes an intercept of $10$ units between the coordinate axes and $y-$ intercept is $\pm \sqrt{k}$, then the value of $k$ is

Answer 1

Given line : $2x+6y-7=0$
Line parallel to this is
$2x+6y+\lambda =0$
Line cuts $x-$axis at $(-\frac{\lambda }{2},0)$ and
$y-$ axis at $(0,-\frac{\lambda }{6})$
Now, the intercept between the coordinate axes is
$\sqrt{{(-\frac{\lambda }{2})}^2+{(-\frac{\lambda }{6})}^2}=10\Rightarrow \frac{\sqrt{40}|\lambda |}{12}=10\Rightarrow |\lambda |=6\sqrt{10}\Rightarrow \lambda =\pm 6\sqrt{10}$

Therefore, the $y-$ intercept
$\pm \sqrt{k}=\pm \sqrt{10}\therefore k=10$