Question

The vertices of a $△OBC$ are $O(0,0),B(3,1),C(1,3)$. A line parallel to $BC$ & intersecting the sides $OB$ & $OC$, has perpendicular distance from the point $(0,0)$ as half. The sum of squares of its intercepts on axes is

Answer 1

Equation of $BC$ is $x+y+4=0$
Equation of the line parallel to $BC$ is
$x+y+\lambda =0$
and its perpendicular distance from the origin is $\frac{1}{2}$.
$\therefore \left|\frac{\lambda }{\sqrt{2}}\right|=\frac{1}{2}\Rightarrow \lambda =\pm \frac{1}{\sqrt{2}}$
but x & y-intercept is negative

$\therefore \lambda =\frac{1}{\sqrt{2}}$
Hence equation is $2x+2y+\sqrt{2}=0$

For x- intercept put $y=0$ then $x=\frac{-1}{\sqrt{2}}$

For y- intercept put $y=0$ then $y=\frac{-1}{\sqrt{2}}$

Sum of squares of its intercepts on axes is $=(\frac{-1}{\sqrt{2}{)}^2}+(\frac{-1}{\sqrt{2}}{)}^2=1$