Question

The square of the distance of the line $2x-3y=4$ from the point $\left(\right.1,1\left.\right)$ measured along the line $x+y=1$ is

Answer 1

Let $\left(\right.h,k\left.\right)$ be the foot of the segment parallel to
$x+y=1$
$\frac{k - 1}{h - 1}=-1$
$\Rightarrow h+k=2.....\left(\right.i\left.\right)$
$2h-3k=4.....\left(\right.ii\left.\right)$
Solving $\left(\right.i\left.\right)$ and $\left(\right.ii\left.\right)$ , we get
$h=2,k=0$
$\Rightarrow $ Required distance $=\sqrt{\left( 2 - 1 \right)^{2} + \left( 0 - 1 \right)^{2}}$
$=\sqrt{2}$
$=1.41$
Hence, square of the distance is $2$