Question

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

Answer 1

Answer: 2

Let $(h,k)$ be the foot of the segment parallel to
$x+y=1$
$\frac{k-1}{h-1}=-1$
$\Rightarrow h+k=\mathrm{2.....}(i)$
$2h-3k=\mathrm{4.....}(ii)$
Solving $(i)$ and $(ii)$ , 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 $2$