Question

Find the perpendicular distance of $(1,1)$ from $x-y+\sqrt{2}=0$.

Answer 1

Perpendicular distance of $p(x_1,y_1)$ from $ax+by+c=0$ is given by,

d=$|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}|$

$\Rightarrow$ $d=|\frac{1-1+\sqrt{2}}{\sqrt{1+1}}|$ = $\frac{\sqrt{2}}{\sqrt{2}}=1$