Question

Find distance of point $A(2,3)$ measured parallel to the line $x-y=5$ from the line $2x+y+6=0$.

Answer 1

Slope of the line, $x-y=5$ is,

$m=1$

Parallel lines have equal slope.

Let the equation of the line passing through $(2,3)$ and parallel to the line $y=x-5$ is,

$y=mx+c$

$y=x+c$

$3=2+c$

$c=1$

$\Rightarrow y=x+1$

Now, the intersection point of $y=x+1$ and $2x+y+6=0$ or $y=-2x-6$ is,

$\Rightarrow (-\frac{7}{3},-\frac{4}{3})$

Therefore, distance between $(2,3)$ and $(-\frac{7}{3},-\frac{4}{3})$ will be

$\Rightarrow \sqrt{{(2+\frac{7}{3})}^2+{(3+\frac{4}{3})}^2}=\frac{1}{3}\sqrt{169+169}=\frac{13\sqrt{2}}{3}units$

Hence, this is the required distance.