Question

Find the distance of the point $(3,-5)$ from the line $3x-7y-4=0$.

Answer 1

As we know that distance $\left(d\right)$ of a point $(h,k)$ from a line $ax+by+c=0$ is given as-

$d=\frac{|ah+bk+c|}{\sqrt{a^2+b^2}}$

Therefore,

Distanvce of the point $(3,-5)$ from the line $3x-7y-4=0$ is-

$d=\frac{|3\left(3\right)-7(-5)-4|}{\sqrt{3^2+{(-7)}^2}}$

$\Rightarrow d=\frac{|9+35-4|}{\sqrt{9+49}}$

$\Rightarrow d=\frac{40}{\sqrt{58}}$

$\Rightarrow d=\frac{20\sqrt{58}}{29}$ units