Question

Find the distance of a point (3, -5) from it
$3x-4y-5=0$

Answer 1

Given line $3x-4y-5=0$

Given point $(3,-5)$

Distance is given by $\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}\frac{|3\left(3\right)-4(-5)-5|}{\sqrt{9+16}}\frac{9+20-5}{5}\frac{24}{5}=4.8units$