Question

Find the distance of a point $(3,4)$ from the line $4x-3y=12$.

Answer 1

$P(3,4)\equiv (x_1,y_1)$

line $\equiv 4x-3y-12=0$

$ax+by+c=0$

the distance of the point from a line

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

$=|\frac{4\times 3+(-3)\times 4+(-12)}{\sqrt{4^2+3^2}}|$

$=|\frac{12-12-12}{\sqrt{25}}|$

$=|\frac{-12}{5}|$

$=\frac{12}{5}units$