Point-to-Line Distance Formula

Let $H(x_H,y_H)$ be the orthogonal projection of $M_0(x_0,y_0)$ onto $\Delta$. Because $\overrightarrow{HM_0}$ is parallel to the normal vector $\vec{n}=(a,b)$, there exists a scalar $t\in\mathbb{R}$ such that $\overrightarrow{HM_0}=t\vec{n}=(ta,tb)$, which gives $x_H=x_0-ta$ and $y_H=y_0-tb$.

Since $H$ lies on $\Delta$, its coordinates satisfy $ax_H+by_H+c=0$. Substituting $x_H$ and $y_H$ yields $a(x_0-ta)+b(y_0-tb)+c=0$, or $ax_0+by_0+c-t(a^2+b^2)=0$.

Solving for $t$ gives $t=\dfrac{ax_0+by_0+c}{a^2+b^2}$. Therefore the distance is $d(M_0,\Delta)=|\overrightarrow{HM_0}|=|t|\,|\vec{n}|=\dfrac{|ax_0+by_0+c|}{a^2+b^2}\sqrt{a^2+b^2}=\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$.