Distance from a point to a line

For a line $Ax+By+C=0$ with $A^2+B^2>0$, its unit normal is $(A,B)/\sqrt{A^2+B^2}$. Choose any point on the line and project the displacement to $(x_0,y_0)$ onto this normal. The absolute value of that projection gives the title formula. This is the shortest Euclidean distance, because any displacement to a different point on the line has the same normal component and an additional orthogonal tangential component.