Distance from a point to a line ($d=|A{x}_0+B{y}_0+C|/\sqrt{A^2+B^2}$)

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.

For distinct points ${\mathbf{r}}_1,{\mathbf{r}}_2$, the cross product in the numerator has magnitude twice the area of the triangle with third vertex ${\mathbf{r}}_0$. The same double area is the base length $|{\mathbf{r}}_2-{\mathbf{r}}_1|$ times the perpendicular height from ${\mathbf{r}}_0$, proving the displayed Euclidean distance. Equivalently, let $\mathbf{u}={\mathbf{r}}_2-{\mathbf{r}}_1$ and project ${\mathbf{r}}_0-{\mathbf{r}}_1$ onto the line direction. The squared distance is $|{\mathbf{r}}_0-{\mathbf{r}}_1{|}^2-[\mathbf{u}\cdot ({\mathbf{r}}_0-{\mathbf{r}}_1){]}^2/|\mathbf{u}{|}^2$; Lagrange's identity gives the same cross-product formula.

For two planes $r\cdot b=\lambda$ and $r\cdot c=\mu$ with linearly independent normal vectors, put $B=b\cdot b$, $C=c\cdot c$, $D=b\cdot c$ and $\Delta =BC-D^2>0$. Their intersection is a line parallel to $b\times c$. Its closest point to the origin is orthogonal to that direction, hence has the form $a=\alpha b+\beta c$. The constraints give $B\alpha +D\beta =\lambda$, $D\alpha +C\beta =\mu$, soFor every other point $a+t(b\times c)$, orthogonality gives squared distance $|a{|}^2+t^2|b\times c{|}^2$, proving the global minimum and its uniqueness. The positive determinant $\Delta =|b\times c{|}^2$ also proves that an intersection exists for every real $\lambda ,\mu$. Parallel normals require a separate consistency check and do not define a unique intersection line.