Euclidean plane/(4,5)/Distance to 2x-3y is 7/Example

We want to determine in the Euclidean plane the distance between the point ${}P={\begin{pmatrix}4\\5\end{pmatrix}}$ and the line ${}G$ given by ${}2x-3y=7$. The line has the form

${}G={\left\{{\begin{pmatrix}{\frac {7}{2}}\\0\end{pmatrix}}+t{\begin{pmatrix}3\\2\end{pmatrix}}\mid t\in \mathbb {R} \right\}}\,,$

and ${}{\begin{pmatrix}2\\-3\end{pmatrix}}$ is a vector perpendicular to ${}G$. We have

${}{\begin{pmatrix}{\frac {7}{2}}\\0\end{pmatrix}}-{\begin{pmatrix}4\\5\end{pmatrix}}={\begin{pmatrix}-{\frac {1}{2}}\\-5\end{pmatrix}}=-{\frac {23}{26}}{\begin{pmatrix}3\\2\end{pmatrix}}+{\frac {14}{13}}{\begin{pmatrix}2\\-3\end{pmatrix}}\,.$

Therefore, the dropped perpendicular foot is

${}{\begin{pmatrix}4\\5\end{pmatrix}}+{\frac {14}{13}}{\begin{pmatrix}2\\-3\end{pmatrix}}={\begin{pmatrix}{\frac {80}{13}}\\{\frac {23}{13}}\end{pmatrix}}\,,$

and the distance is

${\frac {14}{13}}{\sqrt {13}}.$