<s>[INST] Orthogonal projection of a point onto a line How would I go about solving the following problem?

Find an orthogonal projection of a point T$(-4,5)$ onto a line $\frac{x}{3}+\frac{y}{-5}=1$.

 [/INST] The slope of the line $r$, with equation $\frac{x}{3}+\frac{y}{-5}=1$, is $m_{r}=\frac{5}{3}$ (because $\frac{x}{3}+\frac{y}{-5}=1$ is equivalent to $y=\frac{5}{3}x-5$). The slope of the line $s$ orthogonal to $r$ is $m_{s}=-\frac{3}{5}$ (because $m_{r}m_{s}=-1$). Hence the equation of $s$ is of the form
$$y=-\frac{3}{5}x+b_{s}.$$
Since $T(-4,5)$ is a point of $s$, we have 
$$5=-\frac{3}{5}\left( -4\right) +b_{s},$$ 
which means that $b_{s}=\frac{13}{5}$. So the equation of $s$ is
$$y=-\frac{3}{5}x+\frac{13}{5}.$$

The coordinates of the orthogonal projection of $T$ onto $r$ are the solutions of the system
$$\left\{ 
\begin{array}{c}
y=\frac{5}{3}x-5 \\ 
y=-\frac{3}{5}x+\frac{13}{5},
\end{array}
\right. $$
which are $(x,y)=\left( \frac{57}{17},\frac{10}{17}\right) $.
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