How would I go about solving the following problem?
Find an orthogonal projection of a point T$(-4,5)$ onto a line $\frac{x}{3}+\frac{y}{-5}=1$.
How would I go about solving the following problem?
Find an orthogonal projection of a point T$(-4,5)$ onto a line $\frac{x}{3}+\frac{y}{-5}=1$.
It is the intersection of the line $\frac{x+4}{5}=\frac{y-5}{-3}$ and the line you gave.
I'd like to solve the problem using orthogonal projection matrices.
If the line is passing through the origin, it will be very simple to find the orthogonal projection. Suppose $p$ is the given point, $v$ is the given line (passing through the origin, so represented by a vector). Then the orthogonal projection point is $\frac{vv^T}{v^Tv}p$
Now the given line does not pass through the origin. But it can be convert to the simple problem above. Choose a point $p_0=(x_0,y_0)^T$ on the given line. Move the origin to $p_0$ (later move back). Then the line can be represented by a vector $v$, and the original given point becomes $p_1=p-p_0$. Now compute $\frac{vv^T}{v^Tv}p_1$. Then move the origin back, we get the orthogonal projection in the original coordinate system is $\frac{vv^T}{v^Tv}(p-p_0)+p_0=\frac{vv^T}{v^Tv}p+\left(I-\frac{vv^T}{v^Tv}\right)p_0$.
Specifically, for your problem, $p=(-4,5)^T$. Choose $p_0=(0,-5)^T$, then $\frac{vv^T}{v^Tv}p+\left(I-\frac{vv^T}{v^Tv}\right)p_0=(\frac{57}{17},\frac{10}{17})^T$
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) $.