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 $\frac{x}{3}+\frac{y}{-5}=1$.

Solution 1:

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 $$\frac{vv^T}{v^Tv}p+\left(I-\frac{vv^T}{v^Tv}\right)p_0=(\frac{57}{17},\frac{10}{17})^T$$

Solution 2:

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) $.

Solution 3:

I wanted to find a direct equation for the orthogonal projection of a point (X,Y) onto a line (y=mx+b). I will refer to the point of projection as as $(X_p,Y_p)$.

Using the same observation, that two orthogonal slopes multiplied together make -1, the slope of the projection line is -1/m and it is also the rise over run for the arbitrary point (X,Y) and the point of projection $(X_p,Y_p)$.

Hence, $\frac {-1} {m} = \frac{Y-Y_p} {X-X_p}$

The unknown is the projection point, so some simple manipulation gets that on one side: $X_p + mY_p = X + mY$

Knowing that $(X_p,Y_p)$ is on the line, means that $Y_p = mX_p + b$, giving us:

$X_p + m(mX_p + b) = X + mY$

Finally, isolating $X_p$ as:

$X_p = \frac {X + mY - mb} {1+m^2}$

And, through substitution:

$Y_p = \frac {mX + m^2Y + b} {1+m^2}$

Using the point (-4,5) and line definition in the original example we get (noting that the slope of the original line $\frac 5 3$ is approximately 1.667 and the intercept (b) is -5):

$X_p = \frac { -4 + (1.667)(5) - (1.667)(-5)} {1+1.667^2} = 3.3528$

$Y_p = \frac {1.667(-4)+(1.667^2)(5)+(-5)} {1+1.667^2} = 0.589$

which is essentially the same as: $\left(\frac {57}{17},\frac {10}{17}\right)$

The length of the line of projection from (X,Y) to the line y=mx+b can easily be calculated from the two points:

$\sqrt{(Y-Y_p)^2 + (X-X_p)^2}$

But to do it without calculating the point of projection ($X_p$,$Y_p$) first, we can substitute the above equations for $X_p$ and $Y_p$ and obtain:

$\text{length} = \sqrt{\frac {m^2X^2 + Y^2 + b^2 -2bY - 2mXY + 2mbX} {1+m^2}}$