Orthogonal projection of a point onto a line

Question

How would I go about solving the following problem?

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

Answer 1

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) $.

Answer 2

It is the intersection of the line $\frac{x+4}{5}=\frac{y-5}{-3}$ and the line you gave.

Answer 3

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

Answer 4

Let $p_0$ be a point a the line, $v$ be a vector parallel to the line. Then the orthogonal projection from $p$ to the line is $p_0$ + projection from $\vec{p_0p}$ to $v$.

Therefore a general solution is

$$p_0+\frac{(p-p_0)^Tv}{v^Tv}v$$

In you specific case, $v=[3,5]^T$, $p=[-4,5]^T$, choose $p_0=[0,-5]^T$. Hence,

$$ \newcommand\colv[1]{\begin{bmatrix}#1\end{bmatrix}} \text{projection point} = \colv{0\\-5} +\frac{\colv{-4&10} \colv{3\\5}} {\colv{3&5} \colv{3\\5}} \colv{3\\5} = \colv{0\\-5}+\frac{38}{34}\colv{3\\5} = \colv{\frac{57}{17}\\\frac{10}{17}} $$

This solution does not involve matrix-vector multiplication and IMO is a better solution for who would like to implement in computer program.

For example, this python code can solve this question.

def vec_scale(v: tuple[float, float], a: float):
    return (a * v[0], a * v[1])


def vec_mul_add(v1: tuple[float, float], v2: tuple[float, float], a: float = 1):
    """calc v1+a*v2"""
    return (v1[0] + a * v2[0], v1[1] + a * v2[1])


def inner_prod(v1: tuple[float, float], v2: tuple[float, float]):
    """v1^T * v2"""
    return v1[0] * v2[0] + v1[1] * v2[1]


def solve(coef: tuple[float, float, float], p: tuple[float, float]):
    """
    Params:
      coef: 3-tuple (a,b,c) to represent the linear equation ax+by+c=0.
      p: 2-tuple (px,py) to represent the point.
    """
    a, b, c = coef
    # find p0 at the line
    if b == 0:
        p0 = (-c / a, 0)
    else:
        p0 = (0, -c / b)

    # find v
    if a == 0:
        v = (1, 0)
    elif b == 0:
        v = (0, 1)
    else:
        v = (b, -a)
    return vec_mul_add(
        p0, vec_scale(v, inner_prod(vec_mul_add(p, p0, -1), v) / inner_prod(v, v))
    )


print(solve((-5, 3, 15), (-4, 5)))
# output: (3.3529411764705883, 0.5882352941176467)