Distance of a point from a straight line in $\Bbb R^2$: are there other simple proofs?

Question

To demonstrate the distance of a point from a straight line I generally use these two proofs. I apologize with all users if they are handwritten and not in MathJaX. I ask you if anyone knows any other simple way to get other proofs.

Answer 1

Using vectors a simple way is this. The equation of the line in vector notation is $$ (a, b) \begin{pmatrix} x\\y \end{pmatrix}=-c \tag 1$$ that say:

the projection of a point of the line in the direction of $(a,b)^{\intercal}$ is $$-c/\sqrt{a^2+b^2} \tag 2$$

That is: the (signed) distance of the line from the origin is $-c/\sqrt{a^2+b^2}$. Now take a point $P=(x_0,y_0)$. The projection of its distance from the origin in the same direction is

$$ (a, b) \begin{pmatrix} x_0\\y_0 \end{pmatrix} \frac{1}{\sqrt{a^2+b^2}} \tag 3 $$

so the distance of the point $P$ from the line is the absolute value of the difference: $$ \frac{ax_0+by_0}{\sqrt{a^2+b^2}} - \frac{ -c}{\sqrt{a^2+b^2}} $$

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In the picture: $ OB$ is the vector $(a,b) ^T$ (I use the Transpose because a vector is a column). The length of $\overline{OC}$ is $-c/\sqrt{a^2+b^2}$, so the line $DC$ is $ ax+by=-c$. $P$ is the point $(x_0,y_0)$ and $\overline{OP'}= (ax_0+by_0)/\sqrt{a^2+b^2}$. From these we can find $\overline{P'C}$

Answer 2

This method is not too much shorter, but it's quite neat, so I thought I'd share it.

The problem can be written as

\begin{cases} \text{Minimize } & (x-x_0)^2 + (y-y_0)^2 \\ \text{Subject to } & ax+by+c=0 \end{cases}

Thus, we use Lagrange Multipliers by considering the function

$$F(x,y,\lambda) = (x-x_0)^2 + (y-y_0)^2-\lambda(ax+by+c)$$

Setting $\nabla F=\mathbf 0$, we obtain

\begin{align} 2(x-x_0)-\lambda a & = 0 \\ 2(y-y_0) - \lambda b & = 0 \\ ax+by+c & = 0 \end{align}

Rewrite the equations as

\begin{align} (x-x_0) & = \frac{\lambda a}{2} & \\ (y-y_0) & = \frac{\lambda b}{2} \\ a(x-x_0)+b(y-y_0) & = -(ax_0+by_0+c) \end{align}

Plug the first two equations into the third to obtain

$$\frac{\lambda a^2}{2} + \frac{\lambda b^2}{2} = -(ax_0+by_0+c) \implies \lambda = \frac{-2(ax_0+by_0+c)}{a^2+b^2}$$

From this, we obtain the required distance to be

$$\sqrt{(x-x_0)^2 + (y-y_0)^2} = \sqrt{\frac{\lambda^2 a^2}{4} + \frac{\lambda^2 b^2}{4}} = \frac{|\lambda|}{2}\sqrt{a^2+b^2} = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$$

Answer 3

Given a circle representing the distance from $(x_0,y_0$) as

$$ (x-x_0)^2+(y-y_0)^2-r^2=0 $$

and a line as

$$ a x+b y+c=0 $$

assuming $b\ne 0$, taking from the line

$$ y = -\frac{a x+c}{b} $$

and substituting into the distance equation we have

$$ \left(\frac{a x+c}{b}+y_0\right)^2-r^2+(x-x_0)^2=0 $$

solving for $x$ we have

$$ x = \frac{b \left(b x_0\pm\sqrt{r^2 \left(a^2+b^2\right)-(a x_0+b y_0+c)^2}\right)-a (b y_0+c)}{a^2+b^2} $$

now if $x$ represents the tangency point between the circle and the line (orthogonal projection), it should have an unique value hence we need

$$ r^2 \left(a^2+b^2\right)-(a x_0+b y_0+c)^2=0 $$

or in other words

$$ r = \frac{|a x_0+b y_0+c|}{\sqrt{a^2+b^2}} $$

Answer 4

This is what I used during high school. Let us say that we want to find minimum distance from point $A$ with coordinate $(x_0 , y_0)$ to line $ax+by+c=0$.

Say that point $C$ lie on the line such that $|AC|$ is minimum. Draw vertical line through $A$ and intersect the line at $B$ with coordinate $(x_0 , y_1)$.

Key take away:

$\frac{|AC|}{|AB|}=|\frac{b}{\sqrt{a^{2}+b^{2}}}|$

$|AB|=|y_{0}-y_{1}|=\frac{|ax_0 + by_0 + c|}{|b|}$

substitute to the first equation to obtain Your result.

In case it is not clear enough, $ax_0 + by_0 + c = ax_0 + by_1 + c + b(y_0 - y_1) =b(y_0 - y_1)$ since $(x_0 , y_1)$ is on the line.

If the line is vertical, we can draw horizontal line through $A$ instead and proceed with similar steps with the same result.

Answer 5

If You are familiar with vectors, i think this is a good food for thought.

We want the shortest distance between point $A$ and line $\nabla{L}\cdot\vec{X}+c=0$. Suppose $B$ lies on the line, we can project vector $\vec{A}-\vec{B}$ to vector $\frac{\nabla{L}}{|\nabla{L}|}$ to obtain the shortest distance.

$$ d_{min}=\left|\frac{\nabla{L}}{|\nabla{L}|}\cdot(\vec{A}-\vec{B})\right|=\frac{\left|\nabla{L}\cdot{\vec{A}}+c\right|}{|\nabla{L}|} $$