# Difference between revisions of "Distance formula"

The distance formula is a direct application of the Pythagorean Theorem in the setting of a Cartesian coordinate system. In the two-dimensional case, it says that the distance between two points $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$ is given by $d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$. In the $n$-dimensional case, the distance between $(a_1,a_2,...,a_n)$ and $(b_1,b_2,...,b_n)$ is $\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+\cdots+(a_n-b_n)^2}$

## Shortest distance from a point to a line

the distance between the line $ax+by+c = 0$ and point $(x_1,y_1)$ is $$\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$$

### Proof

The equation $ax + by + c = 0$ can be written as $y = -\dfrac{a}{b}x - \dfrac{c}{a}$ Thus, the perpendicular line through $(x_1,y_1)$ is: $$\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{t}{\sqrt{a^2+b^2}}$$ where $t$ is the parameter. $t$ will be the distance from the point $(x_1,y_1)$ along the perpendicular line to $(x,y)$. So $$x = x_1 + a \cdot \dfrac{t}{\sqrt{a^2+b^2}}$$ and $$y = y_1 + b \cdot \dfrac{t}{\sqrt{a^2+b^2}}$$

This meets the given line $ax+by+c = 0$, where: $$a\left(x_1 + a \cdot \dfrac{t}{\sqrt{a^2+b^2}}\right) + b\left(y_1 + b \cdot \dfrac{t}{\sqrt{a^2+b^2}}\right) + c = 0$$ $$\implies ax_1 + by_1 + c + \dfrac{t(a^2+b^2)}{\sqrt{a^2+b^2}} + c = 0$$ $$\implies ax_1 + by_1 + c + t \cdot \sqrt{a^2+b^2} = 0$$

, so: $$t \cdot \sqrt{a^2+b^2} = -(ax_1+by_1+c)$$ $$\implies t = \dfrac{-(ax_1+by_1+c)}{\sqrt{a^2+b^2}}$$

Therefore the perpendicular distance from $(x_1,y_1)$ to the line $ax+by+c = 0$ is: $$|t| = \dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2+b^2}}$$