Point-to-line distance: |ax₀ + by₀ + c| / √(a² + b²)

The formula

For a point $M(x_0, y_0)$ and a line $(d)$ with Cartesian equation $a x + b y + c = 0$:

$d(M, (d)) = \dfrac{|a x_0 + b y_0 + c|}{\sqrt{a^2 + b^2}}$

Interpretation

• Numerator: signed distance (can be positive or negative), take absolute value.
• Denominator: norm of the normal vector $\vec{n} = (a, b)$.

Typical example

Distance from $M(3, 4)$ to line $(d): 3x - 4y + 5 = 0$.

$d = \dfrac{|3 \times 3 - 4 \times 4 + 5|}{\sqrt{9 + 16}} = \dfrac{|9 - 16 + 5|}{\sqrt{25}} = \dfrac{2}{5} = 0.4$.

Special cases

Horizontal line $y = k$ ⇒ equation $0 \cdot x + 1 \cdot y - k = 0$.

$d(M, (d)) = |y_0 - k|$.

Vertical line $x = k$ ⇒ $d(M, (d)) = |x_0 - k|$.

Origin to line: $d(O, (d)) = \dfrac{|c|}{\sqrt{a^2 + b^2}}$.

Setting the equation

If the line is given by $y = m x + p$ (slope form), convert to Cartesian:

$m x - y + p = 0$

Here $a = m$, $b = -1$, $c = p$.

Common mistakes

1. Forgetting the absolute value in the numerator.
2. Forgetting $\sqrt{a^2 + b^2}$ in the denominator.
3. Using $y = mx + p$ as-is without converting to $ax + by + c = 0$.

?À toi de jouer

Distance from O(0, 0) to line 3x + 4y − 10 = 0?

A.2B.10C.5D.1/2

Valider