Distance from a point to a line, and half-planes – Points, Straight Lines and Shapes in 2D (Senior 5 Mathematics) | Ganzaa

Points, Straight Lines and Shapes in 2D

Lessons

Senior 5 · Mathematics

Points, Straight Lines and Shapes in 2D

4 lessons

Lesson 3 of 4

Distance from a point to a line, and half-planes

25 min

The shortest way to a line

A borehole at point P is to be joined to a straight water pipe by the shortest possible branch (illustrative). Common sense says: go straight across, at right angles to the pipe. Every other route is the hypotenuse of a right-angled triangle, and so is longer. This lesson turns that idea into a formula for the distance from a point to a line, and adds the sign test that tells on which side of a line a point lies.

Definition

Foot of the perpendicular from a point $P$ to a line $d$ : the point $H$ of $d$ such that $PH \perp d$ .

Distance from a point to a line: the length $P H$ , the shortest distance from $P$ to any point of $d$ . It is 0 exactly when $P$ lies on $d$ .

Example: the distance from $(3, 5)$ to the $x$ -axis is 5, and the foot of the perpendicular is $(3, 0)$ .

The distance formula

Take the line $d : a x + b y + c = 0$ with normal $\vec{n} = (a, b)$ , a point $A$ on $d$ , and $P (x_{0}, y_{0})$ . The distance $P H$ is the length of the projection of $\overrightarrow{AP}$ on the normal:

PH = \frac{|\vec{n}\cdot\overrightarrow{AP}|}{\|\vec{n}\|} = \frac{|a(x_0 - x_A) + b(y_0 - y_A)|}{\sqrt{a^2 + b^2}} = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}},

because $a x_{A} + b y_{A} = - c$ for a point of the line. To find $H$ itself, move from $P$ along the normal: $H = P - \lambda\vec{n}$ with $\lambda = \frac{ax_0 + by_0 + c}{a^2 + b^2}$ .

Important

Distance from $P (x_{0}, y_{0})$ to $a x + b y + c = 0$

d(P, d) = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}.

Distance between parallel lines $a x + b y + c_{1} = 0$ and $a x + b y + c_{2} = 0$ (same $a$ and $b$ ): $\frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$ .

Worked example 1: distance and foot

Find the distance from $P (4, 7)$ to $3 x + 4 y - 15 = 0$ , and the foot of the perpendicular.

Solution

d = \frac{|3(4) + 4(7) - 15|}{\sqrt{9 + 16}} = \frac{25}{5} = 5.

$\lambda = \frac{25}{25} = 1$ , so $H = (4, 7) - 1 \times (3, 4) = (1, 3)$ . Check: $3 + 12 - 15 = 0$ , so $H$ is on the line, and $PH = \sqrt{3^2 + 4^2} = 5$ . ✓

Figure 1. The distance from

P

to the line is the length of the perpendicular

P H

. The sign of

3 x + 4 y - 15

tells which side of the line a point is on:

P

gives

+ 25

Q

gives

- 14

Which side of the line?

The line $a x + b y + c = 0$ cuts the plane into two open half-planes: one where $ax + by + c \gt 0$ and one where $ax + by + c \lt 0$ . The normal $(a, b)$ points into the positive one. So:

two points are on the same side of the line when $a x + b y + c$ has the same sign at both;
they are on opposite sides, and the segment joining them crosses the line, when the signs differ.

A region defined by several linear inequalities, as in S3 linear programming, is the common part of several half-planes.

Definition

Half-plane: one of the two regions into which a line divides the plane. It is open if the line itself is excluded (strict inequality) and closed if it is included ( $\leq \le$ or $\geq \ge$ ).

Example: $y \gt 2$ is the open half-plane above the line $y = 2$ .

Worked example 2: same side or not?

Are $A (1, 5)$ and $B (4, - 2)$ on the same side of $2 x - y + 1 = 0$ ?

Solution At $A$ : $2 - 5 + 1 = -2 \lt 0$ . At $B$ : $8 + 2 + 1 = 11 \gt 0$ . The signs differ, so $A$ and $B$ are on opposite sides, and the segment $A B$ crosses the line.

Worked example 3: two parallel lines

Find the distance between $3 x - 4 y + 2 = 0$ and $6 x - 8 y - 21 = 0$ .

Solution Divide the second by 2 so that $a$ and $b$ match: $3 x - 4 y - 10.5 = 0$ . Then the distance is $\frac{|2 - (-10.5)|}{\sqrt{9 + 16}} = \frac{12.5}{5} = 2.5$ . Check: $\left(-\frac{2}{3}, 0\right)$ is on the first line, and its distance to the second is $\frac{|6(-\frac{2}{3}) - 21|}{\sqrt{36 + 64}} = \frac{25}{10} = 2.5$ . ✓

Worked example 4: angle bisectors and an area

(a) Find the bisectors of the angles between $x + 2 y - 3 = 0$ and $2 x + y - 3 = 0$ . (b) Find the area of the triangle $A (1, 1)$ , $B (7, 3)$ , $C (3, 6)$ .

Solution

A bisector is the set of points equally far from both lines: $\frac{|x + 2y - 3|}{\sqrt{5}} = \frac{|2x + y - 3|}{\sqrt{5}}$ . So $x + 2y - 3 = \pm(2x + y - 3)$ : the bisectors are $x - y = 0$ and $x + y - 2 = 0$ . They are perpendicular, as angle bisectors must be.
Line $A B$ : direction $(6, 2) \sim (3, 1)$ , so $x - 3 y + 2 = 0$ . Height from $C$ : $\frac{|3 - 18 + 2|}{\sqrt{10}} = \frac{13}{\sqrt{10}}$ . Base $AB = \sqrt{40} = 2\sqrt{10}$ . Area $= \frac{1}{2} \times 2\sqrt{10} \times \frac{13}{\sqrt{10}} = 13$ .

Warning

Make the equation tidy first. The formula needs the line as $a x + b y + c = 0$ , everything on one side: write $y = 2 x + 3$ as $2 x - y + 3 = 0$ . For two parallel lines, make $a$ and $b$ identical (not just proportional) before using $\frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$ . And keep the absolute value: a distance is never negative.

Tip

The sign is free information. When you compute $a x_{0} + b y_{0} + c$ for the distance, its sign also tells you the side of the line. Two points of a triangle on opposite sides of a line means the line cuts that side of the triangle.

Khan Academy India: Distance between a line and a point. Watch how the perpendicular distance is found and how the formula with |ax₀ + by₀ + c| over √(a² + b²) appears.

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Practice

The last two are harder.

Find the distance from $(2, - 1)$ to $5 x - 12 y + 4 = 0$ .
Find the distance from the origin to $x + y - 4 = 0$ .
Find the foot of the perpendicular from $P (6, 1)$ to $x - 2 y + 1 = 0$ , and the distance.
Are $(0, 0)$ and $(3, 2)$ on the same side of $2 x - 3 y + 1 = 0$ ?
Find the distance between $4 x + 3 y - 6 = 0$ and $4 x + 3 y + 9 = 0$ .
Find $c$ so that the line $3 x - 4 y + c = 0$ is at distance 2 from $(1, 1)$ .
Find the area of the triangle with vertices $(0, 0)$ , $(4, 1)$ and $(1, 5)$ .
Find the bisectors of the angles between $3 x - 4 y + 1 = 0$ and $4 x + 3 y - 7 = 0$ , and check that they pass through the point where the two lines meet.

Show answersHide answers

$\frac{|10 + 12 + 4|}{13} = 2$ .
$\frac{4}{\sqrt{2}} = 2\sqrt{2} \approx 2.83$ .
$\lambda = \frac{6 - 2 + 1}{5} = 1$ , so $H = (6, 1) - (1, - 2) = (5, 3)$ ; distance $\sqrt{1 + 4} = \sqrt{5}$ .
Both give $+ 1$ : yes, the same side.
$\frac{|-6 - 9|}{5} = 3$ .
$\frac{|3 - 4 + c|}{5} = 2$ , so $∣ c - 1 ∣ = 10$ : $c = 11$ or $c = - 9$ .
Base from $(0, 0)$ to $(4, 1)$ : line $x - 4 y = 0$ , length $\sqrt{17}$ ; height $\frac{|1 - 20|}{\sqrt{17}}$ . Area $= \frac{1}{2} \times 19 = 9.5$ .
$3x - 4y + 1 = \pm(4x + 3y - 7)$ : $x + 7 y - 8 = 0$ and $7 x - y - 6 = 0$ . The lines meet at $(1, 1)$ , and $1 + 7 - 8 = 0$ , $7 - 1 - 6 = 0$ . ✓

Sources and credits

Written by Ganzaa: the explanations, the derivation, the diagram, the worked examples, the practice items and the quiz questions.
Video: Distance between a line and a point, Khan Academy India, embedded from YouTube.

1The shortest way to a line

2The distance formula

3Which side of the line?

The shortest way to a line

The distance formula

Which side of the line?