Parallel and perpendicular lines – Linear and Quadratic Functions (Senior 3 Mathematics) | Ganzaa

Linear and Quadratic Functions

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Senior 3 · Mathematics

Linear and Quadratic Functions

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Lesson 3 of 5

Parallel and perpendicular lines

25 min

Lines that never meet, and lines that meet square

Look down on a street in Kigali: the kerbs and the lane markings run side by side and never meet, while the zebra crossing runs straight across them at a right angle. These two relationships, parallel and perpendicular, can be read straight from the gradients of the lines, without drawing anything.

Definition

Parallel lines: lines in the same plane that never meet. Notation: $L_1 \parallel L_2$ .

Perpendicular lines: lines that meet at a right angle. Notation: $L_1 \perp L_2$ .

Example: $y = 2x + 1 \parallel y = 2x - 3$ ; the $x$ -axis $\perp$ the $y$ -axis.

Important

Parallel lines have equal gradients. Two non-vertical lines with gradients $m_{1}$ and $m_{2}$ are parallel exactly when

m_{1} = m_{2} .

Why: equal gradients mean both lines rise by the same amount for every unit across, so the vertical gap between them never changes and they never meet. (If they also have the same intercept, they are the same line.) All vertical lines are parallel to each other.

Important

Perpendicular lines: the product of the gradients is $- 1$ . Two lines with gradients $m_{1}$ and $m_{2}$ , neither horizontal nor vertical, are perpendicular exactly when

m_1 \times m_2 = -1, \qquad\text{that is,}\qquad m_2 = -\frac{1}{m_1}.

Why: a gradient triangle with run 3 and rise 1 gives $m_1 = \frac{1}{3}$ . Turn the triangle through $90^\circ$ about O: the run of 3 becomes a rise of 3, and the rise of 1 becomes a run of $- 1$ . The turned line has $m_2 = \frac{3}{-1} = -3$ , and $\frac{1}{3} \times (-3) = -1$ . The same turn works for any triangle with run $p$ and rise $q$ : the gradient $\frac{q}{p}$ becomes $\frac{p}{-q} = -\frac{p}{q}$ .

Turning the blue gradient triangle through a right angle gives the red one: the gradient ⅓ becomes −3.

A horizontal line ( $m = 0$ ) and a vertical line are also perpendicular, but the product rule cannot be used, because a vertical line has no gradient.

Definition

Negative reciprocal of a non-zero number $m$ : the number $-\frac{1}{m}$ . Turn the fraction upside down and change the sign.

Example: the negative reciprocal of $\frac{2}{3}$ is $-\frac{3}{2}$ ; of $- 4$ it is $\frac{1}{4}$ .

Worked example 1: which lines are parallel or perpendicular?

$L_{1} : y = 3 x - 2$ , $L_{2} : 6 x - 2 y + 5 = 0$ , $L_{3} : x + 3 y = 4$ .

Solution Find each gradient. $m_{1} = 3$ . For $L_{2}$ , $m_2 = -\frac{6}{-2} = 3$ . For $L_{3}$ , $3 y = - x + 4$ , so $m_3 = -\frac{1}{3}$ .

$m_{1} = m_{2} = 3$ , so $L_1 \parallel L_2$ . $m_1 \times m_3 = 3 \times \left(-\frac{1}{3}\right) = -1$ , so $L_3 \perp L_1$ , and so $L_3 \perp L_2$ too.

Worked example 2: a parallel line through a point

Find the line through $(2, - 1)$ parallel to $4 x - 2 y + 7 = 0$ .

Solution The given line has $m = -\frac{4}{-2} = 2$ . A parallel line has the same gradient:

y + 1 = 2(x - 2) \;\Longrightarrow\; y = 2x - 5, \qquad\text{or}\qquad 2x - y - 5 = 0.

Shortcut: a line parallel to $4 x - 2 y + 7 = 0$ has the form $4 x - 2 y + k = 0$ . Put in $(2, - 1)$ : $8 + 2 + k = 0$ , so $k = - 10$ , giving $4 x - 2 y - 10 = 0$ , the same line.

Worked example 3: the perpendicular from a point to a line

Find the line through $P (5, 3)$ perpendicular to $2 x + 3 y = 6$ , the point $F$ where the two lines meet, and the distance $P F$ .

Solution The given line has $m_1 = -\frac{2}{3}$ , so the perpendicular has $m_2 = \frac{3}{2}$ :

y - 3 = \frac{3}{2}(x - 5) \;\Longrightarrow\; 2y - 6 = 3x - 15 \;\Longrightarrow\; 3x - 2y - 9 = 0.

Solve $2 x + 3 y = 6$ and $3 x - 2 y = 9$ together: multiply by 2 and by 3 and add, $4 x + 6 y + 9 x - 6 y = 12 + 27$ , so $13 x = 39$ , $x = 3$ , and then $y = 0$ . So $F = (3, 0)$ .

$PF = \sqrt{(5 - 3)^2 + (3 - 0)^2} = \sqrt{13} \approx 3.61$ units. This is the shortest distance from $P$ to the line.

The perpendicular from P meets the line at F, at a right angle.

Worked example 4: a water point fair to two villages

On a map, two villages are at $A (- 2, 1)$ and $B (4, 5)$ . A water point must be the same distance from both, so it must lie on the perpendicular bisector of $A B$ : the line through the midpoint of $A B$ perpendicular to $A B$ . Find its equation.

Solution Midpoint: $M = \left(\frac{-2 + 4}{2}, \frac{1 + 5}{2}\right) = (1, 3)$ . Gradient of $A B$ : $\frac{5 - 1}{4 - (-2)} = \frac{4}{6} = \frac{2}{3}$ , so the bisector has gradient $-\frac{3}{2}$ :

y - 3 = -\frac{3}{2}(x - 1) \;\Longrightarrow\; 2y - 6 = -3x + 3 \;\Longrightarrow\; 3x + 2y - 9 = 0.

Check: the point $(3, 0)$ is on it, and its distances are $\sqrt{25 + 1} = \sqrt{26}$ from $A$ and $\sqrt{1 + 25} = \sqrt{26}$ from $B$ ✓.

Warning

Do both steps of the negative reciprocal. Perpendicular to $m = \frac{2}{3}$ is $-\frac{3}{2}$ : not $\frac{3}{2}$ (sign forgotten) and not $-\frac{2}{3}$ (fraction not turned over). Check: $\frac{2}{3} \times \left(-\frac{3}{2}\right) = -1$ ✓.

Find the gradient before comparing. $6 x - 2 y + 5 = 0$ and $y = 6 x$ are not parallel: the first has gradient 3.

Tip

Parallel: keep $a$ and $b$ . A line parallel to $a x + b y + c = 0$ is $a x + b y + k = 0$ . Perpendicular: swap them and change one sign: $b x - a y + k = 0$ . Find $k$ from the given point.

Khan Academy: Equations of parallel and perpendicular lines. Watch how the gradient of the given line is used to write a parallel line and a perpendicular line through a chosen point.

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Practice

The last three are harder.

Give the gradient of a line (a) parallel to (b) perpendicular to $y = - 4 x + 1$ .
Which two of these lines are parallel, and which is perpendicular to them? $L_{1} : 2 x - y + 3 = 0$ , $L_{2} : x + 2 y = 6$ , $L_{3} : y = 2 x - 5$ .
Find the line through $(1, 4)$ parallel to $y = 3 x - 7$ .
Find the line through $(- 2, 1)$ perpendicular to $y = \frac{1}{2}x + 3$ .
Find the line through $(3, - 2)$ perpendicular to $5 x - 2 y + 1 = 0$ .
Find $k$ if $y = k x + 1$ is perpendicular to $3 x + 4 y = 12$ .
Find the perpendicular bisector of $A (1, 2)$ and $B (5, 8)$ .
$A (0, 0)$ , $B (4, 2)$ and $C (3, 4)$ form a triangle. Show that it has a right angle, and say where.

Show answersHide answers

(a) $- 4$ (b) $\frac{1}{4}$ .
$m_{1} = 2$ , $m_2 = -\frac{1}{2}$ , $m_{3} = 2$ : $L_1 \parallel L_3$ , and $L_{2}$ is perpendicular to both.
$y - 4 = 3 (x - 1)$ : $y = 3 x + 1$ .
Gradient $- 2$ : $y - 1 = - 2 (x + 2)$ , so $y = - 2 x - 3$ .
Given gradient $\frac{5}{2}$ , so $m = -\frac{2}{5}$ : $y + 2 = -\frac{2}{5}(x - 3)$ , that is, $2 x + 5 y + 4 = 0$ .
The given gradient is $-\frac{3}{4}$ , so $k = \frac{4}{3}$ .
Midpoint $(3, 5)$ ; $m_{AB} = \frac{6}{4} = \frac{3}{2}$ , so the gradient is $-\frac{2}{3}$ : $y - 5 = -\frac{2}{3}(x - 3)$ , that is, $2 x + 3 y - 21 = 0$ .
$m_{AB} = \frac{2}{4} = \frac{1}{2}$ , $m_{BC} = \frac{4 - 2}{3 - 4} = -2$ , and $\frac{1}{2} \times (-2) = -1$ : the right angle is at $B$ .

Sources and credits

Written by Ganzaa.
Photo: Zebra crossing in road transport in kigali.jpg, Ndahiro derrick, Wikimedia Commons, licensed under CC BY-SA 4.0. Used unchanged.
Video: Equations of parallel and perpendicular lines, Khan Academy, embedded from YouTube.

The diagrams, worked examples, practice items and quiz questions were written by Ganzaa; the village map is illustrative. This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Lines that never meet, and lines that meet square

Lines that never meet, and lines that meet square