Before we begin: Line $A$ has gradient $2$. Line $B$ is parallel to $A$. Line $C$ is perpendicular to $A$. What can you say about the gradients of $B$ and $C$?
Come back to this after you have worked through the lesson.
Parallel and perpendicular lines are everywhere in geometry and design. Parallel lines never meet because they share the same gradient. Perpendicular lines meet at right angles because their gradients have a special multiplicative relationship. Once you know these properties, you can find the equation of any line parallel or perpendicular to a given line through any point.
- State that parallel lines have the same gradient.
- Find the perpendicular gradient as the negative reciprocal.
- Write the equation of a parallel or perpendicular line through a given point.
- Classify the relationship between two given lines as parallel, perpendicular, or neither.
Wrong: Thinking perpendicular gradients are just negatives. If $m = 2$, the perpendicular gradient is not $-2$.
Right: Perpendicular gradient is $-\dfrac{1}{2}$ (flip to $\dfrac{1}{2}$, then change sign).
Wrong: Forgetting to take the negative of the reciprocal. $m = -\dfrac{2}{3}$ gives perpendicular $m = \dfrac{2}{3}$ (wrong).
Right: Perpendicular to $m = -\dfrac{2}{3}$ is $m = \dfrac{3}{2}$. Flip to $-\dfrac{3}{2}$, then negate to $\dfrac{3}{2}$.
Parallel lines never meet. They point in the same direction, so they share the same gradient. If two lines are parallel, their gradients are equal. To find the equation of a line parallel to a given line through a point: copy the gradient, substitute the point, and solve for $c$.
$m_1 = m_2$. Same direction. Never meet.
Perpendicular lines meet at right angles. Their gradients have a special multiplicative relationship: the product of their gradients equals $-1$. Equivalently, the perpendicular gradient is the negative reciprocal of the original. Flip the fraction and change the sign.
$m_1 \times m_2 = -1$. Flip and negate. $m_2 = -\dfrac{1}{m_1}$.
Given two lines, you can determine if they are parallel, perpendicular, or neither by comparing their gradients. First convert both equations to gradient-intercept form $y = mx + c$ if necessary, then extract the gradients and apply the tests.
Parallel: $m_1 = m_2$. Perpendicular: $m_1 \times m_2 = -1$. Neither: neither test passes.
Many geometry problems require you to combine parallel and perpendicular properties. Rectangles have opposite sides parallel and adjacent sides perpendicular. Squares and rhombuses have similar relationships. When asked to find equations in a geometric figure, identify which sides are parallel or perpendicular, then apply the gradient rules.
Opposite sides parallel. Adjacent sides perpendicular. Use gradient rules.
Gradient: Parallel lines share the same gradient, so $m = 3$.
Write: $y = 3x + c$
Substitute $(2, -1)$: $-1 = 3(2) + c$ → $-1 = 6 + c$ → $c = -7$
Answer: $y = 3x - 7$
Perpendicular gradient: $m_1 = 2$, so $m_2 = -\dfrac{1}{2}$
Write: $y = -\frac{1}{2}x + c$
Substitute $(4, 3)$: $3 = -\frac{1}{2}(4) + c$ → $3 = -2 + c$ → $c = 5$
Answer: $y = -\frac{1}{2}x + 5$
Convert to gradient-intercept form:
$2x + 3y = 6$ → $3y = -2x + 6$ → $y = -\frac{2}{3}x + 2$, so $m_1 = -\frac{2}{3}$
$3x - 2y = 4$ → $-2y = -3x + 4$ → $y = \frac{3}{2}x - 2$, so $m_2 = \frac{3}{2}$
Test: $m_1 \times m_2 = -\frac{2}{3} \times \frac{3}{2} = -1$. The lines are perpendicular.
Brain Trainer
4 quick-fire drills. Beat the clock.
5 MCQs and 3 short-answer questions. Target: 80% accuracy.
Your answer:
Parallel gradient: $m = -2$
$y = -2x + c$
$4 = -2(-1) + c$ → $4 = 2 + c$ → $c = 2$
$y = -2x + 2$
(a) $y = 3x + 2$ and $y = -\dfrac{1}{3}x - 1$
(b) $2x - 5y = 10$ and $4x - 10y = 20$
(c) $x + 2y = 6$ and $2x - y = 3$
Your answer:
(a) $m_1 = 3$, $m_2 = -\frac{1}{3}$. $3 \times (-\frac{1}{3}) = -1$. Perpendicular.
(b) $2x - 5y = 10$ → $y = \frac{2}{5}x - 2$. $4x - 10y = 20$ → $y = \frac{4}{10}x - 2 = \frac{2}{5}x - 2$. Same gradient. Parallel.
(c) $x + 2y = 6$ → $y = -\frac{1}{2}x + 3$. $2x - y = 3$ → $y = 2x - 3$. $(-\frac{1}{2}) \times 2 = -1$. Perpendicular.
(a) Find the equation of side $AB$.
(b) Find the equation of the line through $D$ (the fourth vertex) that is perpendicular to $AB$.
(c) Explain why adjacent sides of a rectangle must be perpendicular, and verify this for two adjacent sides of this rectangle.
Your answer:
(a) $AB$ is horizontal: $y = 2$
(b) $D = (1, 6)$. Perpendicular to $AB$ (horizontal) is vertical: $x = 1$.
(c) Adjacent sides of a rectangle meet at right angles, so they must be perpendicular. $AB$ is horizontal ($m = 0$) and $BC$ is vertical (undefined gradient). A horizontal line and a vertical line are perpendicular because they meet at $90°$.
Consolidate and reflect before moving on.
Find the equation of the line through $(2, -3)$ that is perpendicular to $3x + 4y = 12$. Give your answer in general form. (Hint: first find the gradient of $3x + 4y = 12$ by rearranging to $y = mx + c$.)
Parallel lines have equal gradients ($m_1 = m_2$). Perpendicular lines have gradients that multiply to $-1$ ($m_2 = -\frac{1}{m_1}$). To classify two lines, convert to $y = mx + c$ and compare.
Thinking perpendicular means just negative. If $m = 2$, the perpendicular is $-\frac{1}{2}$, not $-2$. Always flip the fraction first, then change the sign.
This lesson builds on equations of lines by adding geometric relationships. The final lesson brings everything together with modelling and problem solving.
Given $y = \frac{2}{3}x - 4$, find (a) the equation of the parallel line through $(3, 2)$, and (b) the equation of the perpendicular line through $(3, 2)$. Time yourself, can you do both in under 2 minutes?
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