Parallel & Perpendicular Gradients

Video lesson · from 25:372 worked examples

One lesson video covers all of The Straight Line, so it opens at 25:37 for this topic — not from the beginning.

Parallel lines have equal gradients.

If $m_{AB} \times m_{CD} = -1$ then AB & CD are perpendicular.

If AB & CD are perpendicular, then $m_{AB} \times m_{CD} = -1$ .

Negative and reciprocal: incorrect perpendicular gradients are one of the most common slips in this whole section. Doing only half the operation — flipping the fraction but keeping the sign, or vice versa — is the usual cause. From $m = \frac{2}{3}$ the perpendicular gradient is $-\frac{3}{2}$ .
Whole numbers are fractions too: perpendicular to $m = 4$ is $-\frac{1}{4}$ , not $-4$ .
Get the gradient from the equation first: if you are given a line as $3x + 2y = 12$ , rearrange to find $m$ before doing anything else.
Horizontal and vertical are the exception: the rule $m_1 m_2 = -1$ does not apply. A horizontal line has $m = 0$ and its perpendicular is vertical, whose gradient is undefined.

Example 1

Given that T is the point $(1, -2)$ and S is $(-4, 5)$ , find the gradient of a line perpendicular to ST.

\begin{aligned} m_{ST} &= \frac{5 - (-2)}{-4 - 1} \\ &= \frac{7}{-5} \\ &= -\frac{7}{5} \end{aligned}

For perpendicular lines, $m_1 \times m_2 = -1$ .

m_{\perp} = \frac{5}{7}

Example 2

Triangle MOP has vertices $M(-3, 9)$ , $O(0, 0)$ and $P(12, 4)$ . Show that the triangle is right-angled.

\begin{aligned} m_{MO} &= \frac{0 - 9}{0 - (-3)} \\ &= \frac{-9}{3} \\ &= -3 \end{aligned}

\begin{aligned} m_{OP} &= \frac{4 - 0}{12 - 0} \\ &= \frac{4}{12} \\ &= \frac{1}{3} \end{aligned}

\begin{aligned} m_{MO} \times m_{OP} &= -3 \times \frac{1}{3} \\ &= -1 \end{aligned}

Since the product of the gradients is -1, MO is perpendicular to OP, so the triangle is right-angled at O.