3.2 Parallel and perpendicular lines — Practise with help

Work each one first, closed book. Read the gradient of the given line before you write anything down, and rearrange the line if it is not in the form $y = mx + c$ .

Warm-up: remember

Work out the gradient of the line through $(0, 5)$ and $(4, 13)$ .
Rearrange $2x + 3y = 6$ to make $y$ the subject.
Work out the midpoint of the segment from $(2, 4)$ to $(8, 10)$ .

Answers

$m = \dfrac{13 - 5}{4 - 0} = \dfrac{8}{4} = 2$ .
$3y = 6 - 2x$ , so $y = 2 - \dfrac{2}{3}x$ . The gradient is $-\dfrac{2}{3}$ .
$M = \left(\dfrac{2 + 8}{2}, \dfrac{4 + 10}{2}\right) = (5, 7)$ .

If question 2 was hard, revisit rearranging a formula before you continue.

Practise with help

1. Decide whether $y = 5x + 1$ and $y = 5x - 9$ are parallel. Give a reason.

Hint

Parallel lines point in the same direction. Which number in $y = mx + c$ sets the direction?

Next step

Read the coefficient of $x$ in each equation. Both are $5$ .

Full working

Both lines have gradient $5$ , and parallel lines have equal gradients. So the lines are parallel. They are not the same line, because their intercepts, $1$ and $-9$ , are different.

2. A line has gradient $\frac{3}{4}$ . Work out the gradient of a line perpendicular to it.

Hint

Turn the fraction upside down and change its sign.

Next step

The reciprocal of $\frac{3}{4}$ is $\frac{4}{3}$ . Change the sign.

Full working

m = -\frac{4}{3}.

Check: $\frac{3}{4} \times \left(-\frac{4}{3}\right) = -1$ , so the two lines are perpendicular.

3. Find the equation of the line through $(2, 7)$ perpendicular to $y = \frac{1}{2}x + 3$ .

Hint

The given gradient is $\frac{1}{2}$ . Work out the negative reciprocal first.

Next step

The perpendicular gradient is $-2$ . Now substitute the point $(2, 7)$ into $y - y_1 = m(x - x_1)$ .

Full working

The perpendicular gradient is $-2$ , because $\frac{1}{2} \times (-2) = -1$ .

y - 7 = -2(x - 2) = -2x + 4.

Add $7$ :

y = -2x + 11.

Check the point $(2, 7)$ : $-2 \times 2 + 11 = 7$ .

4. Find the equation of the line through $(4, -1)$ parallel to $3x - y = 5$ .

Hint

Rearrange $3x - y = 5$ to read off its gradient. A parallel line has the same gradient.

Next step

Rearranging gives $y = 3x - 5$ , so the gradient is $3$ . Substitute the point $(4, -1)$ into $y - y_1 = m(x - x_1)$ with $m = 3$ .

Full working

y - (-1) = 3(x - 4) = 3x - 12.

y + 1 = 3x - 12, \qquad y = 3x - 13.

The gradient is $3$ , the same as the given line, so the lines are parallel.

5. Work out the shortest distance from $(0, 0)$ to the line $3x + 4y = 20$ .

Hint

The shortest distance is measured along the perpendicular. You can find the perpendicular line and intersect it, or use the distance formula.

Next step

Write the line as $3x + 4y - 20 = 0$ . Then use $\dfrac{|a x_1 + b y_1 + c|}{\sqrt{a^2 + b^2}}$ with $a = 3$ , $b = 4$ , $c = -20$ and $(x_1, y_1) = (0, 0)$ .

Full working

\frac{|3 \times 0 + 4 \times 0 - 20|}{\sqrt{3^2 + 4^2}} = \frac{20}{\sqrt{25}} = \frac{20}{5} = 4.

The shortest distance is $4$ units. The foot of the perpendicular is the point $(2.4, 3.2)$ , which is $\sqrt{2.4^2 + 3.2^2} = 4$ from the origin.

6. Show that $y = 4x - 3$ and $y = -\frac{1}{4}x + 2$ are perpendicular.

Hint

The test for perpendicular lines is the product of the two gradients.

Next step

Multiply the two gradients, $4$ and $-\frac{1}{4}$ , and compare the answer with $-1$ .

Full working

4 \times \left(-\frac{1}{4}\right) = -1.

The product of the gradients is $-1$ , so the lines meet at a right angle.