3.2 Parallel and perpendicular lines — Learn

Prerequisites: the gradient and the equation of a line from 3.1; rearranging a formula; solving two linear equations at once.

By the end of this section you can: decide from their gradients whether two lines are parallel or perpendicular; find the equation of a line parallel or perpendicular to a given line through a point; find the midpoint of a line segment; find the shortest distance from a point to a line.

1. Parallel lines

Two lines are parallel when they point in the same direction. Direction is the gradient, so parallel lines have the same gradient.

m_1 = m_2.

Example 1. Find the equation of the line through $(3, -2)$ parallel to $y = 4x - 1$ .

The gradient is $4$ , the same as the given line. Use $y - y_1 = m(x - x_1)$ with the point $(3, -2)$ :

y - (-2) = 4(x - 3).

Expand and simplify:

y + 2 = 4x - 12, \qquad y = 4x - 14.

The lines are parallel because both have gradient $4$ . They are different lines because the intercepts are different.

2. Perpendicular lines

Two lines are perpendicular when they meet at a right angle. Their gradients multiply to $-1$ :

m_1 m_2 = -1, \qquad \text{so} \qquad m_2 = -\frac{1}{m_1}.

The second gradient is the negative reciprocal of the first. Turn the fraction upside down, and change the sign.

Example 2. A line has gradient $\frac{2}{3}$ . Find the gradient of a line perpendicular to it.

The reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$ . Change the sign:

m_2 = -\frac{3}{2}.

Check: $\frac{2}{3} \times \left(-\frac{3}{2}\right) = -1$ . The product is $-1$ , so the lines are perpendicular.

A gradient of $0$ is the one case that needs care. A horizontal line has gradient $0$ , and a vertical line has no gradient at all. They are perpendicular, and the rule $m_1 m_2 = -1$ cannot express that. Handle vertical lines by writing their equation as $x = a$ .

Example 3. Find the equation of the line through $(3, -2)$ perpendicular to $2x + 3y = 6$ .

Rearrange the given line to see its gradient:

3y = -2x + 6, \qquad y = -\frac{2}{3}x + 2.

The gradient is $-\frac{2}{3}$ . The perpendicular gradient is the negative reciprocal:

m = -\frac{1}{-2/3} = \frac{3}{2}.

Now use the point $(3, -2)$ :

y - (-2) = \frac{3}{2}(x - 3).

Expand and simplify:

y + 2 = \frac{3}{2}x - \frac{9}{2}, \qquad y = \frac{3}{2}x - \frac{13}{2}.

Multiplying by $2$ gives the whole-number form $3x - 2y - 13 = 0$ .

3. The midpoint of a segment

The midpoint of the segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is the average of the two endpoints:

M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right).

Example 4. Find the midpoint of the segment from $(1, -3)$ to $(7, 5)$ .

M = \left(\frac{1 + 7}{2}, \frac{-3 + 5}{2}\right) = (4, 1).

The midpoint sits on the perpendicular bisector of the segment: the line through the midpoint at right angles to the segment. Every point on that line is the same distance from the two endpoints.

4. The shortest distance from a point to a line

The shortest path from a point to a line meets the line at a right angle. So the distance is measured along the perpendicular.

To find it, follow three steps. Write the perpendicular line through the point. Find where it meets the original line. Then work out the distance between the two points.

Example 5. Find the shortest distance from $(6, 8)$ to the line $x + y = 19$ .

The line has gradient $-1$ , so the perpendicular has gradient $1$ . The perpendicular through $(6, 8)$ is

y - 8 = 1(x - 6), \qquad y = x + 2.

Find the intersection of $y = x + 2$ with $x + y = 19$ . Substitute:

x + (x + 2) = 19, \qquad 2x = 17, \qquad x = \frac{17}{2}.

Then $y = \frac{17}{2} + 2 = \frac{21}{2}$ . The foot of the perpendicular is $\left(\frac{17}{2}, \frac{21}{2}\right)$ . The distance is

\sqrt{\left(\frac{17}{2} - 6\right)^2 + \left(\frac{21}{2} - 8\right)^2} = \sqrt{\left(\frac{5}{2}\right)^2 + \left(\frac{5}{2}\right)^2} = \frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2}.

The same answer comes from one formula. For the line $ax + by + c = 0$ and the point $(x_1, y_1)$ , the shortest distance is

\frac{\left|a x_1 + b y_1 + c\right|}{\sqrt{a^2 + b^2}}.

Here $x + y - 19 = 0$ , so $a = 1$ , $b = 1$ and $c = -19$ . Then

\frac{|6 + 8 - 19|}{\sqrt{1 + 1}} = \frac{5}{\sqrt{2}}.

Both routes agree. Use the three steps when a question asks for the foot of the perpendicular, and the formula when it asks only for the distance.

Watch out. Two traps come from the signs.

The negative reciprocal, not the reciprocal. The perpendicular to a line of gradient $\frac{2}{3}$ has gradient $-\frac{3}{2}$ , not $\frac{3}{2}$ . The sign is what makes the product $-1$ , so check the product before you carry on.

A line in the form $ax + by + c = 0$ hides its gradient. For $2x + 3y = 6$ the gradient is $-\frac{2}{3}$ , not $\frac{2}{3}$ and not $2$ . Re-check: rearrange to make $y$ the subject first, then read the coefficient of $x$ .

You should now be able to

decide whether two lines are parallel from their gradients;
find a perpendicular gradient as the negative reciprocal;
find the equation of a line parallel or perpendicular to a given line through a point;
handle the vertical and horizontal case separately;
find the midpoint of a segment;
find the shortest distance from a point to a line, and the foot of the perpendicular.