Gradient and the equation of a line · Coordinate geometry

The gradient of a line measures how steep it is: how far it goes up for every 1 across. Between two points $A(x_1, y_1)$ and $B(x_2, y_2)$ on it:

m = \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}

A positive gradient goes up from left to right; a negative one goes down; a horizontal line has gradient 0; a vertical line has no gradient. A line at an angle $\theta$ to the positive $x$ -axis has gradient $\tan\theta$ .

Every non-vertical straight line has an equation $y = mx + c$ , where $m$ is the gradient and $c$ is the $y$ -intercept, where the line crosses the $y$ -axis. A point lies on the line exactly when its coordinates fit the equation.

Gradientm = rise ÷ run = (y₂ − y₁) ÷ (x₂ − x₁)

y = mx + cm is the rise for 1 across; c is where it crosses the y-axis

Try it

Two pointsDrag the gold points

4rise, y₂ − y₁6run, x₂ − x₁2/3gradient, rise ÷ run

Gradient = rise ÷ run = 4 ÷ 6 = 2/3. Positive: the line goes up from left to right. Take both differences in the same order, B minus A.

In “Gradient”, move $B$ below $A$ and watch the sign change, then line them up vertically. In “Equation”, the equation updates as you drag, and the green point shows $c$ .

To read the gradient from an equation like $3x + 4y = 12$ , make $y$ the subject first.

Finding the equation of a line

With the gradient $m$ and one point $(x_1, y_1)$ on the line:

y - y_1 = m(x - x_1)

With two points, find the gradient first, then use either point. Tidy the answer into the form the question asks for, such as $y = mx + c$ or $ax + by + c = 0$ .

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q2 (a)

Find the equation of the line which passes through the points $A(-2, 7)$ and $B(2, -3)$ .

The gradient
$m = \frac{-3 - 7}{2 - (-2)} = \frac{-10}{4} = -\frac52$ .
Think first. Rise over run from A to B. Watch the signs.
Use a point
$y - 7 = -\frac52(x + 2)$ .
Think first. Which formula gives the line through A with this gradient?
Clear the fraction
$2y - 14 = -5x - 10$ , so $2y + 5x - 4 = 0$ . Check with $B(2, -3)$ : $-6 + 10 - 4 = 0$ ✓.
Think first. Multiply every term by 2.

Parallel and perpendicular lines

Parallel lines have equal gradients.
Perpendicular lines have gradients that multiply to $-1$ : $m_1 m_2 = -1$ . So the perpendicular gradient is $-\frac{1}{m}$ : flip the fraction and change the sign.

ParallelSame gradient m

Perpendicularm × (−1/m) = −1

Switch the board to “Parallel & perpendicular” and drag $C$ : both lines through it keep their gradients wherever $C$ goes.

Worked example · WAEC 2023

WAEC 2023 · Paper 2 · Q7 (b)

Given the points $P(3, 5)$ and $Q(-5, 7)$ on the Cartesian plane such that $R$ is the midpoint of $\overline{PQ}$ , find the equation of the line that passes through $R$ and is perpendicular to $\overline{PQ}$ .

The midpoint R
$R = \left(\frac{3 + (-5)}{2}, \frac{5 + 7}{2}\right) = (-1, 6)$ .
Think first. Average the coordinates of P and Q.
The gradient of PQ
$\frac{7 - 5}{-5 - 3} = \frac{2}{-8} = -\frac14$ .
Think first. Rise over run.
The perpendicular gradient
$m = 4$ , since $4 \times \left(-\frac14\right) = -1$ .
Think first. Flip and change the sign.
The line through R
$y - 6 = 4(x + 1)$ , so $y = 4x + 10$ .
Think first. Use y − y₁ = m(x − x₁).

The angle between two lines

A line at an angle $\theta$ to the positive $x$ -axis rises $\tan\theta$ for every 1 across, so its gradient is

m = \tan\theta

Gradient and anglem = rise ÷ run = tan θ

To find the angle, find the gradient and use $\tan^{-1}$ (see the tangent ratio). For example, for the line $y = \sqrt3\,x + 1$ :

The gradient is ${m = \sqrt3}$ .
So ${\tan\theta = \sqrt3}$ .
So ${\theta = 60^\circ}$ .

For two lines with gradients $m_1$ and $m_2$ , the acute angle $\theta$ between them satisfies

\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|

The angle between two linestan θ = |(m₁ − m₂) ÷ (1 + m₁m₂)|

The bars (take the value without its sign) make the answer the acute angle. Two special cases fit the rules above:

Parallel lines have $m_1 = m_2$ : the top is 0, so $\theta = 0^\circ$ .
Perpendicular lines have $m_1 m_2 = -1$ : the bottom is 0, so there is no $\tan\theta$ and $\theta = 90^\circ$ .

Worked example · NECO 2022

NECO 2022 · Paper 1 · Q38

Find the acute angle between the lines $x + 4y = 12$ and $2y - x = -6$ .

The first gradient
- Take $x$ from both sides: ${4y = -x + 12}$ .
- Divide by 4: ${y = -\frac14x + 3}$ , so ${m_1 = -\frac14}$ .
Think first. Make y the subject of x + 4y = 12.
The second gradient
- Add $x$ to both sides: ${2y = x - 6}$ .
- Divide by 2: ${y = \frac12x - 3}$ , so ${m_2 = \frac12}$ .
Think first. Make y the subject of 2y − x = −6.
The top and the bottom
- Top: ${-\frac14 - \frac12 = -\frac34}$ .
- Bottom: ${1 + \left(-\frac14\right)\left(\frac12\right) = 1 - \frac18 = \frac78}$ .
Think first. Work out m₁ − m₂ and 1 + m₁m₂ separately.
The angle
- Divide: ${\tan\theta = \frac34 \div \frac78 = \frac34 \times \frac87 = \frac67}$ .
- Use $\tan^{-1}$ : ${\theta \approx 40.6^\circ}$ , option B.
Think first. Divide, drop the sign, then use tan⁻¹.

More: the angle between two lines

NECO 2022 · Paper 2 · Q3Calculate the value of $p$ if the distance between the points $(p, 5)$ and $(5, 9)$ is 5. (Enter both values.)

Your turn

WAEC 2020 · Paper 2 · Q11 (b)

(b)
Find the equation of the line which has the same gradient as $8y + 4x = 24$ and passes through the point $(-8, 12)$ .

Worked solution (try it first)

(b)

$8y + 4x = 24$ gives $y = -\frac12x + 3$ , so the gradient is $-\frac12$ .
Through $(-8, 12)$ : $y - 12 = -\frac12(x + 8)$ , so $y = -\frac12x - 4 + 12 = -\frac12x + 8$ .

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