Straight-Line Graphs

Grade 4

The equation of a straight line

Every straight line (that is not vertical) has the equation y = mx + c, where m is the gradient (steepness) and c is the y-intercept (where the line crosses the y-axis).

To draw the line: plot the y-intercept at (0, c), then use the gradient to find a second point — go 1 across, m up — and draw through both.

Finding the gradient

gradient = rise / run = (y₂ − y₁) / (x₂ − x₁)

Pick any two points on the line. Divide the vertical change by the horizontal change. A positive gradient goes up left-to-right; a negative gradient goes down.

Worked example — find the equation from two points

EXAMPLE
Find the equation of the line through (1, 3) and (4, 9).
Gradient: (9 − 3) / (4 − 1) = 6/3 = 2. Use y = 2x + c. Substitute (1, 3): 3 = 2(1) + c, so c = 1. Equation: y = 2x + 1.

Parallel and perpendicular lines

Parallel lines have the same gradient. Perpendicular lines have gradients that multiply to −1 — so the perpendicular gradient is the negative reciprocal: if m = 3, the perpendicular gradient is −1/3.

To find the perpendicular gradient: flip the fraction and change the sign. Gradient 2 → perpendicular is −1/2.
Nova
Try this one

The diagram shows points \(R\) and \(S\) on a coordinate grid.

xy0123-1-2123-1-2R(-1, 2)S(2, -2)

(a) Find the midpoint of \(RS\). [1]

(b) Find the equation of the line through \(R\) and \(S\). [3]

Nova's hint:

This question covers midpoints and linear graph equations.

Ask yourself: 'how do I find the midpoint of two points, and once I have the gradient, how do I use a point to find \(c\)?' For the midpoint, average the \(x\)-coordinates and average the \(y\)-coordinates separately.

Try to Solve it with Nova?