Determining and using the equation of a straight line in the forms y equals mx plus c and y minus b equals m times x minus a, including the gradient, the y-intercept, and parallel lines.

What this dot point is asking

The SQA wants you to find and use the equation of a straight line. You should be able to read the gradient and y-intercept from $y = mx + c$, build an equation from a gradient and a point using the point-gradient form, and recognise that parallel lines share a gradient.

The equation y = mx + c

Every non-vertical straight line can be written as $y = mx + c$. The number $m$ in front of $x$ is the gradient and the number $c$ on its own is the y-intercept, the $y$ value where the line crosses the y-axis (at $x = 0$).

So $y = 2x + 5$ has gradient $2$ and crosses the y-axis at $(0, 5)$, while $y = -x + 3$ has gradient $-1$ and crosses at $(0, 3)$.

Finding the equation from a gradient and a point

When you know the gradient and one point on the line, the quickest route is the point-gradient form.

If you are given two points instead, first find the gradient with $m = \dfrac{y_2 - y_1}{x_2 - x_1}$, then use either point in the point-gradient form.

Parallel lines

Two lines are parallel when they have the same gradient and never meet. To find a line parallel to a given one, keep the gradient and use the new point.

Rearranging into the standard form

Equations are not always given in $y = mx + c$ form. When a line is written as, say, $2x + y = 8$ or $4x - 2y = 6$, rearrange it to $y = mx + c$ to read off the gradient and intercept. Make $y$ the subject by moving the $x$ term across and dividing if needed.

This is also how you find where two lines meet: rearrange both into $y = mx + c$ and set them equal, or solve them as simultaneous equations.

Examples in context

The straight line models any constant rate. A taxi that charges a $\pounds 3$ flag fall plus $\pounds 2$ per mile costs $C = 2m + 3$ pounds for $m$ miles: the gradient $2$ is the price per mile and the intercept $3$ is the fixed charge. Reading the graph of cost against distance, the steepness gives the rate and the starting height gives the standing charge. The same idea links a distance-time graph's gradient to speed.

Try this

Q1. State the gradient and y-intercept of $y = -2x + 9$. [2 marks]

• Cue. Gradient $-2$, y-intercept $9$.

Q2. Find the equation of the line through $(0, -1)$ with gradient $5$. [2 marks]

• Cue. $y = 5x - 1$.

Q3. Find the line through $(2, 7)$ parallel to $y = 4x - 3$. [2 marks]

• Cue. $y = 4x - 1$.