Straight line graphs: plotting lines, finding the gradient and y-intercept, using the equation y = mx + c, finding the equation of a line through two points, and parallel and perpendicular lines (Higher tier).

What this dot point is asking

Straight line graphs link algebra and geometry. Edexcel expects you to plot lines, find a line's gradient and y-intercept, use the equation $y = mx + c$, find the equation of a line through two points, and at Higher tier handle parallel and perpendicular lines. The equation of a line is a rule connecting every $x$ on the line to its $y$, and the gradient measures its steepness.

Gradient and intercept

The gradient measures how steep a line is and in which direction it slopes.

The y-intercept $c$ is the y-coordinate where the line crosses the y-axis, that is when $x = 0$. In $y = 4x - 7$, the gradient is $4$ and the y-intercept is $-7$. A line of the form $y = c$ (such as $y = 3$) is horizontal with gradient $0$; a line $x = a$ is vertical with an undefined gradient.

Using y = mx + c

The equation $y = mx + c$ is the key tool. To find a line's equation you need the gradient and one point.

Reading values from a graph works the same way: pick two clear points, find the gradient, and read the intercept off the y-axis.

Parallel lines (Higher and Foundation)

Parallel lines never meet, which happens exactly when they have the same gradient. So $y = 2x + 1$ and $y = 2x - 5$ are parallel, because both have gradient $2$; only the intercept differs. To find a line parallel to a given one through a point, keep the gradient and find the new $c$.

Perpendicular lines (Higher)

Perpendicular lines cross at a right angle. Their gradients are negative reciprocals.

To find the perpendicular to $y = 4x + 1$ through $(8, 3)$, the new gradient is $-\tfrac{1}{4}$; substitute the point into $y = -\tfrac{1}{4}x + c$ to get $c = 5$, so $y = -\tfrac{1}{4}x + 5$.

Rearranging into y = mx + c

Lines are not always given in the form $y = mx + c$. An equation like $2x + 3y = 12$ hides its gradient until you rearrange it. Make $y$ the subject: $3y = -2x + 12$, so $y = -\tfrac{2}{3}x + 4$. Now the gradient is $-\tfrac{2}{3}$ and the y-intercept is $4$. This matters when a question asks whether two lines are parallel or perpendicular but gives them in different forms, because you must compare gradients, and the gradient is only visible once each equation is in $y = mx + c$ form.

Try this

Q1. Find the gradient of the line through $(2, 1)$ and $(6, 9)$. [2 marks]

• Cue. $m = \dfrac{9 - 1}{6 - 2} = \dfrac{8}{4} = 2$.

Q2. Write $4x + 2y = 10$ in the form $y = mx + c$ and state the gradient. [2 marks]

• Cue. $2y = -4x + 10$, so $y = -2x + 5$; the gradient is $-2$.