Plotting straight lines, finding gradient and intercept from the equation, writing equations of lines, and the conditions for parallel and perpendicular lines.

What this dot point is asking

AQA wants you to plot a straight line, read its gradient and $y$-intercept from the equation $y = mx + c$, find the equation of a line through given points, and at Higher tier apply the parallel and perpendicular gradient conditions. Straight lines connect algebra to geometry and feed directly into rates of change and simultaneous equations, so the skills here are reused across the paper.

Gradient and intercept from the equation

In the form $y = mx + c$, the number multiplying $x$ is the gradient and the constant is the $y$-intercept. For $y = 3x - 2$ the gradient is $3$ (the line rises $3$ for every $1$ across) and it crosses the $y$-axis at $(0, -2)$. A positive gradient slopes upward to the right; a negative gradient slopes downward. If an equation is given as, say, $2y = 4x + 6$, divide through by $2$ first to get $y = 2x + 3$ before reading off $m$ and $c$.

Finding the gradient between two points

For the points $(2, 1)$ and $(6, 9)$, the gradient is $\dfrac{9 - 1}{6 - 2} = \dfrac{8}{4} = 2$. Keep the coordinates in the same order top and bottom; reversing one but not the other flips the sign.

Finding the equation of a line

Parallel and perpendicular lines at Higher tier

To find a line parallel to $y = 4x + 1$ through $(2, 3)$, keep the gradient $4$ and find $c$: $3 = 4(2) + c$ gives $c = -5$, so $y = 4x - 5$. To find a perpendicular line through the same point, use the negative reciprocal gradient $-\tfrac{1}{4}$: $3 = -\tfrac{1}{4}(2) + c$ gives $c = 3.5$, so $y = -\tfrac{1}{4}x + 3.5$.

The midpoint of a segment from $(x_1, y_1)$ to $(x_2, y_2)$ is $\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$, which often appears in coordinate geometry questions alongside the gradient work.

Plotting a line from its equation

To plot a line such as $y = 2x - 1$, build a small table of values: pick a few $x$ values, substitute to find each $y$, then plot and join the points. For $x = 0, 1, 2$ you get $y = -1, 1, 3$, three points that lie on a straight line. For an equation given implicitly, such as $2x + 3y = 6$, the quickest plot uses the intercepts: set $x = 0$ to find $y = 2$, and set $y = 0$ to find $x = 3$, then join $(0, 2)$ and $(3, 0)$. Two points fix a line, but plotting a third is a sensible check.

Finding where a line meets the axes

The intercepts carry useful information in applied problems. The $y$-intercept is the starting value (the value when $x = 0$), such as a fixed charge before any usage. The $x$-intercept is where $y$ reaches zero, such as the time a quantity runs out. In a context like $C = 5 + 2t$ for the cost of a taxi after $t$ minutes, the $y$-intercept $5$ is the flat fare and the gradient $2$ is the cost per minute, showing how the algebra of a line maps directly onto a real situation.