Plot and recognise straight-line graphs, find the gradient and y-intercept, use $y = mx + c$, find the equation of a line through given points, and use parallel and perpendicular gradient conditions.

What this dot point is asking

Straight-line graphs link algebra to coordinate geometry. In the CCEA Algebra strand you must plot and recognise straight lines, find the gradient and $y$-intercept, use the equation $y = mx + c$, find the equation of a line through given points, and at Higher tier use the gradient conditions for parallel and perpendicular lines. These skills connect to simultaneous equations (where lines cross), to sequences (the common difference is a gradient) and to real-life rate graphs.

Plotting and recognising lines

A straight-line graph can be drawn from its equation by making a small table of values, working out $y$ for a few $x$ values, plotting the points and joining them. Lines of the form $x = a$ are vertical, $y = b$ are horizontal, and $y = mx$ pass through the origin. Recognising the shape and key features from the equation saves time.

Gradient and intercept

The gradient measures steepness: how much $y$ changes for each unit increase in $x$. A positive gradient rises to the right, a negative gradient falls.

So through $(1, 2)$ and $(4, 11)$ the gradient is $\tfrac{11 - 2}{4 - 1} = \tfrac{9}{3} = 3$.

The equation y = mx + c

Once you have the gradient $m$ and the intercept $c$, the equation is simply $y = mx + c$. If you know the gradient and a point but not the intercept, substitute the point's coordinates to find $c$.

Parallel and perpendicular lines (Higher)

Two lines are parallel when they have the same gradient, so any line parallel to $y = 4x - 1$ has gradient $4$. Two lines are perpendicular (meeting at a right angle) when the product of their gradients is $-1$; equivalently, one gradient is the negative reciprocal of the other. If a line has gradient $\tfrac{2}{3}$, a perpendicular line has gradient $-\tfrac{3}{2}$.

To find the equation of a parallel or perpendicular line through a given point, work out the required gradient, then substitute the point to find $c$, exactly as above.

A worked perpendicular example makes the steps concrete. Find the line perpendicular to $y = \tfrac{1}{2}x + 3$ through the point $(2, 7)$. The given gradient is $\tfrac{1}{2}$, so the perpendicular gradient is its negative reciprocal, $-2$. Substitute the point into $y = -2x + c$: $7 = -2(2) + c$, so $7 = -4 + c$ and $c = 11$. The equation is $y = -2x + 11$. The same two-step routine, find the gradient then find the intercept, handles every "find the equation of the line" question, whether the line is parallel, perpendicular, or simply through two points. When a question gives an equation in the form $ax + by = c$, rearrange it into $y = mx + c$ first so the gradient can be read off the coefficient of $x$.

Why this matters

Straight-line graphs are the bridge between algebra and geometry and the basis for the rate-of-change ideas that run through measures and applied maths. The intersection of two lines is the graphical meaning of simultaneous equations, and the gradient and intercept appear in real contexts such as fixed-plus-variable costs and conversion graphs. CCEA rewards reading the gradient and intercept accurately and forming equations cleanly.