Plotting Straight Lines Revision Notes | GCSE AQA Higher Further Maths

Worked example: Drawing $y = 2x + 1$ Using a Table of Values

Draw the straight line $y = 2x + 1$ for values of $x$ from $-2$ to $3$.

1. $x = -2,\; -1,\; 0,\; 1,\; 2,\; 3$ — Spread your $x$-values across the given range. Including $x = 0$ makes the $y$-intercept visible immediately. [M1 — setting up the table]
2. $y = 2 \times (-2) + 1 = -4 + 1 = -3$ Point: $(-2,\; -3)$ — Multiply first, then add $c$. Watch the sign: $2 \times (-2) = -4$, not $+4$.
3. $y = 2 \times 0 + 1 = 0 + 1 = 1$ Point: $(0,\; 1)$ — This confirms the $y$-intercept is $(0, 1)$, matching $c = 1$ in the equation.
4. $y = 2 \times 3 + 1 = 6 + 1 = 7$ Point: $(3,\; 7)$ — A point at the far end of the range helps you draw the line across the full grid. [M1 — correct substitution]
5. $x$: $-2,\; -1,\; 0,\; 1,\; 2,\; 3$ $y$: $-3,\; -1,\; 1,\; 3,\; 5,\; 7$ — Each $y$-value increases by 2 as $x$ increases by 1 — this matches the gradient $m = 2$. Use this as a self-check.
6. Plot $(-2, -3)$, $(-1, -1)$, $(0, 1)$, $(1, 3)$, $(2, 5)$, $(3, 7)$. Join with a single straight line using a ruler, extending to the edges of the grid. — All six points must lie on the same straight line. If one does not, recheck that substitution. [A1 — correct line drawn]

Answer: A straight line passing through $(-2, -3)$, $(0, 1)$, and $(3, 7)$, with gradient $2$ and $y$-intercept $(0, 1)$.

Worked example: Drawing $y = -\frac{1}{2}x + 3$ Using Gradient and $y$-Intercept

Draw the straight line with equation $y = -\frac{1}{2}x + 3$.

1. $m = -\frac{1}{2}$, $\quad c = 3$ — Read directly from the equation. The gradient is negative, so the line slopes downward from left to right.
2. Mark the point $(0, 3)$ on the $y$-axis. — This is always your starting point when using this method. [M1 — correct $y$-intercept]
3. $m = -\frac{1}{2}$ means: move $2$ right, move $1$ down. From $(0, 3)$: go to $(2, 2)$. From $(2, 2)$: go to $(4, 1)$. — A gradient of $-\frac{1}{2}$ means for every 2 units across, the line drops 1 unit. Writing the gradient as a fraction makes the rise and run explicit.
4. Check $(4, 1)$: $y = -\frac{1}{2} \times 4 + 3 = -2 + 3 = 1$ ✓ — Substituting back confirms the point lies on the line. This is your built-in error check. [A1 — correct line]
5. Join $(0, 3)$, $(2, 2)$, and $(4, 1)$ with a ruler. Extend the line to the edges of the grid. — Always use a ruler. A freehand line through three correct points will still lose marks if it is visibly curved or does not pass through all three.

Answer: A straight line with negative gradient passing through $(0, 3)$, $(2, 2)$, and $(4, 1)$.

Worked example: Drawing a Line Through Two Points and Finding Its Equation

Plot the straight line passing through $A(1, 5)$ and $B(4, -1)$. Find the equation of the line.

1. Mark $A(1, 5)$ and $B(4, -1)$ on the coordinate grid. — Accurate plotting is essential. Double-check each coordinate before marking — swap $x$ and $y$ and you will draw the wrong line.
2. Join $A$ and $B$ with a ruler and extend the line to the edges of the grid. — Extending the line shows the full picture and makes the $y$-intercept visible for the next step. [M1 — correct line drawn]
3. $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 5}{4 - 1} = \frac{-6}{3} = -2$ — Label which point is $(x_1, y_1)$ and which is $(x_2, y_2)$ before substituting. The order does not matter as long as you are consistent in numerator and denominator. [M1 — correct gradient]
4. Use point $A(1, 5)$ and $m = -2$: $5 = -2 \times 1 + c$ $5 = -2 + c$ $c = 7$ — Either point works here. Substituting $A$ gives $c = 7$. You can verify by substituting $B$: $-1 = -2 \times 4 + 7 = -8 + 7 = -1$ ✓
5. $y = -2x + 7$ — State the full equation clearly. [A1 — correct equation]

Answer: $y = -2x + 7$