Straight Lines – A Level Maths Year 1

1 Gradient: the Number That Describes a Line

Everything about a straight line follows from two facts: its gradient (how steep it is) and one point it passes through. This page is DfE content area C · Coordinate geometry, and it builds directly on the $y = mx + c$ work you met at GCSE.

Gradient between two points

$m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{\text{change in } y}{\text{change in } x}$

A positive gradient slopes up from left to right, a negative gradient slopes down. A horizontal line $y = k$ has gradient $0$; a vertical line $x = k$ has no gradient defined (you would be dividing by zero).

Gradient is a rate. In context the gradient always has units: "metres per second", "pounds per hour", "cm per gram". Saying what the gradient means is a common one-mark question.

2 Three Forms of the Equation

Form	When to use it	What you read off
$y = mx + c$	Sketching; comparing gradients	Gradient $m$, $y$-intercept $(0, c)$
$y - y_1 = m(x - x_1)$	You know a gradient and a point	Passes through $(x_1, y_1)$ with gradient $m$
$ax + by + c = 0$	Final answers "in the form…" with integer $a, b, c$	Rearrange to find $m = -\dfrac{a}{b}$

The point–gradient form — the one to use by default

$y - y_1 = m(x - x_1)$

Worked Example 1 — The line through two points

Find the equation of the line through $A(2, -1)$ and $B(6, 7)$, giving it in the form $ax + by + c = 0$.

①Gradient: $m = \dfrac{7 - (-1)}{6 - 2} = \dfrac{8}{4} = 2$

②Use the point $A$: $y - (-1) = 2(x - 2)$, i.e. $y + 1 = 2x - 4$

③So $y = 2x - 5$

④Rearrange: $2x - y - 5 = 0$

Check with the other point: $2(6) - 7 - 5 = 0$ ✓. It takes five seconds and catches nearly every sign slip.

Worked Example 2 — Reading a line given as $ax + by + c = 0$

The line $L$ has equation $3x + 4y - 12 = 0$. Find its gradient and the points where it crosses the axes.

①Make $y$ the subject: $4y = -3x + 12$, so $y = -\tfrac34 x + 3$

②Gradient $-\tfrac34$; $y$-intercept $(0, 3)$

③On the $x$-axis $y = 0$: $3x = 12$, so $(4, 0)$

For a quick sketch, the two intercepts are all you need: put $x = 0$, then $y = 0$.

"Integer coefficients" means integers. If the question asks for $ax + by + c = 0$ with $a$, $b$, $c$ integers, an answer like $y = \tfrac23 x - 4$ loses the final mark. Multiply through: $2x - 3y - 12 = 0$.

3 Parallel and Perpendicular Lines

Gradient conditions

Parallel: $m_1 = m_2$ · Perpendicular: $m_1 \times m_2 = -1$, i.e. $m_2 = -\dfrac{1}{m_1}$

Why $-1$? Rotating a line through $90°$ turns a "run" of $1$ and "rise" of $m$ into a run of $-m$ and a rise of $1$, so the new gradient is $\dfrac{1}{-m}$. "Flip the fraction and change the sign."

Worked Example 3 — A perpendicular line

Find the equation of the line through $(4, 1)$ that is perpendicular to $2x - 3y + 6 = 0$.

①Rearrange the given line: $3y = 2x + 6$, so $y = \tfrac23 x + 2$. Gradient $\tfrac23$.

②Perpendicular gradient: $-\tfrac32$

③$y - 1 = -\tfrac32(x - 4)$

④Multiply by $2$: $2y - 2 = -3x + 12$

⑤$3x + 2y - 14 = 0$

Worked Example 4 — Proving a right angle

$P(1, 2)$, $Q(5, 4)$ and $R(3, 8)$. Show that triangle $PQR$ is right-angled and find its area.

①$m_{PQ} = \dfrac{4-2}{5-1} = \dfrac12$ and $m_{QR} = \dfrac{8-4}{3-5} = -2$

②$\tfrac12 \times (-2) = -1$, so $PQ \perp QR$: the right angle is at $Q$.

③$PQ = \sqrt{4^2 + 2^2} = \sqrt{20}$ and $QR = \sqrt{(-2)^2 + 4^2} = \sqrt{20}$

④Area $= \tfrac12 \times PQ \times QR = \tfrac12 \times \sqrt{20} \times \sqrt{20} = 10$

The two perpendicular sides are the base and height, so no extra height calculation is needed. The triangle is also isosceles, since $PQ = QR$.

4 Midpoints, Distances and Intersections

Midpoint and distance of $(x_1, y_1)$ and $(x_2, y_2)$

$M = \left(\dfrac{x_1 + x_2}{2},\ \dfrac{y_1 + y_2}{2}\right)$ $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

The distance formula is just Pythagoras: the horizontal and vertical changes are the two shorter sides of a right-angled triangle.

Worked Example 5 — A perpendicular bisector

$A(-3, 2)$ and $B(5, -4)$. Find the length $AB$, and the equation of the perpendicular bisector of $AB$.

①$AB = \sqrt{8^2 + (-6)^2} = \sqrt{100} = 10$

②Midpoint $M = \left(\dfrac{-3+5}{2}, \dfrac{2-4}{2}\right) = (1, -1)$

③$m_{AB} = \dfrac{-6}{8} = -\tfrac34$, so the perpendicular gradient is $\tfrac43$.

④$y + 1 = \tfrac43(x - 1) \;\Rightarrow\; 3y + 3 = 4x - 4$

⑤$4x - 3y - 7 = 0$

Every point on the perpendicular bisector is the same distance from $A$ and $B$. That is why it is the key tool for finding the centre of a circle — see Circles.

Worked Example 6 — Where two lines meet

Find the point of intersection of $x + 2y = 7$ and $3x - y = 7$.

①From the first equation, $x = 7 - 2y$.

②Substitute: $3(7 - 2y) - y = 7 \;\Rightarrow\; 21 - 7y = 7 \;\Rightarrow\; y = 2$

③$x = 7 - 4 = 3$, so the lines meet at $(3, 2)$.

④Check in the second equation: $9 - 2 = 7$ ✓

Worked Example 7 — Collinear points

Show that $(-1, -4)$, $(2, 2)$ and $(5, 8)$ lie on a straight line.

①Gradient from the first to the second point: $\dfrac{2 - (-4)}{2 - (-1)} = \dfrac{6}{3} = 2$

②Gradient from the second to the third: $\dfrac{8 - 2}{5 - 2} = \dfrac{6}{3} = 2$

③Equal gradients and a common point $(2, 2)$, so the three points are collinear.

Equal gradients alone only prove the segments are parallel. You must mention the shared point to conclude they lie on one line.

5 Straight-Line Models

The subject content asks you to use straight-line models in a variety of contexts. A linear model says one quantity changes at a constant rate with respect to another.

Worked Example 8 — A spring

The length of a spring is $12$ cm when a $100$ g mass hangs from it, and $15$ cm with a $250$ g mass. Assume the length $L$ cm is a linear function of the mass $m$ g.

(a) Find $L$ in terms of $m$. (b) Interpret the gradient and intercept. (c) Predict $L$ for a $400$ g mass and comment.

①Gradient $= \dfrac{15 - 12}{250 - 100} = \dfrac{3}{150} = 0.02$

②$L - 12 = 0.02(m - 100)$, so $L = 0.02m + 10$.

③The spring stretches $0.02$ cm for every extra gram. The intercept $10$ cm is the natural (unstretched) length.

④$m = 400$: $L = 8 + 10 = 18$ cm.

⑤$400$ g is outside the data range ($100$–$250$ g), so this is extrapolation. A large mass might overstretch the spring beyond its elastic limit, where the linear model fails.

Evaluating a model. Good comments are specific: say which assumption might fail and why (a quantity cannot become negative; a rate is unlikely to stay constant; the prediction is far outside the data).

6 Quick Reference

Gradient

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

Point–gradient

$y - y_1 = m(x - x_1)$

General form

$ax + by + c = 0$, gradient $-\dfrac{a}{b}$

Parallel

Same gradient.

Perpendicular

$m_1 m_2 = -1$: flip and change sign.

Midpoint

Average the $x$'s, average the $y$'s.

Distance

$\sqrt{(\Delta x)^2 + (\Delta y)^2}$

Perpendicular bisector

Through the midpoint, perpendicular gradient.

Intersection

Solve the equations simultaneously.

Models

Gradient = rate of change, with units.

7 Practice Questions

Question 1

Find the equation of the line through $(-2, 5)$ and $(4, -7)$ in the form $y = mx + c$.

▶ Show solution

$m = \dfrac{-7 - 5}{4 - (-2)} = \dfrac{-12}{6} = -2$

$y - 5 = -2(x + 2) = -2x - 4$

$y = -2x + 1$. Check: $-2(4) + 1 = -7$ ✓

Question 2

Find the gradient and $y$-intercept of the line $5x - 2y + 8 = 0$.

▶ Show solution

$2y = 5x + 8$, so $y = \tfrac52 x + 4$.

Gradient $\tfrac52$, $y$-intercept $(0, 4)$.

Question 3

Find the equation of the line through $(3, -2)$ parallel to $y = 4x + 1$.

▶ Show solution

Parallel, so the gradient is $4$.

$y + 2 = 4(x - 3)$, so $y = 4x - 14$.

Question 4

Find the equation of the line through $(-1, 4)$ perpendicular to $2x + 5y = 3$. Give your answer in the form $ax + by + c = 0$ where $a$, $b$ and $c$ are integers.

▶ Show solution

$5y = -2x + 3$, so the given gradient is $-\tfrac25$.

Perpendicular gradient: $\tfrac52$.

$y - 4 = \tfrac52(x + 1) \;\Rightarrow\; 2y - 8 = 5x + 5$

$5x - 2y + 13 = 0$

Question 5

$A(2, -3)$ and $B(-4, 5)$. Find the midpoint of $AB$ and the length $AB$.

▶ Show solution

Midpoint $\left(\dfrac{2 - 4}{2}, \dfrac{-3 + 5}{2}\right) = (-1, 1)$

$AB = \sqrt{(-6)^2 + 8^2} = \sqrt{100} = 10$

Question 6

Find the perpendicular bisector of $P(1, 6)$ and $Q(7, 2)$, giving your answer in the form $ax + by + c = 0$.

▶ Show solution

Midpoint $(4, 4)$. $m_{PQ} = \dfrac{2 - 6}{7 - 1} = -\tfrac23$, so the perpendicular gradient is $\tfrac32$.

$y - 4 = \tfrac32(x - 4) \;\Rightarrow\; 2y - 8 = 3x - 12$

$3x - 2y - 4 = 0$

Question 7

Find the point where $y = 3x - 1$ meets $2x + 3y = 19$.

▶ Show solution

Substitute: $2x + 3(3x - 1) = 19 \;\Rightarrow\; 11x - 3 = 19 \;\Rightarrow\; x = 2$.

$y = 3(2) - 1 = 5$. The lines meet at $(2, 5)$.

Check: $2(2) + 3(5) = 19$ ✓

Question 8

The points $(1, k)$, $(3, 7)$ and $(6, 13)$ are collinear. Find $k$.

▶ Show solution

Gradient between $(3, 7)$ and $(6, 13)$: $\dfrac{6}{3} = 2$.

The gradient from $(1, k)$ to $(3, 7)$ must also be $2$: $\dfrac{7 - k}{2} = 2$.

$7 - k = 4$, so $k = 3$.

Question 9

$A(-2, 1)$, $B(4, 4)$ and $C(2, 8)$. Show that angle $ABC$ is a right angle and find the exact area of triangle $ABC$.

▶ Show solution

$m_{AB} = \dfrac{4 - 1}{4 + 2} = \dfrac12$, $m_{BC} = \dfrac{8 - 4}{2 - 4} = -2$.

$\tfrac12 \times (-2) = -1$, so $AB \perp BC$ and angle $ABC = 90°$.

$AB = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt5$, $BC = \sqrt{(-2)^2 + 4^2} = \sqrt{20} = 2\sqrt5$.

Area $= \tfrac12 \times 3\sqrt5 \times 2\sqrt5 = \tfrac12 \times 30 = 15$.

Question 10

A candle burns at a steady rate. After $2$ hours it is $24$ cm tall; after $5$ hours it is $15$ cm tall. Let $h$ cm be the height after $t$ hours.

(a) Find a linear model for $h$ in terms of $t$. (b) Interpret the gradient and the intercept in context. (c) Find when the candle burns out. (d) State a suitable domain for the model, and give one reason why the model may not be accurate near the end.

▶ Show solution

(a) Gradient $= \dfrac{15 - 24}{5 - 2} = -3$.

$h - 24 = -3(t - 2)$, so $h = 30 - 3t$.

(b) The gradient $-3$ means the candle gets $3$ cm shorter every hour. The intercept $30$ means it was $30$ cm tall when first lit.

(c) $h = 0$: $30 - 3t = 0$, so $t = 10$ hours.

(d) $0 \leqslant t \leqslant 10$ — outside this range the model gives a height above the original or a negative height, which are impossible.

Near the end the wick may sit in a pool of wax or the candle may be in a holder, so it may not burn at the same constant rate (or may go out before $h$ reaches $0$).

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