Year 1 Pure · Chapter 4

Coordinate Geometry of Straight Lines

Specification: Edexcel Pure 3 — Coordinate geometry · AQA section C

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In this chapter

1. Gradient and the equation of a line
2. Midpoints and distance
3. Parallel and perpendicular lines
4. Modelling with straight lines
Practice

Key facts — this chapter

Gradient between two points: $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ — rise over run.
Equation of a line: $y = mx + c$ (gradient $m$ , $y$ -intercept $c$ ), or $y - y_1 = m(x - x_1)$ when you know a point on it.
Parallel lines have equal gradients; perpendicular lines have gradients multiplying to $-1$ : $m_2 = -\frac{1}{m_1}$ .
Midpoint of $A(x_1, y_1)$ and $B(x_2, y_2)$ : $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ — average the coordinates.
Distance: $AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ — Pythagoras in disguise.

Straight lines are the workhorses of coordinate geometry. Master the handful of tools in this chapter — gradient, equation, midpoint, distance, perpendicularity — and you’ve already done most of the work for circles (Chapter 5) and tangents and normals in differentiation later on.

1. Gradient and the equation of a line

The gradient measures steepness: how much the line climbs for every unit it moves right.

Key result

Between any two points $(x_1, y_1)$ and $(x_2, y_2)$ on the line:

m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}}

A negative gradient falls from left to right; a horizontal line has $m = 0$ ; a vertical line has no gradient (its equation is $x = a$ ).

A straight line through points A at one one and B at seven five, with a dashed triangle showing a run of six and a rise of four, giving gradient two thirds — Any two points on the line give the same rise ÷ run — that's what makes it straight.

Drag A and B and watch the gradient triangle do the work — rise over run, the equation, the midpoint and the exact length all come from the same two points.

Drag A and B · the dashed triangle is rise over run · the hollow dot is the midpoint

Two ways to write a line

$y = mx + c$ — perfect when you know the gradient and the $y$ -intercept.
$y - y_1 = m(x - x_1)$ — perfect when you know the gradient and any point $(x_1, y_1)$ ; no need to find $c$ first.

Exam answers sometimes ask for the form $ax + by + c = 0$ with integer coefficients — multiply through to clear fractions and gather everything on one side.

Worked example 4.1 · A line through two points

Find the equation of the line through $(-1, 3)$ and $(2, -3)$ .

Gradient first:
$m = \frac{-3 - 3}{2 - (-1)} = \frac{-6}{3} = -2$
Use $y - y_1 = m(x - x_1)$ with either point — take $(-1, 3)$ :
$y - 3 = -2(x + 1)$
Tidy:
$y = -2x + 1$

(Check with the other point: $x = 2$ gives $y = -3$ . ✓)

Common mistake

Keep the subtraction order consistent: $\frac{y_2 - y_1}{x_2 - x_1}$ , both differences the same way round. Swapping one but not the other flips the sign of the gradient — the single most common slip with lines.

2. Midpoints and distance

Two more tools, both straight from a picture.

Key result

For $A(x_1, y_1)$ and $B(x_2, y_2)$ :

\begin{aligned} \text{midpoint} &= \left(\frac{x_1 + x_2}{2},\; \frac{y_1 + y_2}{2}\right) \\[0.4em] AB &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \end{aligned}

The midpoint just averages the coordinates. The distance is Pythagoras: the horizontal and vertical gaps are the two legs, and $AB$ is the hypotenuse.

The line segment from A at one two to B at five five with a dashed right-angled triangle of legs four and three, giving length five — Distance is Pythagoras: legs 4 and 3, hypotenuse 5.

Worked example 4.2 · Midpoint and length of a segment

For $A(-3, 2)$ and $B(5, 8)$ , find the midpoint of $AB$ and the exact length $AB$ .

Average the coordinates:
$\text{midpoint} = \left(\frac{-3 + 5}{2},\; \frac{2 + 8}{2}\right) = (1, 5)$
Pythagoras on the gaps ( $8$ across, $6$ up):
$AB = \sqrt{8^2 + 6^2} = \sqrt{100} = 10$

3. Parallel and perpendicular lines

Key result

Parallel lines never meet because they climb at the same rate: $m_1 = m_2$ .
Perpendicular lines meet at right angles, and their gradients multiply to $-1$ :
$m_1 m_2 = -1 \qquad\Longleftrightarrow\qquad m_2 = -\frac{1}{m_1}$
— flip the fraction, flip the sign. The perpendicular to gradient $\tfrac{2}{3}$ has gradient $-\tfrac{3}{2}$ .

Why it works

Rotating a line a quarter turn swaps its rise and run — what used to be “along” is now “up” — and sends the line uphill if it was downhill. So a gradient $\frac{a}{b}$ becomes $-\frac{b}{a}$ , and the product is $\frac{a}{b} \times \left(-\frac{b}{a}\right) = -1$ .

Two lines crossing at a right angle, one with gradient two and the other with gradient negative one half, with a small square marking the right angle — Gradients 2 and −½: flip the fraction, flip the sign, product −1.

Worked example 4.3 · A parallel line through a point

Find the equation of the line through $(2, -1)$ that is parallel to the line $3x + y = 7$ .

Rearrange the given line to read off its gradient:
$y = -3x + 7 \;\;\Longrightarrow\;\; m = -3$
Parallel means the same gradient, so build the line through $(2, -1)$ with $m = -3$ :
$y - (-1) = -3(x - 2) \;\;\Longrightarrow\;\; y = -3x + 5$

Worked example 4.4 · A perpendicular line through a point

Find the equation of the line through $(6, 1)$ that is perpendicular to $y = 3x + 2$ . Give your answer in the form $ax + by + c = 0$ , where $a$ , $b$ and $c$ are integers.

The given gradient is $3$ , so the perpendicular gradient is $-\tfrac{1}{3}$ — flip and negate.
Line through $(6, 1)$ :
$y - 1 = -\tfrac{1}{3}(x - 6)$
Multiply through by $3$ to clear the fraction, and gather:
$3y - 3 = -(x - 6) \;\;\Longrightarrow\;\; x + 3y - 9 = 0$

Common mistake

Perpendicular needs both moves — flip the fraction and flip the sign. The perpendicular to $m = 3$ is $-\frac{1}{3}$ , not $\frac{1}{3}$ (sign forgotten) and not $-3$ (flip forgotten).

4. Modelling with straight lines

A linear model is just $y = mx + c$ wearing real-world clothes — and exam questions love asking what $m$ and $c$ mean in context.

Key result

In a linear model $y = mx + c$ :

$m$ is the rate of change — how much $y$ changes for each unit of $x$ (cost per mile, depth per minute, …);
$c$ is the starting value — what $y$ is when $x = 0$ (the fixed charge, the initial depth, …).

Worked example 4.5 · Building and reading a linear model

A taxi journey costs £ $6.80$ for $2$ miles and £ $12.20$ for $5$ miles. Assuming cost $C$ (£) is linear in distance $d$ (miles), find the model and interpret its constants.

The rate per mile is the gradient:
$m = \frac{12.20 - 6.80}{5 - 2} = \frac{5.40}{3} = 1.80$
Use a known point to find the constant: $6.80 = 1.80(2) + c$ gives $c = 3.20$ .
$C = 1.8d + 3.2$
Interpret: each extra mile costs £1.80 ( $m$ ), and the fixed charge just for getting in the taxi is £3.20 ( $c$ ).

Practice

Twelve questions in three tiers. Try each one properly before opening the solution — the solutions show full method, not just answers.

Warm-up

1 [3 marks]

Find the equation of the line through $(1, 2)$ and $(4, 11)$ , giving your answer in the form $y = mx + c$ .

Show solution

Gradient:
$m = \frac{11 - 2}{4 - 1} = \frac{9}{3} = 3$
Line through $(1, 2)$ with gradient $3$ :
$y - 2 = 3(x - 1) \;\;\Longrightarrow\;\; y = 3x - 1$

(Check: $x = 4$ gives $y = 11$ . ✓)

2 [3 marks]

For the points $A(-2, 3)$ and $B(4, 11)$ , find:

the midpoint of $AB$ ,
the length $AB$ .

Show solution

(a) Average the coordinates:

\left(\frac{-2 + 4}{2},\; \frac{3 + 11}{2}\right) = (1, 7)

(b) Pythagoras on the gaps ( $6$ across, $8$ up):

AB = \sqrt{6^2 + 8^2} = \sqrt{100} = 10

3 [2 marks]

Write down the gradient and $y$ -intercept of the line $3x + 4y - 8 = 0$ .

Show solution

Rearrange for $y$ :

4y = -3x + 8 \;\;\Longrightarrow\;\; y = -\tfrac{3}{4}x + 2

Gradient $-\tfrac{3}{4}$ , $y$ -intercept $(0, 2)$ .

4 [2 marks]

Show that the lines $y = 2x - 1$ and $2y - 4x + 5 = 0$ are parallel.
Write down the gradient of any line perpendicular to them.

Show solution

(a) Rearranging the second: $2y = 4x - 5$ , so $y = 2x - \tfrac{5}{2}$ . Both have gradient $2$ — parallel. (Different intercepts, so they’re distinct lines, not the same one.)

(b) Flip and negate: $-\tfrac{1}{2}$ .

Standard

5 [3 marks]

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

Show solution

The given line rearranges to $y = -\tfrac{2}{3}x + 2$ , so $m = -\tfrac{2}{3}$ .
Same gradient through $(3, -1)$ :
$y + 1 = -\tfrac{2}{3}(x - 3) \;\;\Longrightarrow\;\; y = -\tfrac{2}{3}x + 1$

6 [3 marks]

Find the equation of the line through $(4, 5)$ that is perpendicular to $y = 2x - 3$ .

Show solution

Perpendicular gradient: flip and negate $2$ to get $-\tfrac{1}{2}$ .
Through $(4, 5)$ :
$y - 5 = -\tfrac{1}{2}(x - 4) \;\;\Longrightarrow\;\; y = -\tfrac{1}{2}x + 7$

7 [4 marks]

The points $A(2, 3)$ , $B(6, 5)$ and $C(4, 9)$ form a triangle. Show that the triangle is right-angled and isosceles.

Show solution

Gradients of the sides:
$m_{AB} = \frac{5-3}{6-2} = \frac{1}{2} \qquad m_{BC} = \frac{9-5}{4-6} = -2$
$m_{AB} \times m_{BC} = \tfrac{1}{2} \times (-2) = -1$ , so $AB \perp BC$ — the triangle is right-angled at $B$ . ✓
Lengths of those two sides:
$AB = \sqrt{4^2 + 2^2} = \sqrt{20} \qquad BC = \sqrt{2^2 + 4^2} = \sqrt{20}$

Two equal sides, so the triangle is also isosceles. ✓

8 [4 marks]

Find the equation of the perpendicular bisector of the points $A(1, 4)$ and $B(7, 2)$ .

Show solution

The bisector passes through the midpoint:
$\left(\frac{1+7}{2}, \frac{4+2}{2}\right) = (4, 3)$
Gradient of $AB$ : $\dfrac{2-4}{7-1} = -\dfrac{1}{3}$ , so the perpendicular gradient is $3$ .
Through $(4, 3)$ with gradient $3$ :
$y - 3 = 3(x - 4) \;\;\Longrightarrow\;\; y = 3x - 9$

(This is the Chapter 5 workhorse: a circle’s centre always sits on the perpendicular bisector of any chord.)

9 [3 marks]

The line through $P(2, 5)$ and $Q(6, k)$ is perpendicular to the line $y = -2x + 1$ . Find the value of $k$ .

Show solution

Perpendicular to gradient $-2$ means gradient $\tfrac{1}{2}$ .
Set the gradient of $PQ$ equal to it:
$\frac{k - 5}{6 - 2} = \frac{1}{2}$
Solve: $k - 5 = 2$ , so $k = 7$ .

Exam style

10 [6 marks]

The line $l_1$ has equation $y = 2x - 3$ . The line $l_2$ is perpendicular to $l_1$ and passes through the point $(8, 3)$ .

Find the equation of $l_2$ . [3]
Find the coordinates of the point where $l_1$ and $l_2$ intersect. [3]

Show solution

(a)

Perpendicular gradient to $2$ : $-\tfrac{1}{2}$ .
Through $(8, 3)$ :
$y - 3 = -\tfrac{1}{2}(x - 8) \;\;\Longrightarrow\;\; y = -\tfrac{1}{2}x + 7$

(b)

Set the two equations equal:
$2x - 3 = -\tfrac{1}{2}x + 7$
Multiply by 2 and solve: $4x - 6 = -x + 14$ , so $5x = 20$ and $x = 4$ .
Substitute back: $y = 2(4) - 3 = 5$ . The lines meet at $(4, 5)$ .

11 [5 marks]

The points $A(1, 2)$ , $B(5, 4)$ , $C(7, 8)$ and $D(3, 6)$ are the vertices of a quadrilateral. Show that $ABCD$ is a rhombus but not a square.

Show solution

Opposite sides parallel (gradients):
$m_{AB} = \frac{4-2}{5-1} = \frac{1}{2} = m_{DC} \qquad m_{BC} = \frac{8-4}{7-5} = 2 = m_{AD}$
so $ABCD$ is a parallelogram.
Adjacent sides equal:
$AB = \sqrt{4^2 + 2^2} = \sqrt{20} \qquad BC = \sqrt{2^2 + 4^2} = \sqrt{20}$
A parallelogram with equal adjacent sides has all four sides equal — a rhombus. ✓
Not a square: adjacent sides would have to be perpendicular, but
$m_{AB} \times m_{BC} = \tfrac{1}{2} \times 2 = 1 \neq -1$
so the corners aren’t right angles. ✓

12 [5 marks]

A water tank is being drained. After $2$ minutes the depth of water is $46$ cm; after $7$ minutes it is $31$ cm. Assume the depth $d$ cm after $t$ minutes follows a linear model.

Find the model in the form $d = mt + c$ . [3]
Interpret the values of $m$ and $c$ in context. [1]
Find how long it takes for the tank to empty completely. [1]

Show solution

(a)

Gradient:
$m = \frac{31 - 46}{7 - 2} = \frac{-15}{5} = -3$
Use $(2, 46)$ : $46 = -3(2) + c$ , so $c = 52$ .
$d = -3t + 52$

(b) The depth falls by $3$ cm every minute ( $m = -3$ ); the tank started at $52$ cm deep ( $c = 52$ , the depth at $t = 0$ ).

(c) Empty means $d = 0$ :

0 = -3t + 52 \;\;\Longrightarrow\;\; t = \frac{52}{3} \approx 17.3 \text{ minutes}