Linear Graphs · Study Notes · Edexcel IGCSE 4MA1 Higher

Plot, read and find equations of straight lines. Gradient, intercept, parallel, perpendicular. Foundation for coordinate geometry.

At a glance

$y = mx + c$ : $m$ gradient, $c$ y-intercept.
Gradient: $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ .
Point-slope: $y - y_1 = m(x - x_1)$ .
Parallel: same gradient.
Perpendicular: gradients multiply to $-1$ (negative reciprocal).
Midpoint: average of coordinates.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

3.3 — Find the gradient and y-intercept of a line.
3.3 — Find the equation of a line through two points or a point and a gradient.
3.3 — Identify parallel and perpendicular lines.
3.3 — Find the midpoint of two points.

Equation of a straight line

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$y = mx + c$ — gradient $m$ , y-intercept $c$ .

Standard form: $y = mx + c$ .

$m$ = gradient (slope, steepness).
$c$ = y-intercept (where it crosses $y$ -axis).

c is where the line cuts the y-axis; m is rise divided by run.

Reading from equation:

Equation	Gradient	y-intercept
$y = 3x + 5$	$3$	$5$
$y = -2x + 7$	$-2$	$7$
$y = \dfrac{1}{2}x - 3$	$\dfrac{1}{2}$	$-3$
$y = -x$	$-1$	$0$
$y = 4$	$0$	$4$ (horizontal line)

Vertical lines: $x = a$ . No gradient (or 'undefined').

Worked qualitative. What does a negative gradient mean?

Line slopes DOWN from left to right.
$y$ DECREASES as $x$ increases.
E.g. $y = -3x + 5$ : at $x = 0, y = 5$ ; at $x = 1, y = 2$ .

Edexcel tip. Always rewrite the equation in $y = mx + c$ form before reading off $m$ and $c$ .

$y = mx + c$ .
$m$ = gradient.
$c$ = y-intercept.
Negative gradient: down-sloping.

Finding the equation of a line

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From two points, or point + gradient. Use $y - y_1 = m(x - x_1)$ .

Method 1: Two points.

Find gradient: $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ .
Use one point in $y - y_1 = m(x - x_1)$ .
Expand and simplify.

Example. Points $(2, 7)$ and $(5, 16)$ .

$m = (16 - 7) / (5 - 2) = 3$ .
$y - 7 = 3(x - 2)$ .
$y = 3x - 6 + 7 = 3x + 1$ .

Method 2: Point + gradient.

Just plug into $y - y_1 = m(x - x_1)$ .

Example. Through $(3, 4)$ with gradient $-2$ .

$y - 4 = -2(x - 3)$ .
$y = -2x + 6 + 4 = -2x + 10$ .

Method 3: Two intercepts.

If a line crosses both axes at known points, e.g. $(a, 0)$ and $(0, b)$ , equation: $\dfrac{x}{a} + \dfrac{y}{b} = 1$ .

Worked qualitative. Why is the point-slope form so useful?

Works regardless of whether $c$ is a 'nice' number.
Avoids solving for $c$ separately.
Single substitution gets the equation.

Edexcel tip. Show every step. Mark scheme: M1 for gradient, M1 for substituting point, A1 for the equation.

Two points: find gradient, use point-slope.
Point + gradient: direct substitution.
Always state $m$ explicitly.

Parallel and perpendicular

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Parallel: same gradient. Perpendicular: gradients × = $-1$ .

Parallel: SAME gradient. y-intercepts can differ.

If line A has gradient $3$ , any line parallel has gradient $3$ too.

Example. Find the line parallel to $y = 2x - 5$ through $(3, 4)$ .

Same gradient: $m = 2$ .
$y - 4 = 2(x - 3)$ → $y = 2x - 2$ .

Perpendicular: gradients multiply to $-1$ .

If $m_1 \cdot m_2 = -1$ , then $m_2 = -1/m_1$ — the NEGATIVE RECIPROCAL.

Gradient $m_1$	Perpendicular $m_2$
$2$	$-\dfrac{1}{2}$
$3$	$-\dfrac{1}{3}$
$-\dfrac{1}{2}$	$2$
$-4$	$\dfrac{1}{4}$

Example. Find line perpendicular to $y = \dfrac{1}{2}x + 3$ through $(4, -1)$ .

Negative reciprocal of $\dfrac{1}{2}$ : $-2$ .
$y + 1 = -2(x - 4) \to y = -2x + 7$ .

Verify. $\dfrac{1}{2} \times (-2) = -1$ ✓.

Parallel lines share a gradient; perpendicular gradients are negative reciprocals.

Worked qualitative. Why does perpendicular need NEGATIVE reciprocal?

Lines at $90°$ .
Gradient = tan(angle from horizontal).
$\tan(\theta) \times \tan(\theta + 90°) = -1$ (trig identity).
Geometrically: rotate by $90°$ flips and negates.

Edexcel tip. When you say 'gradient is $-2$ because it's perpendicular to $y = \dfrac{1}{2}x + 3$ ', state the reasoning. Mark schemes credit it.

Parallel: same $m$ .
Perpendicular: $m_1 \cdot m_2 = -1$ .
Perpendicular = negative reciprocal.
Verify: product = $-1$ .

Midpoint of two points

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Average of coordinates.

Formula: midpoint of $(x_1, y_1)$ and $(x_2, y_2)$ is:

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

Example. $A = (3, 7)$ , $B = (9, 1)$ .

$M = \left(\dfrac{3 + 9}{2}, \dfrac{7 + 1}{2}\right) = (6, 4)$ .

Common application: perpendicular bisector.

The perpendicular bisector of $AB$ :

Goes through the midpoint of $A$ and $B$ .
Is perpendicular to $AB$ .

Find:

Midpoint $M$ .
Gradient of $AB$ .
Negative reciprocal as gradient of bisector.
Use point-slope at $M$ .

Worked qualitative. Why average the coordinates for midpoint?

Halfway between two values: arithmetic mean.
Same for $x$ AND $y$ separately.
Geometrically: equidistant from both endpoints.

Edexcel tip. Show the formula explicitly: ' $M = ((3+9)/2, (7+1)/2)$ '. Earns the M1.

Midpoint: average each coordinate.
Used for perpendicular bisectors.
Always show the formula.

Quick recap

$y = mx + c$ — read $m$ and $c$ .
Gradient = rise over run.
Point-slope: $y - y_1 = m(x - x_1)$ .
Parallel: same $m$ .
Perpendicular: $m_1 \cdot m_2 = -1$ .
Midpoint: average coordinates.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$y = mx + c$ : $m$ gradient, $c$ y-intercept.
Gradient: $\Delta y / \Delta x$ .
Point-slope: $y - y_1 = m(x - x_1)$ .
Perpendicular: $m_1 m_2 = -1$ .
Midpoint: average coordinates.

How it’s examined

Linear graphs every Higher Tier paper (4-7 marks). Often combined with perpendicular/parallel. Examiner reports flag (1) gradient formula reversed, (2) just reciprocal (no negative) for perpendicular, (3) forgetting to divide for midpoint.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H Jan 2024, 4MA1/2H May/Jun 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on linear graphs, written exactly the way a tutor would explain them at the board.

1Find gradient and y-intercept
Foundation• gradient, intercept
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Question
Find the gradient and y-intercept of $y = 3x - 5$ .
Step-by-step solution
1. Step 1
  Compare with $y = mx + c$ : $m = 3$ (gradient), $c = -5$ (y-intercept).
Answer
Gradient = $3$ , y-intercept = $(0, -5)$ .
2Equation from two points
Higher• Adapted from 4MA1/1H Jan 2024 Q10• gradient, two points
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Question
Find the equation of the line through $(2, 7)$ and $(5, 16)$ .
Step-by-step solution
1. Step 1
  Gradient: $m = \dfrac{16 - 7}{5 - 2} = \dfrac{9}{3} = 3$ .
2. Step 2
  Use $y - y_1 = m(x - x_1)$ with point $(2, 7)$ : $y - 7 = 3(x - 2)$ .
3. Step 3
  Expand: $y = 3x - 6 + 7 = 3x + 1$ .
Answer
$y = 3x + 1$
Examiner tip
ALWAYS show the gradient calculation. Mark schemes give M1 for gradient, M1 for substitution, A1 for the equation.
3Parallel line through a point
Higher• parallel
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Question
Find the equation of the line parallel to $y = 2x - 5$ that passes through $(3, 4)$ .
Step-by-step solution
1. Step 1
  Parallel lines have the SAME gradient: $m = 2$ .
2. Step 2
  Use $y - 4 = 2(x - 3)$ :
  $y = 2x - 6 + 4 = 2x - 2$
Answer
$y = 2x - 2$
4Perpendicular line
Higher• Adapted from 4MA1/2H May/Jun 2024 Q15• perpendicular
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Question
Find the equation of the line perpendicular to $y = \dfrac{1}{2}x + 3$ passing through $(4, -1)$ .
Step-by-step solution
1. Step 1
  Perpendicular gradient: NEGATIVE RECIPROCAL of $\dfrac{1}{2}$ → $-2$ .
2. Step 2
  Use $y - (-1) = -2(x - 4)$ :
  $y + 1 = -2x + 8 \implies y = -2x + 7$
Answer
$y = -2x + 7$
Examiner tip
Perpendicular: gradients multiply to $-1$ . So $\dfrac{1}{2} \times (-2) = -1$ ✓.
5Midpoint of two points
Foundation• midpoint
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Question
Find the midpoint of $A(3, 7)$ and $B(9, 1)$ .
Step-by-step solution
1. Step 1
  Midpoint: average each coordinate.
  $M = \left(\frac{3 + 9}{2}, \frac{7 + 1}{2}\right) = (6, 4)$
Answer
$(6, 4)$
6A line through the origin at 45 degrees [1 mark]
Getting started• Paper 1H/2H style — 1 mark• straight lines, y = x
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Question
A straight line passes through the points $(0, 0)$ , $(3, 3)$ and $(-4, -4)$ .

Write down the equation of the line. [1]
Step-by-step solution
1. Step 1
  At every point the $y$ -coordinate equals the $x$ -coordinate. The gradient is $\dfrac{3 - 0}{3 - 0} = 1$ and the line passes through the origin, so $c = 0$ .
2. Step 2
  B1 Write the equation.
  $y = x$
Answer
$y = x$
Examiner tip
Examiner tip: $y = x$ and $y = -x$ are worth recognising on sight. $y = -x$ goes through $(3, -3)$ and $(-4, 4)$ , sloping the other way.

Common mistake: Writing $y = 3$ or $x = 3$ , which is a horizontal or vertical line through just one of the points. ( $x = y$ is an acceptable way of writing the correct answer.)
7Where a line crosses both axes [2 marks]
Getting started• Paper 1H/2H style — 2 marks• straight lines, intercepts, axes
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Question
The line $L$ has equation $5x + 2y = 20$ .

Find the coordinates of the points where $L$ crosses

(a) the $x$ -axis, [1]

(b) the $y$ -axis. [1]
Step-by-step solution
1. Step 1
  B1 (a) Every point on the $x$ -axis has $y = 0$ .
  $5x = 20 \quad \Rightarrow \quad x = 4 \quad \Rightarrow \quad (4, 0)$
2. Step 2
  B1 (b) Every point on the $y$ -axis has $x = 0$ .
  $2y = 20 \quad \Rightarrow \quad y = 10 \quad \Rightarrow \quad (0, 10)$
Answer
(a) $(4, 0)$ (b) $(0, 10)$
Examiner tip
Examiner tip: For an equation in the form $ax + by = c$ , covering up one term at a time is the quickest way to find both intercepts. It is also the fastest way to draw the line.

Common mistake: Swapping the two: putting $x = 0$ to find where the line crosses the $x$ -axis. On the $x$ -axis it is $y$ that is zero.
8Equation from a gradient and a point, integer form [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• straight lines, equation of a line, gradient
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Question
A straight line has gradient $\dfrac{3}{4}$ and passes through the point $(-8, 1)$ .

Find an equation of the line. Give your answer in the form $ax + by + c = 0$ , where $a$ , $b$ and $c$ are integers. [3]
Step-by-step solution
1. Step 1
  M1 Substitute the gradient and the point into $y - y_{1} = m(x - x_{1})$ .
  $y - 1 = \dfrac{3}{4}(x + 8)$
2. Step 2
  M1 Expand and simplify.
  $y = \dfrac{3}{4}x + 6 + 1 = \dfrac{3}{4}x + 7$
3. Step 3
  A1 Multiply by $4$ to clear the fraction and collect everything on one side.
  $4y = 3x + 28 \quad \Rightarrow \quad 3x - 4y + 28 = 0$
Answer
$3x - 4y + 28 = 0$ (or any integer multiple)
Examiner tip
Examiner tip: Check with the given point: $3(-8) - 4(1) + 28 = -24 - 4 + 28 = 0$ .

Common mistake: Multiplying only the $y$ side by $4$ , giving $4y = \dfrac{3}{4}x + 28$ . Every term must be multiplied.
9A missing coordinate of a point on a line [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• straight lines, gradient, point on a line
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Question
The straight line $L$ passes through the points $A(2, 3)$ and $B(8, -1)$ .

The point $C(-4, k)$ also lies on $L$ .

Find the value of $k$ . [3]
Step-by-step solution
1. Step 1
  M1 Find the gradient of $L$ from $A$ and $B$ .
  $m = \dfrac{-1 - 3}{8 - 2} = -\dfrac{4}{6} = -\dfrac{2}{3}$
2. Step 2
  M1 Use the gradient between $A$ and $C$ , which must be the same.
  $\dfrac{k - 3}{-4 - 2} = -\dfrac{2}{3}$
3. Step 3
  A1 Solve.
  $k - 3 = -\dfrac{2}{3} \times (-6) = 4 \quad \Rightarrow \quad k = 7$
Answer
$k = 7$
Examiner tip
Examiner tip: Finding the equation first, $y = -\dfrac{2}{3}x + \dfrac{13}{3}$ , and substituting $x = -4$ is equally valid. Using gradients avoids the awkward intercept.

Common mistake: Writing $-4 - 2 = -2$ in the denominator. $C$ is $6$ units to the left of $A$ , so the change in $x$ is $-6$ .
10The other two vertices of a square [5 marks]
Stretch• Paper 1H/2H style — 5 marks• perpendicular lines, coordinate geometry, squares
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Question
$ABCD$ is a square. $A$ is the point $(1, 2)$ and $B$ is the point $(4, 6)$ .

The $x$ -coordinate of $D$ is negative.

Find the coordinates of $C$ and the coordinates of $D$ . [5]
Step-by-step solution
1. Step 1
  M1 From $A$ to $B$ the step is $3$ across and $4$ up, so the gradient of $AB$ is $\dfrac{4}{3}$ and its length is found by Pythagoras.
  $AB = \sqrt{3^{2} + 4^{2}} = 5$
2. Step 2
  M1 $BC$ is perpendicular to $AB$ , so its gradient is $-\dfrac{3}{4}$ . A step with this gradient and length $5$ is $4$ across and $3$ down, or $4$ back and $3$ up.
  $(+4, -3) \quad \text{or} \quad (-4, +3)$
3. Step 3
  M1 Apply the same step to $A$ to reach $D$ . Only $(-4, +3)$ gives $D$ a negative $x$ -coordinate: $(1 - 4, 2 + 3)$ .
  $D = (-3, 5)$
4. Step 4
  A1 Apply the same step to $B$ to reach $C$ .
  $C = (4 - 4, 6 + 3) = (0, 9)$
5. Step 5
  A1 Both vertices found and checked: $\overrightarrow{DC} = (3, 4) = \overrightarrow{AB}$ .
  $C(0, 9), \quad D(-3, 5)$
Answer
$C(0, 9)$ and $D(-3, 5)$
Examiner tip
Examiner tip: Turning a step $(a, b)$ through $90°$ gives $(-b, a)$ or $(b, -a)$ . Here $(3, 4)$ becomes $(-4, 3)$ or $(4, -3)$ , and the condition on $D$ picks one.

Common mistake: Using the gradient $-\dfrac{3}{4}$ but a step of $3$ across and $4$ down. That step has gradient $-\dfrac{4}{3}$ , so the angle at $B$ is not $90°$ ; always check that the two gradients multiply to $-1$ .

Question 1
Paper 1H/2H style — 1 mark1 mark
Write down the coordinates of the point where the line $y = -3$ meets the line $y = x$ . [1]
Model answer
On $y = x$ the two coordinates are equal, and on $y = -3$ the $y$ -coordinate is $-3$ .

Answer: $(-3, -3)$
Why this scores
Mark scheme:
- B1 for $(-3, -3)$
Why this scores full marks: The script uses what each equation says about a point on it. No simultaneous-equation working is needed.
Question 2
Paper 1H/2H style — 2 marks2 marks
$P$ is the point $(-2, 5)$ and $Q$ is the point $(4, 5)$ .

(a) Write down the equation of the line $PQ$ . [1]

(b) Write down the equation of the perpendicular bisector of $PQ$ . [1]
Model answer
(a) Both points have $y$ -coordinate $5$ .

Answer: $y = 5$

(b) $PQ$ is horizontal, so its perpendicular bisector is vertical, through the midpoint $\left(\dfrac{-2 + 4}{2}, 5\right) = (1, 5)$ .

Answer: $x = 1$
Why this scores
Mark scheme:
- B1 for $y = 5$
- B1 for $x = 1$
Why this scores full marks: The script sees that a horizontal segment has a vertical perpendicular bisector, so no gradient calculation is needed. A vertical line has no gradient in the form $y = mx + c$ ; its equation is $x =$ the $x$ -coordinate of the midpoint.
Question 3
Paper 1H/2H style — 2 marks2 marks
Here are the equations of four straight lines.

Line A: $y = 3x + 1$

Line B: $2y = 6x - 5$

Line C: $y = 1 - 3x$

Line D: $3y = x + 9$

(a) Write down the letters of the two lines that are parallel. [1]

(b) Write down the letters of the two lines that are perpendicular. [1]
Model answer
Gradients: A $= 3$ , B: $y = 3x - \dfrac{5}{2}$ so $3$ , C $= -3$ , D: $y = \dfrac{1}{3}x + 3$ so $\dfrac{1}{3}$

(a) Answer: A and B

(b) $-3 \times \dfrac{1}{3} = -1$

Answer: C and D
Why this scores
Mark scheme:
- B1 for A and B
- B1 for C and D
Why this scores full marks: The script puts every line into the form $y = mx + c$ before comparing, and tests the perpendicular pair with the product $-1$ . Choosing A and D (gradients $3$ and $\dfrac{1}{3}$ ) is the usual error: reciprocal gradients alone are not perpendicular, the sign must change too.
Question 4
Paper 1H/2H style — 3 marks3 marks
The straight line $L$ passes through the points $(-6, 4)$ and $(2, -8)$ .

$L$ is perpendicular to the line with equation $ky = 2x + 7$ , where $k$ is a constant.

Find the value of $k$ . [3]
Model answer
Gradient of $L$ :

$\dfrac{-8 - 4}{2 - (-6)} = \dfrac{-12}{8} = -\dfrac{3}{2}$

$ky = 2x + 7$ gives $y = \dfrac{2}{k}x + \dfrac{7}{k}$ , so its gradient is $\dfrac{2}{k}$

$\dfrac{2}{k} \times \left(-\dfrac{3}{2}\right) = -1 \quad \Rightarrow \quad \dfrac{2}{k} = \dfrac{2}{3}$

Answer: $k = 3$
Why this scores
Mark scheme:
- M1 for the gradient of $L$ , $\dfrac{-8 - 4}{2 - (-6)}$ oe
- M1 for $\dfrac{2}{k} \times$ their gradient $= -1$ , or $\dfrac{2}{k} = \dfrac{2}{3}$ (ft their gradient)
- A1 for $k = 3$ cao
Why this scores full marks: Brackets round $-6$ keep the denominator at $8$ , and the second line is divided through by $k$ so that its gradient, $\dfrac{2}{k}$ , can be read off. The answer checks: $3y = 2x + 7$ has gradient $\dfrac{2}{3}$ , and $\dfrac{2}{3} \times \left(-\dfrac{3}{2}\right) = -1$ . Taking the gradient of $ky = 2x + 7$ as $2$ is the usual error.
Question 5
Paper 1H/2H style — 3 marks3 marks
The line $L$ has equation $3y - 2x = 12$ .

The line $M$ is parallel to $L$ and passes through the point $(6, -1)$ .

Find the coordinates of the point where $M$ crosses the $x$ -axis. [3]
Model answer
$L$ : $3y = 2x + 12$ , so $y = \dfrac{2}{3}x + 4$ and the gradient is $\dfrac{2}{3}$

$M$ : $y + 1 = \dfrac{2}{3}(x - 6)$ , so $y = \dfrac{2}{3}x - 5$

When $y = 0$ : $\dfrac{2}{3}x = 5$ , $x = 7.5$

Answer: $(7.5, 0)$
Why this scores
Mark scheme:
- M1 for the gradient of $L$ , $\dfrac{2}{3}$
- M1 for an equation of $M$ , e.g. $y + 1 = \dfrac{2}{3}(x - 6)$
- A1 for $(7.5, 0)$ oe
Why this scores full marks: The equation of $L$ is rearranged before reading the gradient, since $3y - 2x = 12$ lists the $y$ term first and the gradient is not $3$ or $-2$ . The answer is given as a point because the question asks for coordinates.
Question 6
Paper 1H/2H style — 3 marks3 marks
$Q$ is the point $(-1, 2)$ and $R$ is the point $(7, 2)$ .

The point $P$ lies on the line $y = 2x - 3$ .

Triangle $PQR$ is isosceles with $PQ = PR$ .

Find the coordinates of $P$ . [3]
Model answer
$P$ is the same distance from $Q$ and $R$ , so it lies on the perpendicular bisector of $QR$ .

$QR$ is horizontal with midpoint $\left(\dfrac{-1 + 7}{2}, 2\right) = (3, 2)$ , so the perpendicular bisector is $x = 3$ .

$y = 2(3) - 3 = 3$

Answer: $(3, 3)$
Why this scores
Mark scheme:
- M1 for identifying that $P$ lies on $x = 3$ (the perpendicular bisector of $QR$ )
- M1 for substituting $x = 3$ into $y = 2x - 3$
- A1 for $(3, 3)$
Why this scores full marks: The script turns " $PQ = PR$ " into a line, $x = 3$ , and then uses the second piece of information. Without the perpendicular bisector the problem needs distance formulas and far more algebra.
Question 7
Paper 1H/2H style — 4 marks4 marks
The line $L_{1}$ passes through $A(-3, -1)$ and $B(3, 3)$ .

The line $L_{2}$ is perpendicular to $L_{1}$ and passes through $B$ .

Find the coordinates of the point where $L_{2}$ crosses the $y$ -axis. [4]
Model answer
Gradient of $L_{1}$ : $\dfrac{3 - (-1)}{3 - (-3)} = \dfrac{4}{6} = \dfrac{2}{3}$

Gradient of $L_{2}$ : $-\dfrac{3}{2}$

$y - 3 = -\dfrac{3}{2}(x - 3)$

$y = -\dfrac{3}{2}x + \dfrac{9}{2} + 3 = -\dfrac{3}{2}x + 7.5$

Answer: $(0, 7.5)$
Why this scores
Mark scheme:
- M1 for the gradient of $L_{1}$ , $\dfrac{2}{3}$ oe
- M1 for the perpendicular gradient $-\dfrac{3}{2}$ (ft their gradient)
- M1 for an equation of $L_{2}$ through $B$ with their perpendicular gradient
- A1 for $(0, 7.5)$ oe
Why this scores full marks: Each stage is shown on its own line, so each method mark is visible. The sign of the product $-\dfrac{3}{2} \times (-3) = +\dfrac{9}{2}$ is handled correctly; writing $-\dfrac{9}{2}$ gives the wrong intercept $-1.5$ .
Question 8
Paper 1H/2H style — 4 marks4 marks
$A$ is the point $(2k, 5)$ and $B$ is the point $(6, 3k - 1)$ .

The midpoint of $AB$ lies on the line $y = x + 1$ .

(a) Find the value of $k$ . [3]

(b) Write down the coordinates of the midpoint of $AB$ . [1]
Model answer
(a) Midpoint: $\left(\dfrac{2k + 6}{2}, \dfrac{5 + 3k - 1}{2}\right) = \left(k + 3, \dfrac{3k + 4}{2}\right)$

$\dfrac{3k + 4}{2} = (k + 3) + 1$

$3k + 4 = 2k + 8$

Answer: $k = 4$

(b) $(4 + 3, \dfrac{12 + 4}{2})$

Answer: $(7, 8)$
Why this scores
Mark scheme:
- M1 for the midpoint in terms of $k$ , e.g. $\left(k + 3, \dfrac{3k + 4}{2}\right)$
- M1 for substituting their midpoint into $y = x + 1$
- A1 for $k = 4$
- B1 ft for $(7, 8)$
Why this scores full marks: The midpoint is simplified before it is substituted, so the equation is short and easy to solve. The answer checks: $A(8, 5)$ , $B(6, 11)$ , midpoint $(7, 8)$ , and $8 = 7 + 1$ .
Question 9
Paper 1H/2H style — 5 marks5 marks
$A(-3, 1)$ and $B(5, 7)$ are the ends of a diameter of a circle.

(a) Find the coordinates of the centre of the circle. [1]

(b) The tangent to the circle at $B$ is perpendicular to the radius at $B$ .

Find an equation of the tangent at $B$ . Give your answer in the form $ax + by = c$ , where $a$ , $b$ and $c$ are integers. [4]
Model answer
(a) $\left(\dfrac{-3 + 5}{2}, \dfrac{1 + 7}{2}\right)$

Answer: $(1, 4)$

(b) Gradient of the radius from $(1, 4)$ to $B(5, 7)$ : $\dfrac{7 - 4}{5 - 1} = \dfrac{3}{4}$

Gradient of the tangent: $-\dfrac{4}{3}$

$y - 7 = -\dfrac{4}{3}(x - 5)$

$3y - 21 = -4x + 20$

Answer: $4x + 3y = 41$
Why this scores
Mark scheme:
- B1 for $(1, 4)$
- M1 for the gradient of the radius (or of $AB$ ), $\dfrac{3}{4}$
- M1 for the perpendicular gradient $-\dfrac{4}{3}$ (ft their gradient)
- M1 for $y - 7 = -\dfrac{4}{3}(x - 5)$ oe
- A1 for $4x + 3y = 41$ oe with integer coefficients
Why this scores full marks: The radius to $B$ lies along the diameter $AB$ , so its gradient is quick to find. Multiplying through by $3$ clears the fraction before rearranging, and the answer checks at $B$ : $4(5) + 3(7) = 41$ .
Question 10
Paper 1H/2H style — 5 marks5 marks
$A$ is the point $(0, 4)$ and $B$ is the point $(6, 0)$ . $O$ is the origin.

The point $C$ lies on the line $y = x$ and angle $ABC = 90°$ .

(a) Find the coordinates of $C$ . [4]

(b) Work out the area of triangle $OBC$ . [1]
Model answer
(a) Gradient of $AB$ : $\dfrac{0 - 4}{6 - 0} = -\dfrac{2}{3}$

Gradient of $BC$ : $\dfrac{3}{2}$

$BC$ : $y = \dfrac{3}{2}(x - 6) = \dfrac{3}{2}x - 9$

$C$ is on $y = x$ : $x = \dfrac{3}{2}x - 9$ , so $\dfrac{1}{2}x = 9$ , $x = 18$

Answer: $(18, 18)$

(b) Base $OB = 6$ , height $= 18$

$\dfrac{1}{2} \times 6 \times 18 = 54$

Answer: $54$
Why this scores
Mark scheme:
- M1 for the gradient of $AB$ , $-\dfrac{2}{3}$ oe
- M1 for the gradient of $BC$ , $\dfrac{3}{2}$ (ft their gradient)
- M1 for an equation of $BC$ and substituting $y = x$
- A1 for $(18, 18)$
- B1 ft for $54$
Why this scores full marks: The right angle is at $B$ , so the perpendicular line goes through $B$ , not $A$ . In (b) the base $OB$ lies on the $x$ -axis, so the height is the $y$ -coordinate of $C$ .

4MA1 Linear Graphs

At a glance

What you’ll learn

Equation of a straight line

Finding the equation of a line

Parallel and perpendicular

Midpoint of two points

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Find gradient and y-intercept

2Equation from two points

3Parallel line through a point

4Perpendicular line

5Midpoint of two points

6A line through the origin at 45 degrees [1 mark]

7Where a line crosses both axes [2 marks]

8Equation from a gradient and a point, integer form [3 marks]

9A missing coordinate of a point on a line [3 marks]

10The other two vertices of a square [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10