Are straight line graphs on the Foundation paper too?

Yes. Spec points A9 and A10 are crossover content, so gradient, y = mx + c and finding the equation through two points appear on both tiers. The Higher version of A9 adds using y = mx + c to identify perpendicular lines as well as parallel ones, and Higher questions are more likely to give the line as ax + by = c or hide the coordinate work inside a longer problem.

Do I get the gradient and midpoint formulae in the exam?

No. Edexcel's formulae sheet for 1MA1 does not include the gradient formula, the midpoint formula or the distance between two points, so all three have to be memorised. The distance one is just Pythagoras applied to the horizontal and vertical gaps, which is worth knowing if you would rather remember one fact than three.

How many marks are straight line graphs worth on Edexcel Higher?

Expect roughly 3-6 marks per paper. A 'write down the gradient' part is 1 mark, finding an equation from two points is usually 3, and a question combining a midpoint or a length with an equation runs to 4 or 5. Across a full series of three 80-mark papers that adds up to a worthwhile block of algebra marks.

How do I find the equation of a line when I am only given two points?

Work out the gradient first, as the difference in y divided by the difference in x. Then substitute one of the two points into y = mx + c and solve for c. Finally write the whole equation out, and check it by putting the other point in; if both sides agree you have it right.

Straight Line Graphs - Edexcel GCSE Higher Maths

Typically 3-6 marks per paper, usually split across two questions · Spec 1MA1

Straight line graphs are the coordinate geometry strand of Edexcel's 1MA1 algebra content. Spec point A9 covers plotting linear equations, using $y = mx + c$ to identify parallel lines, and finding the equation of a line through two given points or through one point with a given gradient. A10 is the interpreting half: reading gradients and intercepts off a graph and out of an equation. Foundation candidates meet the same two spec points, but Higher papers push harder, with coordinates in all four quadrants, fractional gradients, equations handed to you as $3x + 2y = 12$ rather than in $y = mx + c$ form, and midpoint or line-length work buried inside a longer problem.

Edexcel spreads this across all three papers. A one-mark opener asking you to write down the gradient of $y = 5x - 3$ turns up near the front; a three-mark 'find an equation of the line through $(4, -1)$ and $(6, 4)$ ' sits in the middle; the coordinate work that leads into parallel and perpendicular lines comes later. Paper 1 is non-calculator, so a gradient of $\frac{5}{2}$ stays a fraction instead of becoming 2.5 on a screen, and the length of a line segment is often left as a surd. Papers 2 and 3 allow a calculator, which mainly changes how comfortable the Pythagoras step in a length question feels.

The marks disappear in predictable places. Change in $x$ divided by change in $y$ gives the reciprocal of the gradient and earns nothing; a line running downhill with a positive answer means a minus sign went missing; reading the coefficient of $x$ straight off $2y = 6x + 10$ gives 6 when the gradient is 3. On 'find an equation' questions the mark scheme pays a method mark for a correct gradient and a second for substituting a point to find $c$ , so write both lines down even when the arithmetic feels shaky. Then finish with the letter $y$ in front of it, because an answer of $2x - 2$ on its own is an expression, not the equation of a line.

Worked example

The straight line

L

passes through the points

A(-1, -4)

and

B(5, 8)

. The point

C(k, 10)

also lies on

L

. Find an equation of

L

in the form

y = mx + c

, and find the value of

k

[5 marks]

Gradient of $L$ : $m = \frac{8 - (-4)}{5 - (-1)}$ .
M1: a correct gradient expression, the $y$ -difference over the $x$ -difference, with $A$ and $B$ taken in the same order top and bottom
$m = \frac{12}{6} = 2$ .
A1: gradient 2 (a candidate who wrote $\frac{6}{12}$ has inverted the fraction and loses this)
Substitute $A(-1, -4)$ into $y = 2x + c$ : $-4 = 2(-1) + c$ , so $-4 = -2 + c$ and $c = -2$ .
M1: substituting either given point into $y = mx + c$ to find $c$ ; using $B$ works just as well
$L$ has equation $y = 2x - 2$ . Check with $B$ : $2(5) - 2 = 8$ ✓
A1: the equation stated with $y =$ in front, not the bare expression $2x - 2$
$C(k, 10)$ lies on $L$ , so $10 = 2k - 2$ . Then $2k = 12$ and $k = 6$ .
B1: substituting the known $y$ -coordinate and solving for $k$

Practice questions

These are original questions in Edexcel 1MA1 Higher style, covering gradient from two points, $y = mx + c$ , axis intercepts, midpoints and the length of a line segment. Try the gradient and equation questions without a calculator, since Paper 1 sets them that way and you can type fractional answers such as -3/2 or 5/4 directly. The harder set mixes two skills in one question, which is how the later marks are usually earned.

No account needed: the questions below are always free to check, plus 8 extra questions today. Your score so far updates as you go.

Question 1 Warm-up [2 marks]

Work out the gradient of the straight line through $(2, 3)$ and $(6, 11)$ .

Your answer (fractions like 1/2 are fine)
Question 2 Warm-up [1 mark]

A straight line has equation $y = 5x - 3$ . Write down the gradient of the line.

Your answer (fractions like 1/2 are fine)
Question 3 Warm-up [2 marks]

The point $(4, k)$ lies on the straight line with equation $y = 3x - 5$ . Find the value of $k$ .

Your answer (fractions like 1/2 are fine)
Question 4 Warm-up [2 marks]

A straight line has gradient 4 and crosses the $y$ -axis at $(0, -3)$ . Write down the equation of the line, giving $y$ in terms of $x$ .

Your expression (equivalent forms accepted, e.g. 2(x+3) or 2x+6)
Question 5 Warm-up [1 mark]

A straight line has equation $y = 7 - 4x$ . Write down the gradient of the line.

Your answer (fractions like 1/2 are fine)
Question 6 Exam pace [2 marks]

The straight line with equation $y = mx + 4$ passes through the point $(3, 19)$ . Find the value of $m$ .

Your answer (fractions like 1/2 are fine)

Common mistakes examiners see

Dividing the change in $x$ by the change in $y$ . Through $(1, 2)$ and $(3, 8)$ that produces $\frac{2}{6} = \frac{1}{3}$ when the gradient is 3.

Write the fraction out with the $y$ -difference on top before you touch the numbers: $m = \frac{y_2 - y_1}{x_2 - x_1}$ . A quick sanity check follows: a gradient of $\frac{1}{3}$ describes a line that barely rises, and these two points climb 6 in a run of 2.
Treating both differences as positive on a downhill line, so $(4, 7)$ to $(-2, 10)$ gives $\frac{1}{2}$ instead of $-\frac{1}{2}$ .

Subtract the two points in the same order top and bottom, negatives included. Then check the sign against the picture in your head: if $y$ falls as $x$ rises, the gradient must be negative.
Reading the gradient off an equation that has not been rearranged, so $4y - 5x = 2$ is answered as 5 or $-5$ .

Get $y$ on its own first. $4y = 5x + 2$ , then $y = \frac{5}{4}x + \frac{1}{2}$ , so the gradient is $\frac{5}{4}$ . The coefficient of $x$ is only the gradient once the equation is in $y = mx + c$ form.
Answering 'find an equation of the line' with $2x - 2$ , or stopping after finding $c$ and never writing the equation out.

End every one of these with a full equation starting $y =$ . Edexcel's final accuracy mark is for the equation, and an expression with no $y$ does not get it.
Subtracting coordinates to find a midpoint, turning $(-4, 4)$ and $(1, -5)$ into $(-5, 9)$ rather than $(-1.5, -0.5)$ .

Average each pair: add the two $x$ -values and halve, then add the two $y$ -values and halve. Sketch the two points roughly if you are unsure; the midpoint has to sit between them in both directions.

Frequently asked questions

Are straight line graphs on the Foundation paper too?: Yes. Spec points A9 and A10 are crossover content, so gradient, $y = mx + c$ and finding the equation through two points appear on both tiers. The Higher version of A9 adds using $y = mx + c$ to identify perpendicular lines as well as parallel ones, and Higher questions are more likely to give the line as $ax + by = c$ or hide the coordinate work inside a longer problem.
Do I get the gradient and midpoint formulae in the exam?: No. Edexcel's formulae sheet for 1MA1 does not include the gradient formula, the midpoint formula or the distance between two points, so all three have to be memorised. The distance one is just Pythagoras applied to the horizontal and vertical gaps, which is worth knowing if you would rather remember one fact than three.
How many marks are straight line graphs worth on Edexcel Higher?: Expect roughly 3-6 marks per paper. A 'write down the gradient' part is 1 mark, finding an equation from two points is usually 3, and a question combining a midpoint or a length with an equation runs to 4 or 5. Across a full series of three 80-mark papers that adds up to a worthwhile block of algebra marks.
How do I find the equation of a line when I am only given two points?: Work out the gradient first, as the difference in $y$ divided by the difference in $x$ . Then substitute one of the two points into $y = mx + c$ and solve for $c$ . Finally write the whole equation out, and check it by putting the other point in; if both sides agree you have it right.

Straight Line Graphs - Edexcel GCSE Higher Maths

Worked example

Practice questions

Common mistakes examiners see

Frequently asked questions

Related topics