Straight Line Graphs, Gradient and Midpoint
Coordinate geometry is one of the most reliable sources of marks in IGCSE Maths: the formulae are short, the methods repeat, and questions are usually structured so that each part earns marks on its own. Tutors described these as questions where full marks are realistic — provided you keep the coordinates the right way round.
1. Coordinates
A point is written (x, y) — across first, then up.
Order matters, and so does the notation. (6, 10) and (10, 6) are different points. Write coordinates as a pair in brackets, not as two loose numbers — marks are lost for both errors.
2. The three formulae
For two points A(x₁, y₁) and B(x₂, y₂):
| Formula | In words | |
|---|---|---|
| Gradient | m = (y₂ − y₁) / (x₂ − x₁) | change in y over change in x |
| Midpoint | ( (x₁+x₂)/2 , (y₁+y₂)/2 ) | average the x’s, average the y’s |
| Length | √[ (x₂−x₁)² + (y₂−y₁)² ] | Pythagoras |
Check which of these your formula list provides. In lessons, tutors pointed out that the midpoint method is not given — you need to know it. Confirm against the formula sheet for your own exam series.
Gradient is y over x, not x over y. “Rise over run” is the memory hook; the difference in y goes on top.
Keep the subtractions in the same order. If you start with y₂ − y₁ on top, you must use x₂ − x₁ underneath. Mixing the order flips the sign of your gradient — a very common and very costly slip.
Don’t confuse the three. Midpoint adds and halves; gradient and length subtract. Reaching for the distance formula when the question wants the midpoint was one of the most frequent errors in real lessons.
Worked example — A(1, 2) and B(7, 10):
- Gradient = (10 − 2)/(7 − 1) = 8/6 = 4/3
- Midpoint = ((1+7)/2, (2+10)/2) = (4, 6)
- Length = √[(7−1)² + (10−2)²] = √(36 + 64) = √100 = 10
Leave gradients as fractions in their simplest form. 4/3, not 1.33 and not 1⅓. If a question asks for an exact length, leave it as a surd — writing √52 as 7.21 when the question said “exact” or “in surd form” throws the mark away.
3. The equation of a straight line
y = mx + c, where m is the gradient and c is the y-intercept.
- m — how steep, and which way it slopes
- c — where the line crosses the y-axis (the value of y when x = 0)
Reading the gradient from an equation:
Rearrange to y = mx + c first. The gradient is the coefficient of x once y is by itself.
For 2y = 6x + 8, the gradient is not 6. Divide every term by 2:
- y = 3x + 4 → gradient 3, y-intercept 4
For 3x + 2y = 12:
- 2y = −3x + 12
- y = −3/2 x + 6 → gradient −3/2, y-intercept 6
Divide every term, not just the one you’re looking at. This is where most rearranging errors happen.
What the gradient tells you about the shape:
| Gradient | Line |
|---|---|
| Positive | slopes upward left to right |
| Negative | slopes downward left to right |
| Zero | horizontal (y = c) |
| Undefined | vertical (x = a) |
Steeper lines have a larger magnitude of gradient — a gradient of −5 is steeper than one of 2.
4. Finding the equation of a line
Given the gradient and a point
Gradient 3, through (2, 7):
- y = 3x + c
- Substitute the point: 7 = 3(2) + c
- 7 = 6 + c → c = 1
- y = 3x + 1
Given two points
Through A(1, 2) and B(7, 10):
- Gradient = (10 − 2)/(7 − 1) = 4/3
- y = 4/3 x + c
- Substitute either point — use (1, 2): 2 = 4/3 (1) + c
- c = 2 − 4/3 = 2/3
- y = 4/3 x + 2/3
Substitute a point that is actually on the line. Using a point from elsewhere in the question — the midpoint of a different segment, or a point on another line — produces a plausible-looking wrong answer and is a documented error.
Answer both halves. “Find the equation of the line” needs m and c. Stopping once you have the gradient is one of the easiest marks to leave behind in the whole paper.
Write the final answer as an equation, in the form y = mx + c. A bare value of c is not an equation.
5. Parallel and perpendicular lines
Parallel
Parallel lines have the same gradient.
They differ only in c. Any line parallel to y = 3x + 1 has gradient 3: for example y = 3x − 5.
Perpendicular
Perpendicular gradients multiply to −1. m₂ = −1 / m₁ — the negative reciprocal.
| Original gradient | Perpendicular gradient |
|---|---|
| 3 | −1/3 |
| −2 | 1/2 |
| 2/5 | −5/2 |
| −3/4 | 4/3 |
Two steps: flip the fraction, then change the sign. Doing only one of them is the classic error.
A whole number is a fraction too. For gradient 3, think 3/1 → flip to 1/3 → change sign to −1/3.
Example: line perpendicular to y = 2x + 5, passing through (4, 3)
- Original gradient = 2, so perpendicular gradient = −1/2
- y = −1/2 x + c
- 3 = −1/2 (4) + c → 3 = −2 + c → c = 5
- y = −1/2 x + 5
6. Perpendicular bisectors
A perpendicular bisector of AB is the line that passes through the midpoint of AB and is perpendicular to it.
It is a single line — not two lines crossing, a misreading that appeared in lessons.
Method — the two words tell you the two steps:
- “Bisector” → find the midpoint of AB
- “Perpendicular” → find the gradient of AB, then take its negative reciprocal
- Use that gradient and the midpoint to find c
Example: perpendicular bisector of A(1, 2) and B(7, 10)
- Midpoint = (4, 6)
- Gradient of AB = 4/3 → perpendicular gradient = −3/4
- y = −3/4 x + c, substituting (4, 6): 6 = −3 + c → c = 9
- y = −3/4 x + 9
7. Reading and sketching graphs
Reading from a graph:
Check the scale on each axis before you read anything off. Axes are often scaled differently, and assuming one square is one unit is a reliable way to lose every mark that follows.
- The y-intercept is where the line crosses the y-axis — read it where x = 0
- The x-intercept is where it crosses the x-axis — that is where y = 0
- To find the gradient, pick two clear points where the line passes exactly through grid intersections
The y-intercept is only where x = 0. Reading off the value where the line meets the edge of the grid, or where it crosses another line, is a common error.
Sketching a line: you need only two things — the y-intercept (mark it on the y-axis) and the gradient (from that point, go across 1 and up m). Sketching y = −2x + 3: mark 3 on the y-axis, then go across 1 and down 2.
Finding solutions from graphs: where two lines cross, the coordinates satisfy both equations — that is the solution of the simultaneous equations. If a question says “use the graph”, you must read the values off, not solve algebraically.
8. Gradients that mean something
In real-life graphs the gradient carries units and a meaning:
| Graph | Gradient represents |
|---|---|
| Distance–time | Speed |
| Speed–time | Acceleration |
| Cost against quantity | Cost per unit |
On a speed–time graph, a negative gradient is deceleration, a horizontal line is constant speed, and the area under the graph gives the distance travelled.
Give the units. A gradient of 12 on a distance–time graph in km and hours is 12 km/h. Bare numbers lose the final mark.
9. Mistakes that cost marks
Writing coordinates in the wrong order, or without brackets.
Using x over y for the gradient.
Subtracting in a different order on top and bottom, flipping the sign.
Confusing midpoint with distance. Midpoint adds and halves.
Reading the gradient straight off an equation that isn’t in y = mx + c form.
Dividing only some terms when rearranging.
Giving only the gradient when the question asked for the equation.
Changing the sign but not flipping the fraction for a perpendicular gradient (or the reverse).
Substituting a point that isn’t on the line.
Forgetting the midpoint step in a perpendicular bisector.
Ignoring the axis scale when reading from a graph.
Giving decimals when the question asked for a fraction or surd.
Leaving off the units on a real-life gradient.
Frequently asked questions
What is the formula for gradient? m = (y₂ − y₁) / (x₂ − x₁) — the change in y divided by the change in x.
How do you find the midpoint of a line? Average the coordinates: ((x₁+x₂)/2, (y₁+y₂)/2).
How do you find the length of a line segment? Pythagoras: √[(x₂−x₁)² + (y₂−y₁)²].
What do m and c mean in y = mx + c? m is the gradient; c is the y-intercept.
How do you find the equation of a line through two points? Find the gradient, substitute one of the points into y = mx + c to find c, then write the full equation.
What is the gradient of a parallel line? The same as the original.
How do you find a perpendicular gradient? Take the negative reciprocal — flip the fraction and change the sign. The two gradients multiply to −1.
What is a perpendicular bisector? The line through the midpoint of a segment, perpendicular to it.
What does the gradient of a distance–time graph tell you? The speed. On a speed–time graph it gives acceleration.
Why is my gradient negative when it should be positive? You have almost certainly subtracted the coordinates in a different order on the top and the bottom.
Quick revision checklist
- I write coordinates as (x, y) in brackets, in the right order
- I know the gradient formula and keep the subtraction order consistent
- I know the midpoint formula — and that it may not be on the formula sheet
- I can find the length of a line segment with Pythagoras
- I leave gradients as simplified fractions, and exact lengths as surds
- I can rearrange any equation into y = mx + c, dividing every term
- I can find the equation from a gradient and a point
- I can find the equation from two points, and give both m and c
- I know parallel lines have equal gradients
- I can find a perpendicular gradient — flip and change sign
- I can find a perpendicular bisector in both steps
- I check the axis scales before reading a graph
- I can sketch a line from its gradient and intercept
- I know what gradient means on distance–time and speed–time graphs
- I give units where the context has them
These notes cover straight line graphs, gradient and midpoint in the Cambridge IGCSE Mathematics (0580) syllabus, and are written for Grade 9–11 / Year 10–11 students. They are based on teaching patterns observed across a large set of one-to-one IGCSE Maths lessons, with particular attention to the errors students make most often and the wording examiners reward. Always check the current syllabus and formula list for your own exam series.
Finished this topic?
Saved on this device — no account needed.