Straight line graphs at A-Level
You already know how to find the gradient and equation of a line, the conditions for lines to be parallel or perpendicular, and how to find a midpoint and a length, all from GCSE. A-Level combines these into genuinely new problems: constructing a perpendicular bisector from scratch, using coordinates to find the area of a shape, and formally modelling real situations with a linear equation.
Three skills make up this topic:
- Perpendicular bisectors — combining the midpoint and the perpendicular gradient condition into one construction
- Area from coordinates — using points or intersecting lines to find the area of a triangle
- Modelling — interpreting the gradient and intercept of a linear model in a real-world context
Equations of lines: a quick recap
Two forms are used throughout this page: \(y = mx + c\) (gradient \(m\), y-intercept \(c\)), and \(y - y_1 = m(x - x_1)\) (gradient \(m\), passing through \((x_1, y_1)\)) — the second is usually faster when you're given a point rather than the intercept.
Worked Example 1
Find the equation of the line through \((2, 5)\) with gradient 3.
1
Use \(y - y_1 = m(x - x_1)\): \(y - 5 = 3(x - 2)\)
2
Expand and simplify: \(y - 5 = 3x - 6\)
Answer\(y = 3x - 1\)
Worked Example 2
Find the equation of the line perpendicular to \(y = 2x - 1\) that passes through \((4, 3)\).
1
The gradient of \(y = 2x - 1\) is 2, so the perpendicular gradient is \(-\dfrac{1}{2}\)
2
Use \(y - y_1 = m(x - x_1)\): \(y - 3 = -\dfrac{1}{2}(x - 4)\)
3
Expand: \(y - 3 = -\dfrac{1}{2}x + 2\)
Answer\(y = -\dfrac{1}{2}x + 5\)
Perpendicular bisectors
The perpendicular bisector of a line segment passes through its midpoint at a right angle to it. Finding one always follows the same two-step pattern: find the midpoint, then find the perpendicular gradient, and combine them.
Worked Example 3
Find the equation of the perpendicular bisector of the line segment joining \(A(1, 2)\) and \(B(7, 8)\).
1
Midpoint of \(AB\): \(\left(\dfrac{1+7}{2}, \dfrac{2+8}{2}\right) = (4, 5)\)
2
Gradient of \(AB\): \(\dfrac{8 - 2}{7 - 1} = \dfrac{6}{6} = 1\)
3
Perpendicular gradient: \(-1\)
4
Equation through \((4, 5)\): \(y - 5 = -1(x - 4)\)
Answer\(y = -x + 9\)
The perpendicular bisector of \(AB\) passes through the midpoint \(M(4, 5)\) at a right angle to \(AB\) — the construction found in Worked Example 3.
Finding the area of a triangle from coordinates
When a triangle's vertices are horizontal/vertical distances apart, its area follows directly from the base and height. When it's formed by lines instead of given points, find the vertices first — by finding where the lines cross each other and the axes — then apply the same method.
Worked Example 4
The points \(A(0, 3)\), \(B(4, 3)\) and \(C(4, 7)\) form a right-angled triangle. Find its area.
1
\(AB\) is horizontal (both points have \(y = 3\)): length \(= 4\)
2
\(BC\) is vertical (both points have \(x = 4\)): length \(= 4\)
3
The right angle is at \(B\), so \(AB\) and \(BC\) are the base and height
4
Area \(= \dfrac{1}{2} \times 4 \times 4\)
AnswerArea \(= 8\)
Worked Example 5
Find the area of the triangle formed by the lines \(y = x\), \(y = -x + 6\) and the x-axis.
1
Find where the lines meet: \(x = -x + 6 \implies 2x = 6 \implies x = 3\), so they meet at \((3, 3)\)
2
\(y = x\) meets the x-axis at \((0, 0)\); \(y = -x + 6\) meets the x-axis at \((6, 0)\)
3
Base along the x-axis, from \((0, 0)\) to \((6, 0)\): length \(= 6\)
4
Height \(=\) the y-coordinate of the apex \((3, 3)\): height \(= 3\)
AnswerArea \(= \dfrac{1}{2} \times 6 \times 3 = 9\)
Modelling with straight-line graphs
When a real situation is modelled by \(y = mx + c\), the gradient \(m\) and intercept \(c\) always mean something concrete in context — usually a rate and a fixed starting value. Exam questions often ask you to state what they represent, not just calculate with them.
Worked Example 6
A phone plan's monthly cost \(£C\) for using \(d\) GB of data is modelled by \(C = 15 + 2d\). State what the gradient and the \(C\)-intercept represent, and find the cost of using 12 GB.
1
The gradient, 2, represents the cost per GB of data: £2 per GB
2
The intercept, 15, represents the fixed monthly fee before any data is used: £15
3
Substitute \(d = 12\): \(C = 15 + 2(12) = 15 + 24\)
AnswerGradient = cost per GB (£2); intercept = fixed fee (£15); cost for 12 GB \(= £39\)
Exam tips
💡 Tip 1
Perpendicular bisector = midpoint + perpendicular gradient, always
There's no shortcut around finding both first. Write "midpoint = ..." and "perpendicular gradient = ..." as explicit labelled steps — exam mark schemes usually award a mark for each separately.
💡 Tip 2
For triangle area, look for a right angle first
If two sides are horizontal and vertical, you can read the base and height straight off the coordinates — no need for the full coordinate-geometry area formula.
💡 Tip 3
Find every vertex before calculating anything
For a triangle formed by lines, find all three vertices (intersections with each other and with the axes) as a separate first stage — don't try to combine finding vertices and calculating area in the same step.
💡 Tip 4
Always state units in modelling answers
"The gradient is 2" loses marks compared to "the gradient represents £2 per GB" — modelling questions are marked on interpretation, not just arithmetic.
💡 Tip 5
Check collinearity with gradients, not distances
To show three points lie on the same line, calculate the gradient between the first pair and the second pair and confirm they're equal (and share a common point) — this is faster and less error-prone than working with distances.
Common mistakes
Common Mistake 1
Using the midpoint formula but forgetting the perpendicular gradient
Finding the midpoint of \(AB\) and then using the same gradient as \(AB\) gives the wrong line entirely — a perpendicular bisector needs the negative reciprocal gradient, not the original one.
Common Mistake 2
Using the wrong pair of sides as base and height
For a right-angled triangle, the base and height must be the two sides that meet at the right angle — using the longest side (the hypotenuse) as the base gives an incorrect area.
Common Mistake 3
Finding only some of the triangle's vertices
A triangle formed by two lines and an axis has three vertices: where the two lines meet each other, and where each line meets the axis. Missing one of the axis-intercepts is a common incomplete answer.
Common Mistake 4
Swapping what the gradient and intercept represent
In \(C = 15 + 2d\), the gradient (2) is the rate and the intercept (15) is the fixed value — mixing these up (saying the fixed fee is £2) is a very easy slip under time pressure.
Common Mistake 5
Assuming points are collinear without checking a shared point
Two pairs of points can have equal gradients without lying on the same line unless they also share a common point — always confirm both conditions, not just matching gradients.
Practice questions
Work through each question before checking the worked solution.
Core Skills
Q1Find the equation of the line through \((3, 7)\) with gradient 2.Core Skills
Q2Find the equation of the line through \((1, 4)\) and \((5, 12)\).Core Skills
Q3Find the gradient of the line perpendicular to \(y = 4x - 3\).Core Skills
Q4Find the midpoint of the line segment joining \((2, 5)\) and \((8, 11)\).Core Skills
Q5Find the length of the line segment joining \((1, 1)\) and \((4, 5)\).Core Skills
Exam-Style
Q6Find the equation of the line perpendicular to \(2x + y = 5\) that passes through \((3, -1)\).Exam-Style
Q7Find the equation of the perpendicular bisector of the line segment joining \((0, 4)\) and \((6, 0)\).Exam-Style
Q8The triangle \(ABC\) has vertices \(A(1, 1)\), \(B(5, 1)\) and \(C(5, 4)\). Find its area.Exam-Style
Q9Find the area of the triangle formed by the line \(y = 2x - 4\), the x-axis and the y-axis.Exam-Style
Q10A taxi fare \(£F\) for a journey of \(m\) miles is modelled by \(F = 3 + 1.5m\). State what the gradient and intercept represent, and find the fare for a 10-mile journey.Exam-Style
A* Challenge
Q11Find the equation of the perpendicular bisector of the line segment joining \(A(-2, 5)\) and \(B(4, -3)\).A* Challenge
Q12Find the area of the triangle formed by the lines \(y = x + 1\), \(y = -2x + 7\) and the x-axis.A* Challenge
Q13The line \(l_1\) has equation \(y = 3x - 2\). The line \(l_2\) is perpendicular to \(l_1\) and passes through the point where \(l_1\) crosses the y-axis. Find the equation of \(l_2\).A* Challenge
Q14A company's costs \(£C\) for producing \(n\) items is modelled by \(C = 200 + 5n\). Find the number of items produced if the total cost is £950.A* Challenge
Q15Show that the points \(A(1, 2)\), \(B(4, 8)\) and \(C(7, 14)\) are collinear.A* Challenge
Answers — full worked solutions
Core Skills (Q1–Q5)
Q1 — \(y = 2x + 1\)
1
\(y - 7 = 2(x - 3)\)
2
\(y - 7 = 2x - 6\)
\(y = 2x + 1\)
Q2 — \(y = 2x + 2\)
1
Gradient \(= \dfrac{12 - 4}{5 - 1} = \dfrac{8}{4} = 2\)
2
\(y - 4 = 2(x - 1) \implies y = 2x - 2 + 4\)
\(y = 2x + 2\)
Q3 — \(-\dfrac{1}{4}\)
1
Gradient of \(y = 4x - 3\) is 4
2
Perpendicular gradient is the negative reciprocal: \(-\dfrac{1}{4}\)
\(-\dfrac{1}{4}\)
Q4 — \((5, 8)\)
1
\(\left(\dfrac{2+8}{2}, \dfrac{5+11}{2}\right) = (5, 8)\)
\((5, 8)\)
Q5 — 5
1
\(\sqrt{(4-1)^2 + (5-1)^2} = \sqrt{9 + 16} = \sqrt{25}\)
Length \(= 5\)
Exam-Style (Q6–Q10)
Q6 — \(x - 2y = 5\)
1
\(2x + y = 5 \implies y = -2x + 5\), gradient \(-2\), perpendicular gradient \(\dfrac{1}{2}\)
2
\(y - (-1) = \dfrac{1}{2}(x - 3) \implies y + 1 = \dfrac{1}{2}x - \dfrac{3}{2}\)
3
\(y = \dfrac{1}{2}x - \dfrac{5}{2}\)
\(x - 2y = 5\)
Q7 — \(3x - 2y = 5\)
1
Midpoint: \(\left(\dfrac{0+6}{2}, \dfrac{4+0}{2}\right) = (3, 2)\)
2
Gradient of segment: \(\dfrac{0-4}{6-0} = -\dfrac{2}{3}\), perpendicular gradient \(= \dfrac{3}{2}\)
3
\(y - 2 = \dfrac{3}{2}(x - 3) \implies y = \dfrac{3}{2}x - \dfrac{5}{2}\)
\(3x - 2y = 5\)
Q8 — Area \(= 6\)
1
\(AB\) is horizontal, length \(= 4\); \(BC\) is vertical, length \(= 3\); right angle at \(B\)
2
Area \(= \dfrac{1}{2} \times 4 \times 3\)
Area \(= 6\)
Q9 — Area \(= 4\)
1
x-intercept: \(0 = 2x - 4 \implies x = 2\), point \((2, 0)\)
2
y-intercept: \(x = 0 \implies y = -4\), point \((0, -4)\)
3
Right-angled triangle with legs 2 and 4: Area \(= \dfrac{1}{2} \times 2 \times 4\)
Area \(= 4\)
Q10 — Gradient = £1.50/mile; intercept = £3 base fare; fare = £18
1
Gradient (1.5) represents the cost per mile: £1.50 per mile
2
Intercept (3) represents the fixed base fare: £3
3
\(F = 3 + 1.5(10) = 3 + 15\)
Fare \(= £18\)
A* Challenge (Q11–Q15)
Q11 — \(3x - 4y + 1 = 0\)
1
Midpoint: \(\left(\dfrac{-2+4}{2}, \dfrac{5+(-3)}{2}\right) = (1, 1)\)
2
Gradient of \(AB\): \(\dfrac{-3-5}{4-(-2)} = \dfrac{-8}{6} = -\dfrac{4}{3}\), perpendicular gradient \(= \dfrac{3}{4}\)
3
\(y - 1 = \dfrac{3}{4}(x - 1) \implies y = \dfrac{3}{4}x + \dfrac{1}{4}\)
\(3x - 4y + 1 = 0\)
Q12 — Area \(= 6.75\)
1
Lines meet: \(x + 1 = -2x + 7 \implies 3x = 6 \implies x = 2\), point \((2, 3)\)
2
\(y = x + 1\) meets the x-axis at \((-1, 0)\); \(y = -2x + 7\) meets the x-axis at \((3.5, 0)\)
3
Base \(= 3.5 - (-1) = 4.5\); height \(= 3\)
4
Area \(= \dfrac{1}{2} \times 4.5 \times 3\)
Area \(= 6.75\)
Q13 — \(x + 3y + 6 = 0\)
1
\(l_1\) crosses the y-axis at \((0, -2)\)
2
\(l_1\) has gradient 3, so \(l_2\) has gradient \(-\dfrac{1}{3}\)
3
\(y - (-2) = -\dfrac{1}{3}(x - 0) \implies y = -\dfrac{1}{3}x - 2\)
\(x + 3y + 6 = 0\)
Q14 — \(n = 150\)
1
\(950 = 200 + 5n\)
2
\(750 = 5n\)
\(n = 150\)
Q15 — Confirmed collinear
1
Gradient of \(AB\): \(\dfrac{8-2}{4-1} = \dfrac{6}{3} = 2\)
2
Gradient of \(BC\): \(\dfrac{14-8}{7-4} = \dfrac{6}{3} = 2\)
3
Equal gradients, and both pass through \(B\)
\(A\), \(B\) and \(C\) are collinear
Summary
- Use \(y - y_1 = m(x - x_1)\) when you have a point and a gradient; use the two-point gradient formula when you have two points.
- A perpendicular bisector always needs both the midpoint and the perpendicular (negative reciprocal) gradient — find them as two separate labelled steps.
- For triangle area, find every vertex first, then look for a right angle so you can read off the base and height directly.
- In a modelling equation \(y = mx + c\), the gradient is a rate and the intercept is a fixed starting value — always interpret them in the context given.
- Three points are collinear if the gradient between each consecutive pair is equal and they share a common point.
These techniques combine the completing-the-square and simultaneous-equation skills used earlier in Pure Year 1, and the intersection technique from Graphs & Transformations. Next in Pure Year 1: Circles, which reuses the perpendicular gradient condition covered here.
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