The equation of a straight line follows from its gradient, m = (y₂ − y₁)/(x₂ − x₁), and one point on it: y − y₁ = m(x − x₁). Parallel lines have equal gradients, and perpendicular lines have gradients that multiply to −1. Exam questions want answers in the form y = mx + c or ax + by + c = 0 with integer coefficients, so you need to be fluent in both.
Straight line formulae
- Gradient: m = (y₂ − y₁) / (x₂ − x₁)
- Line through (x₁, y₁) with gradient m: y − y₁ = m(x − x₁)
- Parallel lines: m₁ = m₂
- Perpendicular lines: m₁ × m₂ = −1
For a perpendicular gradient, flip the fraction and change the sign: 2/3 becomes −3/2, and 3 becomes −1/3.
How do you find the equation of a line?
Find the gradient, then use any one point on the line in y − y₁ = m(x − x₁). Expand and rearrange into the form the question asks for.
The two standard forms are y = mx + c, which shows the gradient and the y-intercept, and ax + by + c = 0, where a, b and c must be integers. If your gradient is a fraction, multiply the whole equation by the denominator to clear it.
Parallel and perpendicular lines
Parallel lines have the same gradient. Perpendicular lines have gradients whose product is −1, so the perpendicular gradient is the negative reciprocal.
If a line is given as ax + by + c = 0, you cannot read the gradient off directly. Rearrange to y = mx + c first. The rule also works in reverse: to prove two lines are perpendicular, show their gradients multiply to −1.
Where do two lines meet?
Two lines cross where both equations are true, so solve them simultaneously. The answer is often a fraction; leave it exact. A line meets the x-axis where y = 0 and the y-axis where x = 0.
Always substitute your answer back into both equations. It takes seconds and catches slips.
The perpendicular bisector of a line segment
The perpendicular bisector of AB cuts AB in half at right angles. It needs two ingredients: it passes through the midpoint of AB, and its gradient is the negative reciprocal of the gradient of AB.
Every point on it is the same distance from A and B. That property makes it the key to circles: the centre of a circle lies on the perpendicular bisector of every chord, so two bisectors crossing give the centre of the circle through three points.
Areas of triangles formed by lines
A common question asks for the area of a triangle enclosed by two lines and an axis. Find the vertices, choose a base that is horizontal or vertical, find the perpendicular height, then use ½ × base × height. If a side lies along the x-axis, the height is simply the y-coordinate of the opposite vertex. Leave fractional answers exact.
Worked examples
Example 1. Find an equation of the line through A(2, 3) and B(6, 11), in the form y = mx + c.
- Gradient: m = (11 − 3) / (6 − 2) = 8/4 = 2.
- Using A: y − 3 = 2(x − 2).
- Expand and rearrange: y = 2x − 1.
- Check with B: 2(6) − 1 = 11.
Answer: y = 2x − 1.
Marks: M1 for the gradient, M1 for the point form, A1 for the rearranged equation.
Example 2. Find the line perpendicular to y = 3x + 2 through (6, 2), in the form ax + by + c = 0 with integer coefficients.
- The given gradient is 3, so the perpendicular gradient is −1/3.
- Point form: y − 2 = −⅓(x − 6).
- Multiply by 3: 3y − 6 = −x + 6.
- Collect: x + 3y − 12 = 0. Check: 6 + 6 − 12 = 0.
Answer: x + 3y − 12 = 0.
Marks: B1 for the perpendicular gradient, M1 for the point form, A1 for the integer form.
Example 3. Find the perpendicular bisector of A(1, 2) and B(7, 6), in the form ax + by + c = 0.
- Midpoint: (4, 4).
- Gradient of AB: (6 − 2) / (7 − 1) = 2/3, so the perpendicular gradient is −3/2.
- Point form through (4, 4): y − 4 = −3/2 (x − 4).
- Multiply by 2: 2y − 8 = −3x + 12, so 3x + 2y − 20 = 0.
Answer: 3x + 2y − 20 = 0.
Marks: M1 for the midpoint, M1 for the perpendicular gradient, A1 for the integer form.
Example 4. The lines y = x + 2 and y = −x + 8 form a triangle with the x-axis. Find its area.
- Where the lines meet: x + 2 = −x + 8, so x = 3 and y = 5.
- y = x + 2 meets the x-axis at (−2, 0); y = −x + 8 meets it at (8, 0).
- Base along the x-axis: 8 − (−2) = 10. Height: 5.
- Area = ½ × 10 × 5 = 25.
Answer: 25.
Marks: M1 for the intersection, M1 for both intercepts, A1 for the area.
Common mistakes
- Reading the gradient from ax + by + c = 0. In 3x + 2y − 8 = 0 the gradient is −3/2, not 3. Rearrange first.
- Only flipping, or only changing sign. The perpendicular gradient of 2/3 is −3/2. Both steps are needed.
- Leaving fractions in ax + by + c = 0. The coefficients must be integers. Multiply through to clear denominators.
- Using the wrong point for the perpendicular bisector. It passes through the midpoint of AB, not through A or B.
- Rounding coordinates of an intersection. Leave answers like 4/9 exact unless the question asks for decimals.
How straight line questions are examined
Straight lines are Pure 1 content in Edexcel International A-Level (paper WMA11). In Edexcel A-Level (9MA0) they are covered in the first year of Pure. Typical questions ask you to:
- find a line through two points, or through a point with a given gradient
- find a parallel or perpendicular line in a required form
- find where lines meet, including the axes
- find a perpendicular bisector, or the area of a triangle formed by lines
The Pure 1 straight line graphs exam questions cover all of these, with mark schemes. For more, try the straight line graphs worksheet or the coordinate geometry revision engine.
This builds on distance and midpoint and leads into tangents and chords of circles.
Frequently asked questions
What is the equation of a straight line?
Any non-vertical straight line can be written as y = mx + c, where m is the gradient and c is the y-intercept. The equivalent form ax + by + c = 0 is used when integer coefficients are required.
How do I know if two lines are perpendicular?
Multiply their gradients. If the product is −1, the lines are perpendicular.
What is the gradient of a vertical line?
It is undefined. A vertical line has equation x = k, and it is perpendicular to any horizontal line y = c.