Worked example 1
Find a line equation
Find the equation of the line through (2, 5) with gradient 3.
- Use y − b = m(x − a).
- Substitute m = 3 and (a, b) = (2, 5)
- Expand if required.
Answer: y − 5 = 3(x − 2), so y = 3x − 1
Back to Higher Mathematics
Higher Mathematics
Straight line methods
Find gradients, equations of lines and parallel or perpendicular relationships.
Before you start
- Subtract coordinates in a consistent order.
- Know y = mx + c and y − b = m(x − a)
- Recall that perpendicular gradients multiply to −1.
Explanation
Straight-line work is the language used across coordinate geometry, tangents and normals. The gradient tells you steepness and direction; the equation fixes the line in the plane.
At Higher, line questions often combine coordinate facts with parallel or perpendicular relationships.
Visual support
Method and rules
- m = y₂ − y₁x₂ − x₁
- y = mx + c
- Through (a, b): y − b = m(x − a)
- Parallel lines have equal gradients; perpendicular gradients are negative reciprocals.
Worked examples
Worked example 2
Use perpendicular gradients
A line has gradient 4. Find the gradient of a perpendicular line.
- Take the negative reciprocal.
- Change the sign and flip the fraction.
So: The perpendicular gradient is −14.
Watch out
- Swapping x-change and y-change in the gradient formula.
- Subtracting coordinates in different orders.
- Using the same gradient for a perpendicular line.
- Losing the sign when finding c.
Exam reminder
Leave line equations in the requested form. If no form is specified, a clear equation such as y = mx + c is acceptable.