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This lesson covers the equations and properties of straight lines, building on GCSE knowledge and extending it to A-Level standard as required by the Edexcel 9MA0 specification. You must be confident with gradient, different forms of the equation of a line, parallel and perpendicular lines, and finding equations from given information.
The gradient (slope) of a line through two points (x₁, y₁) and (x₂, y₂) is:
m = (y₂ − y₁) / (x₂ − x₁)
The gradient measures the rate of change of y with respect to x. A positive gradient means the line slopes upward from left to right; a negative gradient means it slopes downward.
Example: Find the gradient of the line through (2, 5) and (6, 13).
m = (13 − 5) / (6 − 2) = 8/4 = 2
This is often the most useful form when you know the gradient and one point on the line.
Example: Find the equation of the line with gradient 3 passing through (2, 7).
y − 7 = 3(x − 2) y − 7 = 3x − 6 y = 3x + 1
Any linear equation can be written in this form. It is useful when dealing with perpendicular lines and when the question asks for the answer "in the form ax + by + c = 0".
Example: Write y = 2x − 5 in general form.
2x − y − 5 = 0
The midpoint of the line segment joining (x₁, y₁) and (x₂, y₂) is:
M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
d = √((x₂ − x₁)² + (y₂ − y₁)²)
Example: Find the distance between (1, 3) and (4, 7).
d = √((4 − 1)² + (7 − 3)²) = √(9 + 16) = √25 = 5
Two lines are parallel if and only if they have the same gradient.
If line 1 has gradient m₁ and line 2 has gradient m₂, then the lines are parallel when m₁ = m₂.
Two lines are perpendicular if and only if the product of their gradients is −1:
m₁ × m₂ = −1
Equivalently, m₂ = −1/m₁. The gradient of the perpendicular line is the negative reciprocal of the original gradient.
Example: A line has gradient 3/4. The gradient of any line perpendicular to it is −4/3.
Exam Tip: When asked to prove two lines are perpendicular, calculate both gradients and show their product is −1. Simply stating they "look perpendicular" on a sketch earns no marks.
Example: Find the equation of the line parallel to 2x + 3y = 6 that passes through (3, −1).
Step 1: Find the gradient of 2x + 3y = 6. Rearrange: y = −(2/3)x + 2, so m = −2/3.
Step 2: Parallel line has the same gradient: m = −2/3.
Step 3: Use point-gradient form: y − (−1) = (−2/3)(x − 3) y + 1 = (−2/3)x + 2 y = (−2/3)x + 1
Example: Find the equation of the perpendicular bisector of the line segment joining A(1, 4) and B(5, 2).
Step 1: Midpoint of AB = ((1+5)/2, (4+2)/2) = (3, 3).
Step 2: Gradient of AB = (2 − 4)/(5 − 1) = −2/4 = −1/2.
Step 3: Gradient of perpendicular = −1/(−1/2) = 2.
Step 4: Equation: y − 3 = 2(x − 3), so y = 2x − 3.
| Term | Definition |
|---|---|
| Gradient | The rate of change of y with respect to x; the slope of a line |
| y-intercept | The y-coordinate where the line crosses the y-axis |
| Perpendicular bisector | A line that is perpendicular to a given segment and passes through its midpoint |
| Negative reciprocal | If m is the gradient, the negative reciprocal is −1/m |
| Collinear | Points that lie on the same straight line |