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This lesson revises and extends the key results for straight lines that underpin all of coordinate geometry at A-Level. You must be completely fluent with gradients, equations of lines, midpoints, distances, and the conditions for parallel and perpendicular lines. These results are used constantly — in circle problems, parametric questions, and coordinate proofs.
The gradient (or slope) of the straight line passing 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 1: Find the gradient of the line through A(−3, 5) and B(1, −7).
m = (−7 − 5) / (1 − (−3)) = −12 / 4 = −3
The gradient is −3.
Exam Tip: Always be careful with signs when subtracting negative coordinates. Write out the subtraction in full to avoid errors.
The midpoint M of the line segment joining (x1,y1) and (x2,y2) is
M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Example 2: Find the midpoint of the segment from P(2, −4) to Q(8, 6).
M = ((2 + 8)/2, (−4 + 6)/2) = (5, 1)
The distance between (x1,y1) and (x2,y2) is given by
d = √[(x₂ − x₁)² + (y₂ − y₁)²]
This is a direct application of Pythagoras' theorem in the coordinate plane.
Example 3: Find the distance between A(−1, 3) and B(5, −5).
d = √[(5 − (−1))² + (−5 − 3)²]
= √[6² + (−8)²]
= √[36 + 64]
= √100
= 10
This is the gradient-intercept form. Here m is the gradient and c is the y-intercept.
Example 4: A line has gradient 2 and passes through (0, −3). Its equation is y = 2x − 3.
This is useful when you know the gradient m and a point (x1,y1) on the line.
Example 5: Find the equation of the line with gradient −4 through (3, 7).
y − 7 = −4(x − 3)
y − 7 = −4x + 12
y = −4x + 19
This is the general form. At A-Level, you are often asked to give your answer in this form with integer coefficients.
Example 6: Write y = ³⁄₄x − 2 in the form ax + by + c = 0.
y = (3/4)x − 2
4y = 3x − 8
3x − 4y − 8 = 0
Exam Tip: When asked for the equation "in the form ax + by + c = 0", always clear fractions so that a, b, and c are integers.
Two lines are parallel if and only if they have the same gradient.
If line l1 has gradient m1 and line l2 has gradient m2, then
l₁ ∥ l₂ ⟺ m₁ = m₂
Example 7: The line y = 3x − 1 is parallel to the line y = 3x + 5, since both have gradient 3.
Two lines are perpendicular if and only if the product of their gradients is −1.
l₁ ⊥ l₂ ⟺ m₁ × m₂ = −1
Equivalently, the gradient of a line perpendicular to a line with gradient m is −1/m (the negative reciprocal).
Example 8: A line has gradient 2/5. Find the gradient of a perpendicular line.
m₂ = −1 / (2/5) = −5/2
Example 9: Show that the line through A(1, 4) and B(3, 10) is perpendicular to the line through C(2, 1) and D(8, −1).
Gradient of AB = (10 − 4)/(3 − 1) = 6/2 = 3
Gradient of CD = (−1 − 1)/(8 − 2) = −2/6 = −1/3
Product = 3 × (−1/3) = −1 ✓
The product is −1, so the lines are perpendicular.
The perpendicular bisector of a line segment AB is the line that passes through the midpoint of AB and is perpendicular to AB.
Method:
Example 10: Find the equation of the perpendicular bisector of the segment from A(2, 1) to B(6, 5).
Midpoint = ((2 + 6)/2, (1 + 5)/2) = (4, 3)
Gradient of AB = (5 − 1)/(6 − 2) = 4/4 = 1
Perpendicular gradient = −1
Equation: y − 3 = −1(x − 4)
y = −x + 7
To find where two lines intersect, solve their equations simultaneously.
Example 11: Find the intersection of y = 2x + 1 and 3x + y = 16.
Substitute y = 2x + 1 into 3x + y = 16:
3x + (2x + 1) = 16
5x + 1 = 16
5x = 15
x = 3
y = 2(3) + 1 = 7
The lines meet at (3, 7).
Three points are collinear (lie on the same straight line) if the gradient between any two pairs of points is the same.
Example 12: Show that A(1, 3), B(3, 7), and C(6, 13) are collinear.
Gradient of AB = (7 − 3)/(3 − 1) = 4/2 = 2
Gradient of AC = (13 − 3)/(6 − 1) = 10/5 = 2
Since gradient AB = gradient AC, the points are collinear.
Problem: The line l1 passes through A(−2, 5) and B(4, 2). The line l2 is perpendicular to l1 and passes through B.
(a) Find the equation of l1 in the form ax + by + c = 0.
(b) Find the equation of l2.
(c) The line l2 meets the x-axis at point C. Find the coordinates of C.
(d) Find the area of triangle ABC.
Solution:
(a)
Gradient of l₁ = (2 − 5)/(4 − (−2)) = −3/6 = −1/2
y − 2 = −1/2 (x − 4)
2(y − 2) = −(x − 4)
2y − 4 = −x + 4
x + 2y − 8 = 0
(b)
Gradient of l₂ = −1/(−1/2) = 2
y − 2 = 2(x − 4)
y = 2x − 6
(c) Set y = 0:
0 = 2x − 6
x = 3
C = (3, 0).
(d)
AB = √[(4−(−2))² + (2−5)²] = √[36 + 9] = √45 = 3√5
The perpendicular distance from l₁ to C is the distance from C(3, 0) to line x + 2y − 8 = 0:
d = |3 + 0 − 8| / √(1² + 2²) = |−5| / √5 = 5/√5 = √5
Area = ½ × base × height = ½ × 3√5 × √5 = ½ × 15 = 7.5
The area of triangle ABC is 7.5 square units.
Exam Tip: Always show your gradient calculations explicitly. When asked for a line equation in a specific form, make sure you rearrange fully. For perpendicular bisectors, state both the midpoint and the perpendicular gradient clearly before writing the equation — this makes your working easy for the examiner to follow and earns method marks even if you make an arithmetic slip.