Key Idea: A straight line is fully defined by two numbers: its gradient (steepness) and its y-intercept (where it hits the y-axis). Know these two numbers and you can write the equation, spot parallel lines, and find perpendicular ones.
What the IB asks you to do:
📐 Reading the gradient
Parallel lines have the same gradient. They never meet. Perpendicular lines have gradients that multiply to −1. If one line has gradient 3, the perpendicular gradient = −1 ÷ 3 = −⅓. Shortcut: flip the fraction and change the sign.
✏️ Worked examples
Find the equation of a line through two points
Find the equation of the line through (1, 4) and (3, 10).
Step by step:
Gradient: m = (10 − 4) ÷ (3 − 1) = 6 ÷ 2 = 3
Use point-slope with point (1, 4): y − 4 = 3(x − 1)
Expand: y − 4 = 3x − 3
Rearrange: y = 3x + 1
y = 3x + 1
Find a perpendicular line
Line L has equation y = 2x − 5. Find the equation of the perpendicular through (4, 3).
Step by step:
Gradient of L: m = 2
Perpendicular gradient: −1 ÷ 2 = −½
Use point-slope with (4, 3): y − 3 = −½(x − 4)
Rearrange: y = −½x + 5
y = −½x + 5
Interpret in context
A phone plan charges a $20 connection fee plus $0.15 per minute. Write the cost equation and interpret m and c.
Step by step:
Equation: C = 0.15t + 20 (t = minutes)
Gradient m = 0.15 → cost increases by $0.15 per minute
y-intercept c = 20 → fixed connection fee of $20 (cost when t = 0)
C = 0.15t + 20
Show the gradient calculation first — even if you can see it from the graph, write m = (y₂ − y₁)/(x₂ − x₁). That step earns a method mark on its own. Point-slope is your friend — you almost never need to use y = mx + c from scratch. Find m, use point-slope, then rearrange. Perpendicular check: multiply the two gradients together. If the answer is −1, you're right. Context questions: m = rate of change, c = starting value. The units of m come from the units of y divided by the units of x.
IB-style question [7 marks]
Line L passes through the points A(−2, 1) and B(2, 9). (a) Find the gradient of L. (b) Find the equation of L in the form y = mx + c. (c) The line N is perpendicular to L and passes through the point (4, 0). Find the equation of N in the form y = mx + c.
Step by step:
(a) Find the gradient of L from the two given points.
(b) Use point-slope with m = 2 and the point A(−2, 1), then rearrange.
(c) N is perpendicular to L, so its gradient is the negative reciprocal of 2 — flip the fraction and change the sign.
N passes through (4, 0); substitute into y = mx + c to find c.
Write the equation of N.
(a) gradient of L = 2. (b) y = 2x + 5. (c) y = −(1/2)x + 2.