The big idea: Parallel lines have the same gradient — the number in front of x must be identical.
If the gradients match but the y-intercepts differ → the lines are parallel.
If both the gradient and the y-intercept match → it is the same line, not two separate ones.
| Line A | Line B | Parallel? | Reason |
|---|---|---|---|
| y = 2x + 1 | y = 2x + 7 | Yes | m₁ = m₂ = 2, c₁ ≠ c₂ |
| y = 3x − 4 | y = 3x − 4 | No — same line | Same line, not parallel |
| y = 4x + 1 | y = −4x + 1 | No | m₁ = 4 ≠ −4 = m₂ |
| y = (1/2)x + 3 | y = (1/2)x − 9 | Yes | Same gradient 1/2 |
IB wording: IB may say 'the lines are parallel' and ask you to find a missing coefficient.
Set the gradients equal and solve.
Interactive graph: parallel vs perpendicular lines
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The big idea: Two lines are perpendicular if they cross at exactly 90°.
Their gradients always multiply to −1: m₁ × m₂ = −1.
To find the perpendicular gradient: flip the fraction, then change the sign. Both steps are required.
- gradient of the original line
- gradient of the perpendicular line — the negative reciprocal of m₁
| Line gradient m₁ | Perpendicular gradient m₂ = −1/m₁ | Check: m₁ × m₂ |
|---|---|---|
| 2 | −1/2 | 2 × (−1/2) = −1 ✓ |
| −3 | 1/3 | −3 × 1/3 = −1 ✓ |
| 1/4 | −4 | (1/4) × (−4) = −1 ✓ |
| −2/5 | 5/2 | (−2/5) × (5/2) = −1 ✓ |
The most common mistake: Two errors come up constantly:
Forget to flip → gives −m instead of −1/m.
Forget to negate → gives 1/m instead of −1/m.
You must do both: flip the fraction AND change the sign.
Toggle between parallel lines (same gradient) and perpendicular lines (gradients multiply to −1).
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The big idea: Once you have the right gradient — same gradient for parallel, negative reciprocal for perpendicular — the rest works exactly like writing any line equation.
Substitute the given point into y = mx + c, then solve for c.
Parallel line through a point
Find the equation of the line parallel to y = 3x + 1 through (2, 8).
Step by step
- Parallel → same gradient.
- Substitute (2, 8) into y = 3x + c.
- Write the equation.
Final answer
y = 3x + 2
Perpendicular line through a point
Find the equation of the line perpendicular to y = 2x − 5 through (4, 3).
Step by step
- Perpendicular → negative reciprocal of 2.
- Substitute (4, 3) into y = −(1/2)x + c.
- Write the equation.
Final answer
y = −(1/2)x + 5
Marks for working: IB awards marks at two specific steps:
1. Identifying the correct gradient — state it explicitly (same gradient or negative reciprocal).
2. The substitution step — substitute the given point and solve for c.
Both steps need to be written out to collect both marks.
What is a perpendicular bisector?
The perpendicular bisector of a segment does two things at once: it cuts the segment exactly in half (passes through the midpoint) AND it crosses at exactly 90°.
That is it.
Click each step to build the perpendicular bisector one stage at a time.
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4-step method:
- Find the midpoint of the two points.
- Find the gradient of the segment (rise ÷ run).
- Flip and negate the gradient to get the perpendicular gradient.
- Substitute the midpoint into y = mx + c and solve for c.
Find the perpendicular bisector
Find the equation of the perpendicular bisector of the segment joining A(2, 4) and B(6, 8).
Step by step
- Midpoint.
- Gradient of AB.
- Perpendicular gradient — flip and negate.
- Substitute M(4, 6) into y = −x + c.
- Equation:
Final answer
y = −x + 10
IB asks 3 types of problems:
- Are these lines perpendicular? — multiply gradients; check it equals −1.
- Find the perpendicular bisector — midpoint + perp gradient + equation.
- Does this triangle have a right angle? — check if any two sides give m₁ × m₂ = −1.