Parallel and Perpendicular Lines (AA HL)
Two lines are either parallel, perpendicular, or neither - and the whole question comes down to comparing their gradients. This page covers both gradient conditions on their own, separate from the wider Straight Lines topic, with worked examples and the mistake that costs the most marks.
11 questions on this sub-topic.
The two conditions
Covered under IB syllabus reference SL2.1: "Parallel lines \(m_1=m_2\). Perpendicular lines \(m_1m_2=-1\)." Neither result is printed in the formula booklet - both are treated as things you already know.
Parallel condition
\[m_1=m_2\]
Match the gradient of a new line to a given one to guarantee they never meet.
Perpendicular condition
\[m_1m_2=-1\]
Take the negative reciprocal of a gradient, \(-\dfrac1m\), to get a line at right angles to it.
Need the full syllabus wording and formula-booklet reference table? See Straight Lines.
Worked examples
The line \(y=kx+1\) is perpendicular to \(y=2x.\) Find \(k.\)
Worked solution
Perpendicular \(\Rightarrow k\times2=-1.\) M1
\(k=-\tfrac12.\) A1
The line through \((1,2)\) and \((5,k)\) is perpendicular to a line of gradient 4.
Find \(k.\)
Worked solution
The required gradient is \(-\tfrac14.\) M1
\(\dfrac{k-2}{5-1}=-\tfrac14.\) M1
\(k-2=-1.\) A1
\(k=1.\) A1
Line A has gradient \(\tfrac23\) and line B has gradient \(-\tfrac32.\) Show that the lines are perpendicular.
Worked solution
Multiply the gradients: \(\tfrac23\times(-\tfrac32).\) M1
\(=-1.\) A1
Since the product is \(-1\), the lines are perpendicular. R1
Common mistakes
- Flipping the gradient without also negating it. A perpendicular gradient is the negative reciprocal, \(-\dfrac{1}{m}\) - taking only the reciprocal (or only the negative) gives a line that isn't actually at right angles.
- Mixing up which condition is which. \(m_1=m_2\) is parallel; \(m_1m_2=-1\) is perpendicular - under exam pressure it's easy to set up the wrong equation, especially when a question gives one gradient and asks for the other relationship.
- Forgetting the vertical-line exception. A vertical line has no defined gradient, so it can't be checked against \(m_1m_2=-1\) directly - it's perpendicular to any horizontal line (\(m=0\)) and parallel only to another vertical line.
- Substituting coordinates into the gradient formula the wrong way round. \(m=\dfrac{y_2-y_1}{x_2-x_1}\) needs the same point first in both the numerator and the denominator - swapping only one of them flips the sign of the gradient, turning a correct perpendicular check into a wrong one even though every other step in the working was carried out correctly, which makes the mistake easy to overlook when checking back through the answer.
Ready to practise properly?
11 parallel-and-perpendicular-lines questions, marked instantly like the real exam.
Quick answers
How do I find the gradient of a line perpendicular to another?
Take the negative reciprocal of the known gradient: if the given line has gradient \(m\), the perpendicular gradient is \(-\tfrac1m\). Multiplying the two together always gives \(-1\).
How do I show that two lines are parallel?
Rearrange both equations into gradient-intercept form \(y=mx+c\) and compare the two \(m\) values. If \(m_1=m_2\) the lines are parallel; if they also share the same \(c\) they are actually the same line. For a quick refresher on your GDC, see Straight Lines.