Parallel and Perpendicular Lines (AA SL)
Whether two lines are parallel or perpendicular comes down entirely to comparing gradients - no need to sketch anything. This page covers both gradient conditions and how examiners like to disguise them inside a bigger question, with worked examples and the mistakes that lose the most marks. It's part of the broader Straight Lines topic.
22 questions on this sub-topic.
The two gradient conditions
Covered under IB syllabus reference SL2.1. Both conditions are in the formula booklet, so the job in an exam is spotting which one a question actually wants and applying it to the right gradient.
Parallel condition
\[m_1=m_2\]
Parallel lines share exactly the same gradient.
✓ In the formula bookletPerpendicular condition
\[m_1\times m_2=-1\]
Take the negative reciprocal of the first gradient to get the second.
✓ In the formula bookletNeed the full syllabus wording and formula-booklet reference table? See Straight Lines.
Worked examples
Find the equation of the line parallel to \(y=3x+1\) passing through \((0,4).\)
(a)(i) State the gradient.
(a)(ii) State the y-intercept.
Worked solution
(a)(i) Parallel lines have equal gradient: \(m=3.\) M1
(a)(ii) Through \((0,4)\): \(c=4.\) A1 So the equation is \(y=3x+4.\) A1
Find the equation of the line perpendicular to \(y=2x\) passing through \((0,1).\)
(a)(i) State the gradient.
(a)(ii) State the y-intercept.
Worked solution
(a)(i) Perpendicular gradient: \(-\dfrac12.\) M1
(a)(ii) Through \((0,1)\): \(y=-\tfrac12x+1.\) A1A1
Find the equation of the line through \((4,1)\) parallel to the segment joining \((0,0)\) and \((2,3).\)
(a)(i) State the gradient.
(a)(ii) State the y-intercept.
Worked solution
(a)(i) Gradient of segment \(=\tfrac{3}{2}.\) M1A1
(a)(ii) \(y-1=\tfrac32(x-4)\Rightarrow y\) M1
\(=\tfrac32x-5.\) A1
Find the equation of the line through \((1,4)\) and \((3,10).\)
(a)(i) State the gradient.
(a)(ii) State the y-intercept.
Worked solution
(a)(i) \(m=\dfrac{10-4}{3-1}\) M1
\(=3.\) A1
(a)(ii) \(y-4=3(x-1)\Rightarrow y\) M1
\(=3x+1.\) A1
Find the equation of the line perpendicular to \(3x-y=2\) passing through \((6,1).\)
(a)(i) State the gradient.
(a)(ii) State the y-intercept.
Worked solution
(a)(i) Gradient of \(3x-y=2\) is \(3.\) Perpendicular gradient \(=-\tfrac13.\) M1A1
(a)(ii) \(y-1=-\tfrac13(x-6)\Rightarrow y\) M1
\(=-\tfrac13x+3.\) A1
Common mistakes
- Assuming parallel lines are identical. \(m_1=m_2\) only guarantees the lines never meet - they can still have different \(y\)-intercepts and so be distinct, non-overlapping lines.
- Forgetting the negative reciprocal for perpendicular gradients. A gradient of \(\tfrac34\) gives a perpendicular gradient of \(-\tfrac43\), not \(\tfrac43\) - both the sign and the flip matter.
- Comparing gradients before rearranging into \(y=mx+c\). If a line is given as \(2x+y=5\), its gradient is \(-2\), not \(2\) - always isolate \(y\) before reading off \(m\) and applying either condition.
Ready to practise properly?
22 parallel-and-perpendicular questions, marked instantly like the real exam.
Quick answers
How do I know if two lines are parallel?
Two lines are parallel exactly when their gradients are equal, \(m_1=m_2\). They can still have different \(y\)-intercepts and so never actually touch.
How do I find the gradient of a line perpendicular to a given one?
Take the negative reciprocal of the given gradient, since perpendicular gradients satisfy \(m_1\times m_2=-1\). A gradient of \(\tfrac34\) gives a perpendicular gradient of \(-\tfrac43\).