Co-Ordinate Geometry
Parallel and perpendicular lines
In this note
- 3.6Know that parallel lines have equal gradients and find a parallel line through a point.
- 3.7Know that perpendicular gradients multiply to \(-1\) Extended; find a perpendicular line and a perpendicular bisector.
- 3.7Decide whether two lines are parallel, perpendicular or neither.
Before you start: The equation of a line.
Formula bank
Parallel and perpendicular gradients
Parallel lines rise at the same rate, so they have the same gradient. Perpendicular lines meet at 90°; their gradients are negative reciprocals: flip the fraction and change the sign.
| Gradient | Parallel | Perpendicular |
|---|---|---|
| 2 | 2 | \(-\tfrac12\) |
| \(-\tfrac23\) | \(-\tfrac23\) | \(\tfrac32\) |
| −3 | −3 | \(\tfrac13\) |
Try it yourself
Show answer
Answer −1/4
Show answer
Answer −5/2
Finding parallel and perpendicular lines
General methodHow to find a line parallel or perpendicular to a given line
- Get the given line's gradient (rearrange to y = mx + c if needed).
- Parallel: use the same m. Perpendicular: use −1 ÷ m.
- Put the new m and the given point into \(y=mx+c\); solve for c.
- Perpendicular bisector of AB: the point is the MIDPOINT of AB, the gradient is −1 ÷ (gradient of AB).
Example Find the line perpendicular to \(y=2x+1\) through (4, 1).
New gradient \(-\tfrac12\); then c.
- \(1=-\tfrac12(4)+c\)
- \(c=3\)
Answer \(y=-\tfrac12x+3\)
Example 2 Find the perpendicular bisector of A(1, 2) and B(5, 10).
Midpoint, then the perpendicular gradient.
- midpoint (3, 6); \(m_{AB}=\tfrac{8}{4}=2\)
- new gradient \(-\tfrac12\)
- \(6=-\tfrac12(3)+c\Rightarrow c=7.5\)
Answer \(y=-\tfrac12x+7.5\)
Try it yourself
Show answer
Answer \(y=2x-2\)
Show answer
Answer \(y=-3x+1\)
Ways they test it
Type 1 Parallel through a point
"Find the equation of the line parallel to \(y=-3x+7\) that passes through (2, −5)." (2 marks)
Spot it 'parallel'.
- Same gradient.
- c from the point.
For example \(-5=-3(2)+c\)
Answer \(y=-3x+1\)
Type 2 Perpendicular through a point
"Find the line perpendicular to \(3x+2y=8\) through (6, 7)." (3 marks, Extended)
Spot it 'perpendicular'.
- Rearrange: \(m=-\tfrac32\).
- Perpendicular: \(\tfrac23\).
- c from the point.
For example \(7=\tfrac23(6)+c\)
- \(c=3\)
Answer \(y=\tfrac23x+3\)
Type 3 Perpendicular bisector
"Find the perpendicular bisector of A(−2, 3) and B(6, −1)." (4 marks, Extended)
Spot it 'perpendicular bisector'.
- Midpoint.
- Gradient of AB, then −1 ÷ it.
- c from the midpoint.
For example midpoint (2, 1), \(m_{AB}=-\tfrac12\)
- new gradient 2
- \(1=4+c\)
Answer \(y=2x-3\)
Type 4 Parallel, perpendicular or neither?
"Show that \(2y=x+3\) and \(y=-2x+5\) are perpendicular."
Spot it two equations to compare.
- Both into y = mx + c.
- Equal m: parallel; product −1: perpendicular.
For example \(\tfrac12\times(-2)\)
Answer −1: perpendicular
Mistakes to avoid
- ✗ perpendicular to m = 2 is 1/2✓ \(-\tfrac12\)Flip AND change the sign.
- ✗ gradient of \(3x+2y=8\) is 3✓ \(-\tfrac32\)Rearrange first.
- ✗ perpendicular bisector through A✓ through the MIDPOINT of ABBisector means cuts in half.
Summary
- Parallel lines: equal gradients.
- Perpendicular lines: \(m_1\times m_2=-1\) (negative reciprocal).
- New line: gradient first, then c from the point.
- Perpendicular bisector: through the midpoint, perpendicular gradient.
If you see …
- "parallel to"copy the gradient.
- "perpendicular to"flip the gradient and change its sign.
Practice
Practice: easy, medium, hard
Easy
Show answer
Answer 5
Show answer
Answer 1/3
Show answer
Answer yes
Medium
Guide me
What is it asking? an equation
What do we know? gradient of the given line \(-\tfrac32\)
Plan perpendicular \(\tfrac23\); c from (6, 7)
Show working
- \(7=\tfrac23(6)+c=4+c\)
- \(c=3\)
Answer \(y=\tfrac23x+3\)
Guide me
What is it asking? an equation
What do we know? new gradient \(-\tfrac12\)
Plan c from (4, 1)
Show working
- \(1=-2+c\)
- \(c=3\)
Answer \(y=-\tfrac12x+3\)
Guide me
What is it asking? yes or no
What do we know? gradients \(\tfrac12\) and −2
Plan multiply
Show working
- \(\tfrac12\times(-2)=-1\)
Answer yes
Hard
Guide me
What is it asking? an equation
What do we know? midpoint (2, 1), \(m_{AB}=-\tfrac12\)
Plan new gradient 2
Show working
- \(1=2(2)+c\)
- \(c=-3\)
Answer \(y=2x-3\)
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| midpoint (2, 1) | B1 | ||
| \(m_{AB}=-\tfrac12\) | B1 | ||
| gradient 2 | M1 | ||
| \(y=2x-3\) | A1 | ||
| Total: 4 marks | |||
Guide me
What is it asking? an equation
What do we know? midpoint (3, 6), \(m_{AB}=2\)
Plan new gradient \(-\tfrac12\)
Show working
- \(6=-1.5+c\)
- \(c=7.5\)
Answer \(y=-\tfrac12x+7.5\)
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| midpoint (3, 6) | B1 | ||
| gradient \(-\tfrac12\) | M1 | ||
| \(y=-\tfrac12x+7.5\) | A1 | ||
| Total: 3 marks | |||
Guide me
What is it asking? an equation
What do we know? gradient −3
Plan c from (2, −5)
Show working
- \(-5=-6+c\)
- \(c=1\)
Answer \(y=-3x+1\)
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| m = −3 | B1 | ||
| \(y=-3x+1\) | A1 | ||
| Total: 2 marks | |||
Can I do this?
- I know parallel lines have equal gradients and perpendicular gradients multiply to −1.
- I can find a parallel or perpendicular line through a point.
- I can find a perpendicular bisector.
Anything unticked: revise that section first. All ticked: take the quiz.
Quiz
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