Salam Academy
Sign inCreate account
Learn7357Paper 2 · Pure and MechanicsPure content (sections A–J)Parallel and perpendicular lines
Pure content (sections A–J) · Subtopic 7 of 66

Parallel and Perpendicular Lines

Why parallel lines share a gradient and perpendicular gradients multiply to −1-1, and how to use that for perpendicular bisectors, the foot of a perpendicular and right-angled shapes.

Cambridge 9709 Paper 1Edexcel YMA01 WMA11Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 35 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • State the gradient of a line parallel or perpendicular to a given line.
  • Find the equation of a parallel or perpendicular line through a given point.
  • Find the perpendicular bisector of a line segment.
  • Find the foot of a perpendicular, and prove that a triangle or quadrilateral has a right angle.
Pause and think

A line goes 33 across and 22 up. Turn it through 90∘90^\circ anticlockwise. How far across and up does the new line go?

Section 2 of 14

Core idea

Start hereSquare corners from numbers alone

A builder checks a corner is exactly 90∘90^\circ without a protractor, using only coordinates of three marks on the floor. How? There's a remarkable rule: two lines meet at a right angle exactly when their gradients multiply to −1-1. This lesson proves it.

The big idea

Parallel lines have equal gradients: m1=m2m_1=m_2. Perpendicular lines have m1m2=−1m_1m_2=-1, i.e. m2=−1m1m_2=-\dfrac1{m_1} ("flip and change sign").

1. Parallel lines

Parallel lines point the same way, so for each step across they rise the same amount: equal gradients, different intercepts. y=3x+1y=3x+1 and y=3x−4y=3x-4 are parallel.

2. Perpendicular lines: the proof

Turning a gradient triangle through 90°
−22468−2−11234560xygradient 2/3gradient −3/2across 3up 2down 3across 2
The same triangle turned through a right angle: "across 3, up 2" becomes "across 2, down 3". So m1m2=23×(−32)=−1m_1m_2=\tfrac23\times(-\tfrac32)=-1.
Why m1m2=−1m_1m_2=-1
  1. m1=abm_1=\frac ab
    A gradient triangle: across bb, up aa.
  2. Rotate 90∘ anticlockwise: across −a, up b\text{Rotate }90^\circ\text{ anticlockwise: across }-a,\ \text{up }b
    Rotating by 90∘90^\circ turns "across" into "up" and "up" into "backwards".
  3. m2=b−a=−bam_2=\frac{b}{-a}=-\frac ba
    Gradient of the rotated line.
  4. m1m2=ab×(−ba)=−1m_1m_2=\frac ab\times\left(-\frac ba\right)=-1
    Multiply.
Perpendicular ⇔ m1m2=−1m_1m_2=-1 (for non-vertical, non-horizontal lines).
Edge caseHorizontal and vertical y=2y=2 and x=5x=5 are perpendicular, but a vertical gradient is undefined, so the product rule doesn't apply. Treat this case separately.
Common misconceptionOnly changing the sign The perpendicular to gradient 22 is −12-\frac12, not −2-2. Flip and negate.

3. Finding perpendicular lines

Worked examplePerpendicular through a point

Find the line through (2,5)(2,5) perpendicular to y=4x−1y=4x-1.

Goal
Its equation in the form ax+by+c=0ax+by+c=0.
Strategy
Perpendicular gradient, then point–gradient form.
  1. m=−14m=-\tfrac14
    Flip 44 and negate.
  2. y−5=−14(x−2)y-5=-\tfrac14(x-2)
    Point–gradient.
  3. 4y−20=−x+2⇒x+4y−22=04y-20=-x+2\Rightarrow x+4y-22=0
    Multiply by 44, rearrange.
Answerx+4y−22=0x+4y-22=0.
✓ Check(2,5)(2,5): 2+20−22=02+20-22=0 ✓; gradient −14-\frac14 and 4×(−14)=−14\times(-\frac14)=-1 ✓.
Worked examplePerpendicular bisector of A(1,3)A(1,3), B(7,−1)B(7,-1)

Find the perpendicular bisector of ABAB.

Goal
The line through the midpoint at right angles to ABAB.
Strategy
Midpoint for the point; negative reciprocal for the gradient.
  1. M=(1+72,3−12)=(4,1)M=\left(\tfrac{1+7}2,\tfrac{3-1}2\right)=(4,1)
    Midpoint.
  2. mAB=−1−37−1=−23 ⇒ m=32m_{AB}=\frac{-1-3}{7-1}=-\tfrac23\ \Rightarrow\ m=\tfrac32
    Perpendicular gradient.
  3. y−1=32(x−4)⇒2y−2=3x−12⇒3x−2y−10=0y-1=\tfrac32(x-4)\Rightarrow2y-2=3x-12\Rightarrow3x-2y-10=0
    Point–gradient; clear fractions.
Answer3x−2y−10=03x-2y-10=0.
✓ CheckAA and BB should be equidistant from any point on it, e.g. (4,1)(4,1): AM=9+4AM=\sqrt{9+4}, BM=9+4BM=\sqrt{9+4} ✓.
Worked exampleShortest distance from a point to a line

Find the shortest distance from P(5,4)P(5,4) to y=2x−1y=2x-1.

Goal
The perpendicular distance.
Strategy
Drop a perpendicular from PP, find its foot FF, then measure PFPF.
  1. perpendicular through P: y−4=−12(x−5)\text{perpendicular through }P:\ y-4=-\tfrac12(x-5)
    Gradient −12-\frac12.
  2. 2x−1=−12x+6.5⇒2.5x=7.5⇒x=32x-1=-\tfrac12x+6.5\Rightarrow2.5x=7.5\Rightarrow x=3
    Solve with the line.
  3. F=(3,5),PF=(5−3)2+(4−5)2=5F=(3,5),\quad PF=\sqrt{(5-3)^2+(4-5)^2}=\sqrt5
    Distance formula.
Answer5≈2.24\sqrt5\approx2.24.
✓ CheckFF is on the line: 2(3)−1=52(3)-1=5 ✓; PFPF gradient −12\frac{-1}{2} is perpendicular to 22 ✓.
Pause and think

Show that A(1,1)A(1,1), B(3,5)B(3,5), C(7,3)C(7,3) form a right angle at BB.

Summary

  • Parallel: equal gradients.
  • Perpendicular: product −1-1.
  • Perpendicular bisector: midpoint + negative reciprocal.
  • Shortest distance: along the perpendicular.
Key words
Perpendicular
At right angles.
Negative reciprocal
−1m-\frac1m.
HistoryA word from the building site. "Perpendicular" comes from the Latin perpendiculum, a plumb line: a weight hanging on a string. Builders in ancient Egypt and Rome used plumb lines to make walls exactly vertical, that is, perpendicular to level ground.
Keep reading for free

Create a free account to read the rest of this lesson: key formulae, worked examples, explore the graph, common mistakes, exam tip, guided practice and more. Your answers are checked and your progress is saved.

Create free accountSign in