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Cambridge IGCSE0607

Perpendicular lines

International Mathematics 0607 Chapter Notes

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Perpendicular lines
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In this chapter

  1. 1The gradient of a perpendicular line
  2. 2The line through a given point
  3. 3The perpendicular bisector
  4. 4Perpendicular at a point or perpendicular bisector?

1. The gradient of a perpendicular line

Two lines are perpendicular when they meet at a right angle. Their gradients multiply to give -1.

To find the perpendicular gradient, turn the gradient upside down and change its sign. This is the negative reciprocal.

A line such as 2y = 3x + 1 must first be written as y = mx + c. Divide every term by 2 to get y = 3/2x + 1/2. The gradient is 3/2, not 3, so the perpendicular gradient is -2/3.

m₁ × m₂ = -1

m₂ = -1/m₁

Examiner insight

“To calculate the new gradient, some candidates used the reciprocal of their gradient from part (a) instead of the negative reciprocal, while others made errors in rearrangement or transcription.”

Worked example 12 marks

The equation of line L is y = 3x - 4. Find the equation of the line perpendicular to line L that passes through the point (0, 5). Give your answer in the form y = mx + c.

  1. 1

    gradient of L is 3, so the perpendicular gradient is -1/3 [1]

  2. 2

    (0, 5) is on the y-axis, so c = 5 and y = -1/3x + 5 [1]

2. The line through a given point

To find the equation of a perpendicular line through a point, work in three steps. Find the gradient of the given line. Find the negative reciprocal. Then substitute the coordinates of the point into y = mx + c to find c.

Use the coordinates of the point the line must pass through. Do not use any other point from the question.

Examiner insight

“A common error was not substituting the coordinates of N (4, 8) into their y = mx + c, with some candidates using the coordinates of A, C or the mid-point of AC.”

Worked example 14 marks

Point A has coordinates (-2, 7) and point B has coordinates (2, -1). Find the equation of the line perpendicular to AB that passes through B. Give your answer in the form y = mx + c.

  1. 1

    gradient of AB = (-1 - 7)/(2 - (-2)) = (-8)/4 = -2 [1]

  2. 2

    perpendicular gradient = 1/2 [1]

  3. 3

    substitute B: -1 = 1/2(2) + c [1]

  4. 4

    c = -2, so y = 1/2x - 2 [1]

3. The perpendicular bisector

The perpendicular bisector of AB cuts AB in half at right angles. It passes through the midpoint of AB and has the negative reciprocal of the gradient of AB.

So it needs two things: the midpoint and the perpendicular gradient. Substitute the midpoint into y = mx + c. A quick sketch of A, B and the midpoint shows whether the sign of your gradient makes sense.

midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2)

gradient of AB = (y₂ - y₁)/(x₂ - x₁)

Examiner insight

“The most common errors came from slips with arithmetic, signs and cancelling, finding the equation of the line AB, using point A or B instead of the midpoint and errors in finding the midpoint.” “Candidates were expected to find the gradient of AB, then find the negative reciprocal in order to determine the gradient of the perpendicular bisector.”

Worked example 15 marks

A is the point (-4, 3) and B is the point (8, -1). Find the equation of the perpendicular bisector of AB. Give your answer in the form y = mx + c.

  1. 1

    midpoint of AB = ((-4 + 8)/2, (3 + (-1))/2) = (2, 1) [1]

  2. 2

    gradient of AB = (-1 - 3)/(8 - (-4)) = -1/3 [1]

  3. 3

    perpendicular gradient = 3 [1]

  4. 4

    substitute the midpoint: 1 = 3(2) + c [1]

  5. 5

    c = -5, so y = 3x - 5 [1]

4. Perpendicular at a point or perpendicular bisector?

Read the question for where the line passes. A line perpendicular to AB at A goes through A. A perpendicular bisector goes through the midpoint of AB.

Both use the same perpendicular gradient. Only the point you substitute is different.

Give m and c as exact fractions. Do not round them to decimals.

Examiner insight

“It was important that candidates worked out the equation of the line perpendicular to the line AB at point A and not the equation of the perpendicular bisector of AB.” “Some candidates rounded decimal values of m and c.”

Worked example 14 marks

A is the point (-5, 8) and B is the point (1, -1). The line AC is perpendicular to AB. Find the equation of the line AC.

  1. 1

    gradient of AB = (-1 - 8)/(1 - (-5)) = -9/6 = -3/2 [1]

  2. 2

    gradient of AC = 2/3 [1]

  3. 3

    substitute A, not the midpoint: 8 = 2/3(-5) + c [1]

  4. 4

    c = 8 + 10/3 = 34/3, so y = 2/3x + 34/3 [1]

Recap

  • Perpendicular lines have gradients that multiply to -1.
  • The perpendicular gradient is the negative reciprocal: m₂ = -1/m₁.
  • Rearrange to y = mx + c before you read the gradient.
  • Substitute the point the line passes through to find c.
  • A perpendicular bisector passes through the midpoint of AB.
  • Give m and c as exact fractions.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. A line has gradient 4. Write down the gradient of a line perpendicular to it.1 mark
  2. A line has gradient -2/3. Write down the gradient of a line perpendicular to it.1 mark
  3. Do not use a calculator. Find the gradient of a line perpendicular to 2y = 3x + 1.2 marks
  4. Line L has equation y = 2x - 3. Find the equation of the line perpendicular to L that passes through (4, 1). Give your answer in the form y = mx + c.3 marks
  5. Do not use a calculator. Show that the lines 3x + 4y = 8 and 4x - 3y = 5 are perpendicular.3 marks
  6. Do not use a calculator. Find the equation of the line perpendicular to 3x - 5y = 10 that passes through (-2, 5). Give your answer in the form y = mx + c.4 marks
  7. Do not use a calculator. P is the point (-3, 8) and Q is the point (9, -2). Find the equation of the perpendicular bisector of PQ. Give your answer in the form y = mx + c.5 marks
  8. Do not use a calculator. A is the point (-1, 6) and B is the point (5, -2). The line AC is perpendicular to AB. Find the equation of line AC. Give exact values of m and c.4 marks
  9. Do not use a calculator. A is the point (k, 3) and B is the point (5, -1). The line AB is perpendicular to the line 2y = 3x + 1. (a) Find the value of k. (b) Find the equation of the perpendicular bisector of AB. Give your answer in the form y = mx + c.7 marks
  10. On a chart, two buoys are at A (-4, 9) and B (6, -6). A ship sails in a straight line that crosses AB at right angles, halfway between the buoys. (a) Find the equation of the ship's route. Give exact values of m and c. (b) A rock is at R (7, 5.5). Show that the route passes through R. (c) The route crosses the x-axis at D. Find the x-coordinate of D.8 marks

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