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Cambridge IGCSE0580

Perpendicular lines

Mathematics 0580 Chapter Notes

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Perpendicular lines
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In this chapter

  1. 1The gradient of a given line
  2. 2Gradients of perpendicular lines
  3. 3Equation of the perpendicular through a point
  4. 4Horizontal and vertical lines
  5. 5Perpendicular bisector of a line segment

1. The gradient of a given line

A line in the form y = mx + c has gradient m. The equation is often not in that form.

Rearrange it first so that y stands alone. Divide every term by the number in front of y. The gradient is then the number in front of x.

y = mx + c gradient = m

Examiner insight

“While many candidates recognised the need to find the negative reciprocal of the gradient, not all were able to determine the gradient of the given line.”

Worked example 12 marks

Do not use a calculator. Find the gradient of the line 4y + 6x = 5.

  1. 1

    rearrange: 4y = 5 - 6x, so y = 5/4 - 3/2x [1]

  2. 2

    the gradient is the number in front of x: -3/2 [1]

2. Gradients of perpendicular lines

Two lines are perpendicular when they meet at a right angle. If one line has gradient m₁, the perpendicular line has gradient m₂, where the product of the gradients is -1.

So m₂ is the negative reciprocal of m₁. Turn the fraction upside down and change the sign. Do both.

The same test shows whether two given lines are perpendicular: multiply their gradients.

m₁ × m₂ = -1

m₂ = -1/m₁

Examiner insight

“A common error was to only apply half of the relationship between the gradients, and so give either the reciprocal or the negative value of the gradient.” “Some realised the gradient needed to change sign or that they needed the reciprocal of the gradient in part (b) but many did not do both.”

Worked example 13 marks

Do not use a calculator. Find the gradient of a line perpendicular to 2y = 3x + 1.

  1. 1

    rearrange: y = 3/2x + 1/2, so m₁ = 3/2 [1]

  2. 2

    m₁ × m₂ = -1, so 3/2 × m₂ = -1 [1]

  3. 3

    m₂ = -2/3 [1]

3. Equation of the perpendicular through a point

Work in three steps. Find the gradient of the given line. Change it to the perpendicular gradient. Then put the new gradient and the given point into y = mx + c to find c.

The gradient between two points is the change in y divided by the change in x. Subtract in the same order on top and bottom.

m = (y₂ - y₁)/(x₂ - x₁)

Examiner insight

“Many substituted a point into a linear equation using the gradient of the original line rather than the perpendicular, even though they often gave this inverted gradient within the final answer.”

Worked example 14 marks

A is the point (5, 2) and B is the point (-3, 6). Find the equation of the line perpendicular to AB that passes through the point (-2, 5). Give your answer in the form y = mx + c.

  1. 1

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

  2. 2

    perpendicular gradient = 2 [1]

  3. 3

    substitute (-2, 5): 5 = 2 × (-2) + c [1]

  4. 4

    c = 9, so y = 2x + 9 [1]

4. Horizontal and vertical lines

A horizontal line such as y = 2 has gradient 0. The negative reciprocal cannot be used here, because -1/0 has no value.

A line perpendicular to a horizontal line is vertical. A line perpendicular to a vertical line is horizontal.

Lines parallel to the axes have a simpler form of equation than y = mx + c. The perpendicular to y = 2 is x = k, with k the x-coordinate of the given point.

Examiner insight

“Most candidates thought the equation should be in the form y = mx + c, not remembering the simpler form of equation for lines parallel to the axes.”

Worked example 11 mark

Line L has equation y = 2. Find the equation of the line perpendicular to line L that passes through the point (-1, 5).

  1. 1

    y = 2 is horizontal, so the perpendicular is vertical and passes through x = -1: x = -1 [1]

5. Perpendicular bisector of a line segment

The perpendicular bisector of a segment cuts it in half at a right angle. It passes through the midpoint of the segment, and its gradient is the negative reciprocal of the gradient of the segment.

Find the midpoint by adding the coordinates and halving. Find the gradient of the segment. Change it to the perpendicular gradient. Then use the midpoint as the point on the line.

Examiner insight

“A common answer was to find the gradient correctly but then to find the equation of the perpendicular through one of the two given points.”

Worked example 15 marks

Do not use a calculator. Find the equation of the perpendicular bisector of the line joining the points (-3, 8) and (9, -2). Give your answer in the form y = mx + c.

  1. 1

    midpoint = ((-3 + 9)/2, (8 + (-2))/2) = (3, 3) [1]

  2. 2

    gradient of the segment = (-2 - 8)/(9 - (-3)) = -10/12 = -5/6 [1]

  3. 3

    perpendicular gradient = 6/5 [1]

  4. 4

    substitute (3, 3): 3 = 6/5 × 3 + c, so c = -3/5 [1]

  5. 5

    y = 6/5x - 3/5 [1]

Recap

  • Rearrange to y = mx + c to read the gradient.
  • Perpendicular lines: m₁ × m₂ = -1.
  • The perpendicular gradient is the negative reciprocal: upside down and change the sign.
  • Use the perpendicular gradient with the given point to find c.
  • A line perpendicular to a horizontal line is vertical: x = k.
  • Perpendicular bisector: midpoint plus perpendicular gradient.

End-of-chapter exercise

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

  1. Do not use a calculator. Find the gradient of a line perpendicular to y = 4x - 7.1 mark
  2. Do not use a calculator. Find the gradient of a line perpendicular to 3y = 6 - 2x.3 marks
  3. Do not use a calculator. Find the equation of the line perpendicular to y = 3x - 2 that passes through (6, 1). Give your answer in the form y = mx + c.3 marks
  4. Do not use a calculator. Show that the lines 4y = x + 8 and y = 4 - 4x are perpendicular.3 marks
  5. Do not use a calculator. Find the equation of the line perpendicular to x = 3 that passes through (-2, 7).1 mark
  6. A is the point (2, 1) and B is the point (6, 9). Find the equation of the line perpendicular to AB that passes through A. Give your answer in the form y = mx + c.4 marks
  7. Line l has equation 3x + 4y = 12. Line m is perpendicular to l and passes through (0, 5). Find the equation of m.3 marks
  8. The line y = 0.4x + 3 is perpendicular to line n. Line n passes through (5, -2). Find the equation of n.3 marks
  9. Do not use a calculator. A is the point (-5, 1) and B is the point (7, 10). Point D lies on the line AB with AD : DB = 2 : 1. Find the equation of the line perpendicular to AB that passes through D. Give your answer in the form y = mx + c.5 marks
  10. On a map, a straight path joins A at (-3, 2) to B at (5, 6). A second straight path leaves B at right angles to AB and meets the x-axis at C. A straight fence is the perpendicular bisector of BC. Find the equation of the fence in the form y = mx + c.6 marks

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