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Maths · Graphs

Line segments

The midpoint and length of a line segment from its end points, working back from a midpoint to a missing end - and at Higher, the equation of the perpendicular bisector.

  • 4 key terms
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Last Lesson and Before

Answer each one, then check.

  1. 1

    What is the mean of 2 and 8?

    Show answerHide answer

    5

  2. 2

    Distance between \((1, 1)\) and \((4, 5)\)?

    Show answerHide answer

    5

  3. 3

    Work out \(\frac{-3 + 7}{2}\).

    Show answerHide answer

    2

  4. 4

    (Higher) Gradient perpendicular to 2?

    Show answerHide answer

    \(-\frac{1}{2}\)

Learning Objectives

  1. 1Find the midpoint of a line segment.
  2. 2Find the length of a line segment.
  3. 3Find a missing end point from the midpoint.
  4. 4(Higher) Find the equation of the perpendicular bisector of a line segment.

Midpoints

A line segment is the part of a line between two end points. Its midpoint is exactly halfway along.

  • The mean of the ends

    Midpoint of \((x_1, y_1)\) and \((x_2, y_2)\) is \(\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)\).

  • Example

    Midpoint of \((2, 3)\) and \((8, 11)\) is \(\left(\frac{10}{2}, \frac{14}{2}\right) = (5, 7)\).

  • Negatives

    Take care adding negatives: midpoint of \((-3, 5)\) and \((7, 1)\) is \((2, 3)\).

  • Length

    Use Pythagoras with the horizontal and vertical differences.

Finding the Other End

M\((3, 1)\) is the midpoint of the line segment AB. A is \((-1, 4)\). Find the coordinates of B.

Show the solutionHide the solution
  1. 1 From A to M \(x\): \(-1\) to 3 is \(+4\); \(y\): 4 to 1 is \(-3\)
  2. 2 Do the same again from M to B \(x\): \(3 + 4 = 7\); \(y\): \(1 - 3 = -2\)
  3. 3 Check the midpoint of A and B \(\left(\frac{-1 + 7}{2}, \frac{4 + (-2)}{2}\right) = (3, 1)\)

AnswerB\((7, -2)\)

The Perpendicular Bisector

It passes through the midpoint, and is perpendicular to the segment.

  1. 1 Midpoint

    Find the midpoint of the segment.
  2. 2 Gradient of the segment

    \(\dfrac{\text{change in } y}{\text{change in } x}\).
  3. 3 Perpendicular gradient

    Take the negative reciprocal.
  4. 4 Find \(c\)

    Substitute the midpoint into \(y = mx + c\).

A Perpendicular Bisector

A is \((1, 2)\) and B is \((5, 10)\). Find the equation of the perpendicular bisector of AB.

Show the solutionHide the solution
  1. 1 Midpoint \(\left(\frac{1 + 5}{2}, \frac{2 + 10}{2}\right) = (3, 6)\)
  2. 2 Gradient of AB \(\dfrac{10 - 2}{5 - 1} = \dfrac{8}{4} = 2\)
  3. 3 Perpendicular gradient \(-\dfrac{1}{2}\)
  4. 4 Substitute \((3, 6)\) \(6 = -\dfrac{1}{2} \times 3 + c\), so \(c = 7.5\)

Answer\(y = -\dfrac{1}{2}x + 7.5\), or \(x + 2y = 15\)

Hidden Square

Three corners of a square are \((1, 1)\), \((5, 2)\) and \((4, 6)\). (a) Find the midpoint of each diagonal you can draw. (b) Use the fact that the diagonals of a square bisect each other to find the fourth corner. (c) Find the length of one side.

1. Sketch the three points.

2. Decide which points are opposite corners.

3. Use the shared midpoint.

A good answer shows: (a) The diagonal from \((1, 1)\) to \((4, 6)\) has midpoint \((2.5, 3.5)\). (b) The other diagonal has the same midpoint, so the fourth corner is \((2 \times 2.5 - 5, 2 \times 3.5 - 2) = (0, 5)\). (c) \(\sqrt{4^2 + 1^2} = \sqrt{17} = 4.12\).

Can I...?

  1. 1Find the midpoint of a line segment.
  2. 2Find the length of a line segment.
  3. 3Find an end point from the midpoint and the other end.
  4. 4(Higher) Find the gradient perpendicular to a segment.
  5. 5(Higher) Find the equation of a perpendicular bisector.

Summary & Exam Focus

  • Midpoint: the mean of the \(x\)s and the mean of the \(y\)s.
  • Length: Pythagoras on the differences.
  • Missing end: same step again from the midpoint.
  • (Higher) Perpendicular bisector: midpoint plus negative reciprocal gradient.

Exam focus

M\((3, 1)\) is the midpoint of AB. A has coordinates \((-1, 4)\). Find the coordinates of B. (2 marks) (2 marks)

For a missing end point, don't halve anything: find the step from A to M, then take the same step again from M.

Key terms

The vocabulary this lesson expects you to use. Each one is linked from the first place it appears above.

Line segment
The part of a line between two end points.
Midpoint
The point exactly halfway along a line segment.
Bisect
Cut into two equal parts.
Perpendicular bisector
A line that cuts a segment in half at right angles.

Practice questions

Have a go at each one before you open its answer.

  1. Question 1 Non-calculator 2 marks

    Find the midpoint of the line segment joining \((-3, 5)\) and \((7, 1)\).

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    Model answer

    \(\left(\frac{-3 + 7}{2}, \frac{5 + 1}{2}\right) = (2, 3)\)

    Mark scheme

    • One coordinate correct — B1
    • \((2, 3)\) — B1
  2. Question 2 Non-calculator 4 marks

    A is the point \((2, 3)\) and B is the point \((8, 11)\). (a) Find the midpoint of AB. (b) Work out the length of AB.

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    Model answer

    (a) \((5, 7)\) (b) \(\sqrt{6^2 + 8^2} = \sqrt{100} = 10\)

    Mark scheme

    • (a) \((5, 7)\) — B1
    • (b) Differences 6 and 8 — M1
    • (b) \(6^2 + 8^2\) — M1
    • (b) 10 — A1
  3. Question 3 Non-calculator 2 marks

    M\((3, 1)\) is the midpoint of the line segment AB. A has coordinates \((-1, 4)\). Find the coordinates of B.

    Show answerHide answer

    Model answer

    B = \((2 \times 3 - (-1), 2 \times 1 - 4) = (7, -2)\)

    Mark scheme

    • One coordinate correct — B1
    • \((7, -2)\) — B1
  4. Question 4 Non-calculator · Higher 4 marks

    A is the point \((1, 2)\) and B is the point \((5, 10)\). Find an equation of the perpendicular bisector of AB.

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    Model answer

    Midpoint \((3, 6)\). Gradient of AB \(= 2\), so the perpendicular gradient is \(-\frac{1}{2}\). \(6 = -\frac{1}{2} \times 3 + c\), \(c = 7.5\). \(y = -\frac{1}{2}x + 7.5\)

    Mark scheme

    • Midpoint \((3, 6)\) — P1
    • Gradient of AB = 2 — P1
    • Perpendicular gradient \(-\frac{1}{2}\) used with the midpoint — P1
    • \(y = -\frac{1}{2}x + 7.5\), or \(x + 2y = 15\) — A1

Quick check

  1. What is the midpoint of \((0, 4)\) and \((6, 10)\)?

    1. A\((6, 14)\)
    2. B\((3, 7)\)
    3. C\((3, 3)\)
    4. D\((6, 6)\)
    Show answerHide answer

    B: \((3, 7)\)

    \(\left(\frac{0 + 6}{2}, \frac{4 + 10}{2}\right) = (3, 7)\).

  2. How long is the line segment from \((1, 2)\) to \((13, 7)\)?

    1. A17
    2. B7
    3. C169
    4. D13
    Show answerHide answer

    D: 13

    Differences 12 and 5: \(\sqrt{144 + 25} = \sqrt{169} = 13\).

  3. The midpoint of AB is \((4, 4)\) and A is \((1, 6)\). Where is B?

    1. A\((2.5, 5)\)
    2. B\((5, 10)\)
    3. C\((7, 2)\)
    4. D\((3, -2)\)
    Show answerHide answer

    C: \((7, 2)\)

    From A to M: \(+3\) across, \(-2\) up. Again from M: \((7, 2)\).

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