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Maths · Graphs

More linear graphs

Finding the equation of a line from a point and a gradient or from two points, spotting parallel lines, drawing lines like \(3x + 2y = 12\) - and at Higher, perpendicular lines.

  • 4 key terms
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Last Lesson and Before

Answer each one, then check.

  1. 1

    Last lesson: what is the gradient of \(y = 7 - 2x\)?

    Show answerHide answer

    \(-2\)

  2. 2

    Last lesson: gradient between \((0, 1)\) and \((2, 9)\)?

    Show answerHide answer

    4

  3. 3

    Solve \(5 = 3 \times 2 + c\).

    Show answerHide answer

    \(c = -1\)

  4. 4

    What is the reciprocal of 4?

    Show answerHide answer

    \(\frac{1}{4}\)

Learning Objectives

  1. 1Find the equation of a line from its gradient and one point.
  2. 2Find the equation of a line through two points.
  3. 3Recognise and find parallel lines.
  4. 4Draw a line given in the form \(ax + by = c\).
  5. 5(Higher) Recognise and find perpendicular lines.

Parallel Lines

Parallel lines have the same gradient.

  • Same \(m\)

    \(y = 3x + 1\), \(y = 3x - 4\) and \(y = 3x\) are all parallel.

  • Different \(c\)

    Parallel lines cross the \(y\)-axis at different points.

  • Rearrange to compare

    \(2y = 6x + 5\) is \(y = 3x + 2.5\), so it is parallel to \(y = 3x + 1\).

  • Same line

    If \(m\) and \(c\) are both the same, it is the same line.

Equation from a Point and a Gradient

Write \(y = mx + c\), then use the point to find \(c\).

  1. 1 Gradient

    Write down \(m\) - given, or from a parallel line, or worked out from two points.
  2. 2 Substitute

    Put the point's \(x\) and \(y\) into \(y = mx + c\).
  3. 3 Solve

    Solve for \(c\).
  4. 4 Write

    Write the equation in full: \(y = mx + c\) with numbers.

A Parallel Line through a Point

Find the equation of the line parallel to \(y = 3x - 4\) that passes through \((2, 5)\).

Show the solutionHide the solution
  1. 1 Parallel, so the same gradient \(m = 3\), so \(y = 3x + c\)
  2. 2 Substitute \((2, 5)\) \(5 = 3 \times 2 + c\)
  3. 3 Solve \(c = -1\)

Answer\(y = 3x - 1\)

A Line through Two Points

Find the equation of the line through \((1, 4)\) and \((3, 10)\).

Show the solutionHide the solution
  1. 1 Gradient \(\dfrac{10 - 4}{3 - 1} = \dfrac{6}{2} = 3\)
  2. 2 Substitute \((1, 4)\) into \(y = 3x + c\) \(4 = 3 + c\), so \(c = 1\)
  3. 3 Check with the other point \(3 \times 3 + 1 = 10\) - correct

Answer\(y = 3x + 1\)

Drawing 3x + 2y = 12

Find where \(3x + 2y = 12\) crosses the axes, and find its gradient.

Show the solutionHide the solution
  1. 1 Crosses the \(y\)-axis when \(x = 0\) \(2y = 12\), so \((0, 6)\)
  2. 2 Crosses the \(x\)-axis when \(y = 0\) \(3x = 12\), so \((4, 0)\)
  3. 3 Rearrange for the gradient \(2y = -3x + 12\), so \(y = -1.5x + 6\)

AnswerThrough \((0, 6)\) and \((4, 0)\); gradient \(-1.5\)

Perpendicular Gradients

Flip the gradient and change its sign.

  • Negative reciprocal

    Perpendicular to gradient \(m\) is gradient \(-\dfrac{1}{m}\).

  • Examples

    2 goes to \(-\frac{1}{2}\); \(-3\) goes to \(\frac{1}{3}\); \(\frac{2}{5}\) goes to \(-\frac{5}{2}\).

  • The test

    Two lines are perpendicular if \(m_1 \times m_2 = -1\).

  • Then as before

    Substitute the point to find \(c\).

A Perpendicular Line through a Point

Find the equation of the line perpendicular to \(y = 2x + 3\) that passes through \((4, 1)\).

Show the solutionHide the solution
  1. 1 Perpendicular gradient \(m = -\dfrac{1}{2}\)
  2. 2 Substitute \((4, 1)\) \(1 = -\dfrac{1}{2} \times 4 + c = -2 + c\)
  3. 3 Solve \(c = 3\)

Answer\(y = -\dfrac{1}{2}x + 3\)

Build a Square

The line \(y = 2x\) passes through \((0, 0)\) and \((1, 2)\). Find the equations of three more lines that make a square with it, with \((0, 0)\) and \((1, 2)\) as two neighbouring corners. Check your square on squared paper.

1. Parallel sides share a gradient.

2. Neighbouring sides are perpendicular.

3. Check each corner lies on two lines.

A good answer shows: Sides through the origin and \((1, 2)\) must be perpendicular to \(y = 2x\): \(y = -\frac{1}{2}x\) and \(y = -\frac{1}{2}x + 2.5\). The fourth side is parallel to \(y = 2x\) through \((-2, 1)\): \(y = 2x + 5\). (The square on the other side gives \(y = 2x - 5\) instead.)

Can I...?

  1. 1Find the equation of a line from a point and a gradient.
  2. 2Find the equation of a line through two points.
  3. 3Tell whether lines are parallel.
  4. 4Draw a line in the form \(ax + by = c\).
  5. 5(Higher) Find a perpendicular gradient.
  6. 6(Higher) Find the equation of a perpendicular line.

Summary & Exam Focus

  • Parallel lines: same gradient.
  • Find \(c\) by substituting a point into \(y = mx + c\).
  • \(ax + by = c\): set \(x = 0\) then \(y = 0\) to find where it crosses the axes.
  • (Higher) Perpendicular gradients multiply to \(-1\).

Exam focus

Find the equation of the line that is parallel to \(y = 3x - 4\) and passes through \((2, 5)\). (3 marks) (3 marks)

Always check your final equation by substituting the point back in. If the \(y\) doesn't come out right, you've made a slip finding \(c\).

Key terms

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

Parallel
Lines with the same gradient that never meet.
Perpendicular
Lines that meet at a right angle.
Negative reciprocal
\(-\dfrac{1}{m}\): the gradient of a line perpendicular to one with gradient \(m\).
Intercept
Where a line crosses an axis.

Practice questions

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

  1. Question 1 Non-calculator 3 marks

    Find the equation of the line that is parallel to \(y = 3x - 4\) and passes through the point \((2, 5)\).

    Show answerHide answer

    Model answer

    \(m = 3\). \(5 = 3 \times 2 + c\), so \(c = -1\). \(y = 3x - 1\)

    Mark scheme

    • Gradient 3 — M1
    • Substituting \((2, 5)\) into \(y = 3x + c\) — M1
    • \(y = 3x - 1\) — A1
  2. Question 2 Non-calculator 3 marks

    Find the equation of the straight line that passes through \((1, 4)\) and \((3, 10)\).

    Show answerHide answer

    Model answer

    Gradient \(= \dfrac{10 - 4}{3 - 1} = 3\). \(4 = 3 + c\), \(c = 1\). \(y = 3x + 1\)

    Mark scheme

    • Gradient 3 — M1
    • Correct method for \(c\) — M1
    • \(y = 3x + 1\) — A1
  3. Question 3 Non-calculator 2 marks

    Show that the line \(2y = 6x + 5\) is parallel to the line \(y = 3x + 1\).

    Show answerHide answer

    Model answer

    \(2y = 6x + 5\) gives \(y = 3x + 2.5\). Both lines have gradient 3, so they are parallel.

    Mark scheme

    • Rearranging to \(y = 3x + 2.5\) — M1
    • Both gradients 3, so parallel — C1
  4. Question 4 Non-calculator · Higher 3 marks

    Find the equation of the line that is perpendicular to \(y = 2x + 3\) and passes through \((4, 1)\).

    Show answerHide answer

    Model answer

    Perpendicular gradient \(-\frac{1}{2}\). \(1 = -\frac{1}{2} \times 4 + c\), so \(c = 3\). \(y = -\frac{1}{2}x + 3\)

    Mark scheme

    • Gradient \(-\frac{1}{2}\) — M1
    • Substituting \((4, 1)\) — M1
    • \(y = -\frac{1}{2}x + 3\), or \(x + 2y = 6\) — A1

Quick check

  1. Which line is parallel to \(y = 4x + 1\)?

    1. A\(y = x + 4\)
    2. B\(y = -4x + 1\)
    3. C\(y = 4x - 7\)
    4. D\(y = \frac{1}{4}x + 1\)
    Show answerHide answer

    C: \(y = 4x - 7\)

    Parallel lines have the same gradient: 4.

  2. Where does \(2x + 5y = 20\) cross the \(x\)-axis?

    1. A\((0, 4)\)
    2. B\((10, 0)\)
    3. C\((4, 0)\)
    4. D\((20, 0)\)
    Show answerHide answer

    B: \((10, 0)\)

    On the \(x\)-axis \(y = 0\), so \(2x = 20\) and \(x = 10\).

  3. (Higher) What is the gradient of a line perpendicular to \(y = -\frac{2}{3}x + 5\)?

    1. A\(\frac{3}{2}\)
    2. B\(-\frac{3}{2}\)
    3. C\(\frac{2}{3}\)
    4. D\(-5\)
    Show answerHide answer

    A: \(\frac{3}{2}\)

    The negative reciprocal of \(-\frac{2}{3}\) is \(\frac{3}{2}\).

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