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Maths · Graphs

More linear graphs

Find the equation of a line through two points, write the equation of a line parallel to another, and at Higher tier use the perpendicular gradient rule.

  • 6 key terms
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Warm-up

Answer each one, then check.

  1. 1

    What is the gradient of \(y = 5x - 2\)?

    Show answerHide answer

    5

  2. 2

    Work out \(\dfrac{9 - 3}{4 - 1}\).

    Show answerHide answer

    2

  3. 3

    Solve \(7 = 2 \times 3 + c\).

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    \(c = 1\)

  4. 4

    What is the equation of the y-axis?

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    \(x = 0\)

  5. 5

    Work out \(-1 \div 2\).

    Show answerHide answer

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

Learning Objectives

  1. 1Find the equation of a line through two points.
  2. 2Find the equation of a line parallel to a given line.
  3. 3Find the equation of a perpendicular line (Higher).
  4. 4Find where two lines meet.

Line Through Two Points

Gradient first, then the y-intercept.

  1. 1 Find the gradient

    \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\).
  2. 2 Write \(y = mx + c\)

    Put in the value of \(m\).
  3. 3 Substitute one point

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

    And check with the other point.

Through Two Points

Find the equation of the line through \((1, 3)\) and \((4, 12)\).

Show the solutionHide the solution
  1. 1 Gradient \(\dfrac{12 - 3}{4 - 1} = \dfrac{9}{3} = 3\)
  2. 2 Substitute \((1, 3)\) into \(y = 3x + c\) \(3 = 3 + c\)
  3. 3 Solve for c \(c = 0\)
  4. 4 Write the equation \(y = 3x\)

Answer\(y = 3x\)

Parallel and Perpendicular

Parallel lines

  • They never meet.
  • They have the same gradient.
  • \(y = 2x + 1\) and \(y = 2x - 3\) are parallel.

Perpendicular lines (Higher)

  • They meet at a right angle.
  • The gradients multiply to give \(-1\): \(m_1 \times m_2 = -1\).
  • \(y = 2x + 1\) and \(y = -\dfrac{1}{2}x + 4\) are perpendicular.

A Parallel Line

Find the equation of the line parallel to \(y = 2x + 3\) that passes through \((1, 7)\).

Show the solutionHide the solution
  1. 1 Parallel lines have the same gradient \(m = 2\)
  2. 2 Substitute \((1, 7)\) into \(y = 2x + c\) \(7 = 2 + c\)
  3. 3 Solve \(c = 5\)

Answer\(y = 2x + 5\)

A Perpendicular Line

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

Show the solutionHide the solution
  1. 1 The gradient of the original is 2, so the perpendicular gradient is \(-\dfrac{1}{2}\)
  2. 2 Substitute \((4, 3)\) into \(y = -\dfrac{1}{2}x + c\) \(3 = -2 + c\)
  3. 3 Solve \(c = 5\)

Answer\(y = -\dfrac{1}{2}x + 5\)

Where Two Lines Meet

Find the coordinates of the point where \(y = x + 1\) and \(y = -x + 5\) cross.

Show the solutionHide the solution
  1. 1 At the crossing point, the y-values are equal \(x + 1 = -x + 5\)
  2. 2 Solve \(2x = 4\), so \(x = 2\)
  3. 3 Substitute to find y \(y = 2 + 1 = 3\)

Answer\((2, 3)\)

Finding the Perpendicular Gradient

Flip the fraction and change the sign.

  • 2

    Perpendicular gradient: \(-\dfrac{1}{2}\)

  • \(\dfrac{3}{4}\)

    Perpendicular gradient: \(-\dfrac{4}{3}\)

  • \(-5\)

    Perpendicular gradient: \(\dfrac{1}{5}\)

  • \(-\dfrac{2}{7}\)

    Perpendicular gradient: \(\dfrac{7}{2}\)

Which Lines Are Parallel?

Sort these lines into parallel pairs and find one line perpendicular to \(y = 2x\): \(y = 3x + 1\), \(y = 3x - 4\), \(2y = x + 6\), \(y = \dfrac{1}{2}x - 1\), \(y = -\dfrac{1}{2}x + 3\).

1. Rearrange each equation.

2. Compare gradients.

A good answer shows: \(y = 3x + 1\) and \(y = 3x - 4\) are parallel. \(2y = x + 6\) is \(y = \dfrac{1}{2}x + 3\), which is parallel to \(y = \dfrac{1}{2}x - 1\). \(y = -\dfrac{1}{2}x + 3\) is perpendicular to \(y = 2x\).

Can I...?

  1. 1Find the gradient from two points.
  2. 2Find the equation through two points.
  3. 3Write a line parallel to another.
  4. 4Recognise parallel lines from their equations.
  5. 5Find the perpendicular gradient (Higher).
  6. 6Find a perpendicular line through a point (Higher).
  7. 7Find where two lines cross.
  8. 8Check my equation with the other point.

Summary & Exam Focus

  • Two points: find the gradient, then substitute one point to find \(c\).
  • Parallel lines have equal gradients.
  • Perpendicular gradients multiply to \(-1\) (Higher).
  • Where lines cross: set the y-expressions equal.

Exam focus

Find the equation of the line that is parallel to \(y = 3x - 2\) and passes through the point \((2, 9)\). (3 marks) (3 marks)

Parallel means the same gradient, so write \(y = 3x + c\) and substitute the point to find \(c\).

Key terms

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

Parallel
Lines that never meet, with equal gradients.
Perpendicular
Lines that meet at a right angle.
Intersection
The point where two lines cross.
Reciprocal
A number turned upside down: the reciprocal of 2 is \(\dfrac{1}{2}\).
Simultaneous
Happening together; the intersection satisfies both equations.
Coordinates
A pair \((x, y)\) giving a position.

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 straight line that passes through the points \((1, 3)\) and \((4, 12)\).

    Show answerHide answer

    Model answer

    Gradient \(= \dfrac{12 - 3}{4 - 1} = 3\). Substituting \((1, 3)\): \(3 = 3 + c\), so \(c = 0\). The equation is \(y = 3x\).

    Mark scheme

    • Gradient of 3 — M1
    • Substituting a point — M1
    • \(y = 3x\) — A1
  2. Question 2 Non-calculator 3 marks

    Find the equation of the line that is parallel to \(y = 3x - 2\) and passes through the point \((2, 9)\).

    Show answerHide answer

    Model answer

    The gradient is 3, so \(y = 3x + c\). \(9 = 6 + c\), so \(c = 3\). The equation is \(y = 3x + 3\).

    Mark scheme

    • Gradient 3 — M1
    • \(9 = 3 \times 2 + c\) — M1
    • \(y = 3x + 3\) — A1
  3. Question 3 Non-calculator 2 marks

    Are the lines \(y = 4x + 1\) and \(2y = 8x - 6\) parallel? Explain your answer.

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    Model answer

    \(2y = 8x - 6\) rearranges to \(y = 4x - 3\). Both lines have gradient 4, so they are parallel.

    Mark scheme

    • Rearranging to \(y = 4x - 3\) — M1
    • Same gradient, so parallel — C1
  4. Question 4 Non-calculator 3 marks

    Find the coordinates of the point where the lines \(y = x + 1\) and \(y = -x + 5\) cross.

    Show answerHide answer

    Model answer

    \(x + 1 = -x + 5\), so \(2x = 4\) and \(x = 2\). Then \(y = 3\). The point is \((2, 3)\).

    Mark scheme

    • Setting the expressions equal — M1
    • \(x = 2\) — A1
    • \((2, 3)\) — A1
  5. Question 5 Non-calculator 4 marks

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

    Show answerHide answer

    Model answer

    The perpendicular gradient is \(-\dfrac{1}{2}\). \(3 = -\dfrac{1}{2} \times 4 + c\), so \(c = 5\). The equation is \(y = -\dfrac{1}{2}x + 5\).

    Mark scheme

    • Perpendicular gradient \(-\dfrac{1}{2}\) — M1
    • \(3 = -\dfrac{1}{2} \times 4 + c\) — M1
    • \(c = 5\) — A1
    • \(y = -\dfrac{1}{2}x + 5\) — A1
  6. Question 6 Non-calculator 3 marks

    The line \(L_1\) has equation \(3y = 6x - 2\). The line \(L_2\) is perpendicular to \(L_1\). Work out the gradient of \(L_2\).

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    Model answer

    \(y = 2x - \dfrac{2}{3}\), so the gradient of \(L_1\) is 2. The gradient of \(L_2\) is \(-\dfrac{1}{2}\).

    Mark scheme

    • Rearranging to \(y = \dots\) — M1
    • Gradient of \(L_1\) is 2 — A1
    • \(-\dfrac{1}{2}\) — A1

Quick check

  1. Which line is parallel to \(y = 5x + 2\)?

    1. A\(y = 2x + 5\)
    2. B\(y = 5x - 7\)
    3. C\(y = -5x + 2\)
    4. D\(y = \dfrac{1}{5}x\)
    Show answerHide answer

    B: \(y = 5x - 7\)

    Parallel lines have the same gradient: 5.

  2. A line passes through \((0, 2)\) and \((3, 8)\). What is its equation?

    1. A\(y = 3x + 2\)
    2. B\(y = 6x + 2\)
    3. C\(y = 2x + 2\)
    4. D\(y = 2x + 8\)
    Show answerHide answer

    C: \(y = 2x + 2\)

    Gradient \(\dfrac{6}{3} = 2\) and y-intercept 2, so \(y = 2x + 2\).

  3. The lines \(y = x + 4\) and \(y = 2x + 1\) meet where...

    1. A\(x = 3\)
    2. B\(x = 5\)
    3. C\(x = 1\)
    4. DThey never meet
    Show answerHide answer

    A: \(x = 3\)

    \(x + 4 = 2x + 1\), so \(x = 3\) and \(y = 7\).

  4. What is the gradient of a line perpendicular to \(y = 4x + 1\)?

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

    D: \(-\dfrac{1}{4}\)

    The perpendicular gradient is the negative reciprocal: \(-\dfrac{1}{4}\).

  5. Which pair of lines are parallel?

    1. A\(y = 3x\) and \(y = -3x\)
    2. B\(2y = 6x + 1\) and \(y = 3x - 8\)
    3. C\(y = x + 1\) and \(y = 2x + 1\)
    4. D\(y = 2x\) and \(y = \dfrac{1}{2}x\)
    Show answerHide answer

    B: \(2y = 6x + 1\) and \(y = 3x - 8\)

    \(2y = 6x + 1\) is \(y = 3x + \dfrac{1}{2}\), parallel to \(y = 3x - 8\).

  6. The lines \(y = 2x + 3\) and \(y = 2x - 1\) ...

    1. AMeet at \((2, 7)\)
    2. BMeet at \((-1, 1)\)
    3. CNever meet
    4. DAre perpendicular
    Show answerHide answer

    C: Never meet

    They have the same gradient and different intercepts, so they never meet.

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