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Exam questions · Maths · Graphs

Equations of Straight Lines

  • 6 exam questions
  • 20 marks
  • 9 quick checks
  1. 1 Work out [2 marks]

    Work out the coordinates of the midpoint of \((-4, 3)\) and \((6, 9)\).

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    Model answer

    \(\left(\dfrac{-4 + 6}{2}, \dfrac{3 + 9}{2}\right) = (1, 6)\).

    Mark scheme

    • One coordinate correct — M1
    • \((1, 6)\) — A1
  2. 2 Work out [6 marks]

    The diagram shows a straight line through the points \(A\) and \(B\). (a) Work out the gradient of \(AB\). [2 marks] (b) Find the equation of \(AB\). [2 marks] (c) Work out the coordinates of the midpoint of \(AB\). [2 marks]

    A straight line passing through the points A (minus 2, 5) and B (4, minus 1).
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    Model answer

    (a) \(\dfrac{-1 - 5}{4 - (-2)} = \dfrac{-6}{6} = -1\). (b) \(y = -x + c\) with \((4, -1)\) gives \(-1 = -4 + c\), so \(c = 3\) and \(y = -x + 3\). (c) \(\left(\dfrac{-2 + 4}{2}, \dfrac{5 + (-1)}{2}\right) = (1, 2)\).

    Mark scheme

    • (a) \(\dfrac{-1 - 5}{4 - (-2)}\) — M1
    • (a) \(-1\) — A1
    • (b) \(y = -x + c\) with a point substituted — M1
    • (b) \(y = -x + 3\) — A1
    • (c) One coordinate correct — M1
    • (c) \((1, 2)\) — A1
  3. 3 Find [3 marks]

    A line has gradient \(-2\) and passes through the point \((3, 1)\). Find the equation of the line.

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    Model answer

    \(y = -2x + c\). Putting in \((3, 1)\) gives \(1 = -6 + c\), so \(c = 7\) and \(y = -2x + 7\).

    Mark scheme

    • \(y = -2x + c\) — M1
    • \(1 = -2 \times 3 + c\) — M1
    • \(y = -2x + 7\) — A1
  4. 4 Show that [2 marks]

    Show that the lines \(2y = 4x + 5\) and \(y = 2x - 3\) are parallel.

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    Model answer

    Dividing the first equation by 2 gives \(y = 2x + 2.5\), which has gradient 2. The second line also has gradient 2, so they are parallel.

    Mark scheme

    • \(y = 2x + 2.5\) or gradient 2 found — M1
    • Both gradients are 2, so they are parallel — Q1
  5. 5 Find [3 marks]

    The line \(P\) has equation \(y = \dfrac{1}{2}x + 1\). The line \(Q\) is perpendicular to \(P\) and passes through \((0, 4)\). Find an equation of \(Q\).

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    Model answer

    The gradient of \(Q\) is \(-2\), because \(\dfrac{1}{2} \times (-2) = -1\). It crosses the \(y\)-axis at 4, so \(y = -2x + 4\).

    Mark scheme

    • Gradient \(-2\) seen — M1
    • \(y = -2x + c\) or \(c = 4\) seen — M1
    • \(y = -2x + 4\) — A1
  6. 6 Work out [4 marks]

    \(A\) is the point \((-1, 2)\) and \(B\) is the point \((5, 10)\). (a) Work out the length of \(AB\). [3 marks] (b) Work out the midpoint of \(AB\). [1 mark]

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    Model answer

    (a) The differences are 6 and 8, so \(AB = \sqrt{6^2 + 8^2} = \sqrt{100} = 10\). (b) The midpoint is \(\left(\dfrac{-1 + 5}{2}, \dfrac{2 + 10}{2}\right) = (2, 6)\).

    Mark scheme

    • (a) Differences 6 and 8 — M1
    • (a) \(\sqrt{6^2 + 8^2}\) — M1
    • (a) 10 — A1
    • (b) \((2, 6)\) — B1

Quick check

  1. 1

    What is the equation of the line through \((2, 1)\) and \((6, 9)\)?

    1. A\(y = 2x + 3\)
    2. B\(y = \dfrac{1}{2}x - 3\)
    3. C\(y = 2x - 3\)
    4. D\(y = -2x - 3\)
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    C: \(y = 2x - 3\)

    The gradient is \(\dfrac{8}{4} = 2\). Then \(1 = 2 \times 2 + c\) gives \(c = -3\).

  2. 2

    What is the midpoint of \((2, 1)\) and \((6, 9)\)?

    1. A\((8, 10)\)
    2. B\((4, 5)\)
    3. C\((2, 4)\)
    4. D\((4, 8)\)
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    B: \((4, 5)\)

    \(\left(\dfrac{2 + 6}{2}, \dfrac{1 + 9}{2}\right) = (4, 5)\).

  3. 3

    A line is parallel to \(y = 5x - 2\). What is its gradient?

    1. A\(5\)
    2. B\(-5\)
    3. C\(-\dfrac{1}{5}\)
    4. D\(\dfrac{1}{5}\)
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    A: \(5\)

    Parallel lines have equal gradients.

  4. 4

    A line has gradient 3 and passes through \((2, 9)\). What is its equation?

    1. A\(y = 3x + 9\)
    2. B\(y = 3x - 3\)
    3. C\(y = 3x + 6\)
    4. D\(y = 3x + 3\)
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    D: \(y = 3x + 3\)

    \(9 = 3 \times 2 + c\) gives \(c = 3\).

  5. 5

    What is the gradient of the line \(2y - 4x = 6\)?

    1. A\(4\)
    2. B\(3\)
    3. C\(2\)
    4. D\(-2\)
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    C: \(2\)

    Rearranged, \(2y = 4x + 6\), so \(y = 2x + 3\).

  6. 6

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

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

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

  7. 7

    Where does the line \(3x + 2y = 12\) cross the axes?

    1. A\((0, 6)\) and \((4, 0)\)
    2. B\((0, 4)\) and \((6, 0)\)
    3. C\((0, 12)\) and \((12, 0)\)
    4. D\((0, 3)\) and \((2, 0)\)
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    A: \((0, 6)\) and \((4, 0)\)

    Put \(x = 0\) to get \(y = 6\), and \(y = 0\) to get \(x = 4\).

  8. 8

    What is the gradient of a line perpendicular to a line with gradient 4?

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

    Perpendicular gradients multiply to \(-1\), so the new gradient is \(-\dfrac{1}{4}\).

  9. 9

    What is the equation of the line perpendicular to \(y = 2x + 3\) through \((4, 1)\)?

    1. A\(y = -2x + 9\)
    2. B\(y = \dfrac{1}{2}x - 1\)
    3. C\(y = -\dfrac{1}{2}x + 3\)
    4. D\(y = -\dfrac{1}{2}x + 1\)
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    C: \(y = -\dfrac{1}{2}x + 3\)

    The gradient is \(-\dfrac{1}{2}\). Then \(1 = -2 + c\), so \(c = 3\).