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Exam questions · Maths · Graphs

Equations of Straight Lines

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

    Calculate the midpoint of \((-3, -2)\) and \((5, 4)\).

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    Model answer

    \(\left(\dfrac{-3 + 5}{2}, \dfrac{-2 + 4}{2}\right) = (1, 1)\).

    Mark scheme

    • One coordinate correct — M1
    • \((1, 1)\) — A1
  2. 2 Calculate [5 marks]

    The diagram shows a straight line through the points \(A\) and \(B\). (a) Calculate the gradient of \(AB\). [2 marks] (b) Find the equation of \(AB\). [2 marks] (c) Find the coordinates of the midpoint of \(AB\). [1 mark]

    A straight line passing through the points A (0, 6) and B (4, 0).
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    Model answer

    (a) \(\dfrac{0 - 6}{4 - 0} = -\dfrac{3}{2}\). (b) The line crosses the \(y\)-axis at 6, so \(y = -\dfrac{3}{2}x + 6\). (c) \(\left(\dfrac{0 + 4}{2}, \dfrac{6 + 0}{2}\right) = (2, 3)\).

    Mark scheme

    • (a) \(\dfrac{0 - 6}{4 - 0}\) — M1
    • (a) \(-\dfrac{3}{2}\) — A1
    • (b) \(y = -\dfrac{3}{2}x + c\) or \(y = mx + 6\) — M1
    • (b) \(y = -\dfrac{3}{2}x + 6\) — A1
    • (c) \((2, 3)\) — B1
  3. 3 Find [3 marks]

    Find the equation of the line that is parallel to \(y = 4x + 1\) and passes through \((1, 9)\).

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    Model answer

    The gradient is 4, so \(y = 4x + c\). Putting in \((1, 9)\) gives \(9 = 4 + c\), so \(c = 5\) and \(y = 4x + 5\).

    Mark scheme

    • \(y = 4x + c\) — M1
    • \(9 = 4 \times 1 + c\) — M1
    • \(y = 4x + 5\) — A1
  4. 4 Find [3 marks]

    A line has equation \(2x - 3y = 12\). Find the coordinates of the points where the line crosses the \(x\)-axis and the \(y\)-axis.

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    Model answer

    When \(y = 0\), \(2x = 12\) and \(x = 6\), so the point is \((6, 0)\). When \(x = 0\), \(-3y = 12\) and \(y = -4\), so the point is \((0, -4)\).

    Mark scheme

    • \(y = 0\) or \(x = 0\) used — M1
    • \((6, 0)\) — A1
    • \((0, -4)\) — A1
  5. 5 Find [3 marks]

    A line has equation \(3y + x = 6\). Find the gradient of a line that is perpendicular to it.

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    Model answer

    Rearranging, \(3y = -x + 6\), so \(y = -\dfrac{1}{3}x + 2\) and the gradient is \(-\dfrac{1}{3}\). The perpendicular gradient is 3.

    Mark scheme

    • \(y = -\dfrac{1}{3}x + 2\) or the gradient \(-\dfrac{1}{3}\) — M1
    • Negative reciprocal used — M1
    • 3 — A1
  6. 6 Calculate [4 marks]

    \(A\) is the point \((3, -2)\) and \(B\) is the point \((-2, 10)\). (a) Calculate the coordinates of the midpoint of \(AB\). [2 marks] (b) Calculate the length of \(AB\). [2 marks]

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    Model answer

    (a) \(\left(\dfrac{3 + (-2)}{2}, \dfrac{-2 + 10}{2}\right) = (0.5, 4)\). (b) The differences are 5 and 12, so \(AB = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\).

    Mark scheme

    • (a) One coordinate correct — M1
    • (a) \((0.5, 4)\) — A1
    • (b) \(\sqrt{5^2 + 12^2}\) — M1
    • (b) 13 — A1

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 \(c = 3\).