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Exam questions · Maths · Graphs

Straight-Line Graphs

  • 6 exam questions
  • 17 marks
  • 9 quick checks
  1. 1 Write down [2 marks]

    Write down the equation of the line with gradient 3 that crosses the \(y\)-axis at \((0, -4)\).

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    Model answer

    Using \(y = mx + c\) with \(m = 3\) and \(c = -4\), the equation is \(y = 3x - 4\).

    Mark scheme

    • \(y = 3x + c\) or \(y = mx - 4\) — B1
    • \(y = 3x - 4\) — B1
  2. 2 Complete [3 marks]

    (a) Complete the table of values for \(y = 2x + 3\). \(x = -2, -1, 0, 1, 2\) [2 marks] (b) Does the point \((6, 15)\) lie on the line \(y = 2x + 3\)? Show how you know. [1 mark]

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    Model answer

    (a) The values are \(-1, 1, 3, 5, 7\). (b) When \(x = 6\), \(y = 2 \times 6 + 3 = 15\), so the point lies on the line.

    Mark scheme

    • (a) At least three correct values — M1
    • (a) \(-1, 1, 3, 5, 7\) — A1
    • (b) Yes, with \(2 \times 6 + 3 = 15\) — B1
  3. 3 Work out [4 marks]

    The diagram shows a straight line through the points \(P\) and \(Q\). (a) Work out the gradient of the line. [2 marks] (b) Write down the equation of the line. [2 marks]

    A straight line sloping downwards through the points P (2, 3) and Q (6, 1), crossing the y-axis at 4.
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    Model answer

    (a) \(\dfrac{1 - 3}{6 - 2} = \dfrac{-2}{4} = -\dfrac{1}{2}\). (b) The line crosses the \(y\)-axis at 4, so \(y = -\dfrac{1}{2}x + 4\).

    Mark scheme

    • (a) \(\dfrac{1 - 3}{6 - 2}\) — M1
    • (a) \(-\dfrac{1}{2}\) — A1
    • (b) \(y = -\dfrac{1}{2}x + c\) or \(y = mx + 4\) — M1
    • (b) \(y = -\dfrac{1}{2}x + 4\) — A1
  4. 4 Show that [3 marks]

    Line \(A\) has equation \(y = 2x - 1\). Line \(B\) passes through \((0, 5)\) and \((2, 9)\). Show that lines \(A\) and \(B\) are parallel.

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    Model answer

    The gradient of line \(B\) is \(\dfrac{9 - 5}{2 - 0} = 2\). The gradient of line \(A\) is also 2. Parallel lines have the same gradient, so the lines are parallel.

    Mark scheme

    • \(\dfrac{9 - 5}{2 - 0}\) — M1
    • Gradient of \(B\) is 2 — A1
    • States that the gradients are equal so the lines are parallel — Q1
  5. 5 Write down [2 marks]

    The lines \(x = 3\) and \(y = -2\) cross at a point. (a) Write down the coordinates of the point. [1 mark] (b) Which of the two lines is vertical? [1 mark]

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    Model answer

    (a) Where \(x = 3\) and \(y = -2\), the point is \((3, -2)\). (b) The line \(x = 3\) is vertical.

    Mark scheme

    • (a) \((3, -2)\) — B1
    • (b) \(x = 3\) — B1
  6. 6 Work out [3 marks]

    A line has equation \(2x + y = 10\). (a) Work out its gradient. [2 marks] (b) Write down the coordinates of the point where it crosses the \(x\)-axis. [1 mark]

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    Model answer

    (a) \(y = -2x + 10\), so the gradient is \(-2\). (b) When \(y = 0\), \(2x = 10\) and \(x = 5\), so the point is \((5, 0)\).

    Mark scheme

    • (a) \(y = -2x + 10\) — M1
    • (a) \(-2\) — A1
    • (b) \((5, 0)\) — B1

Quick check

  1. 1

    What is the equation of the \(x\)-axis?

    1. A\(x = 0\)
    2. B\(y = 0\)
    3. C\(y = x\)
    4. D\(x + y = 0\)
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    B: \(y = 0\)

    Every point on the \(x\)-axis has \(y = 0\).

  2. 2

    Which point is on the line \(y = 3x - 2\)?

    1. A\((4, 10)\)
    2. B\((2, 6)\)
    3. C\((1, 3)\)
    4. D\((0, 2)\)
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    A: \((4, 10)\)

    Put in the coordinates: \(3 \times 4 - 2 = 10\), so \((4, 10)\) fits.

  3. 3

    Work out the gradient of the line through \((1, 7)\) and \((4, 1)\).

    1. A\(2\)
    2. B\(-\dfrac{1}{2}\)
    3. C\(-6\)
    4. D\(-2\)
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    D: \(-2\)

    \(\dfrac{1 - 7}{4 - 1} = \dfrac{-6}{3} = -2\).

  4. 4

    Which line is parallel to \(y = 3x + 1\)?

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

    Parallel lines have the same gradient, 3.

  5. 5

    What is the equation of the vertical line through 4 on the \(x\)-axis?

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

    Every point on the line has \(x = 4\).

  6. 6

    What is the gradient of the line \(y = 5 - 3x\)?

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

    Written as \(y = -3x + 5\), the number multiplying \(x\) is \(-3\).

  7. 7

    What is the value of \(y\) on the line \(y = 2x - 1\) when \(x = -1\)?

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

    \(2 \times (-1) - 1 = -2 - 1 = -3\).

  8. 8

    Where does the line \(y = 3x + 2\) cross the \(y\)-axis?

    1. A\((2, 0)\)
    2. B\((0, 3)\)
    3. C\((0, 2)\)
    4. D\((0, -2)\)
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    C: \((0, 2)\)

    The number on its own, 2, is the \(y\)-intercept.

  9. 9

    Work out the gradient of the line through \((-2, 3)\) and \((4, -9)\).

    1. A\(2\)
    2. B\(-2\)
    3. C\(-\dfrac{1}{2}\)
    4. D\(-6\)
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    B: \(-2\)

    \(\dfrac{-9 - 3}{4 - (-2)} = \dfrac{-12}{6} = -2\).