GCSE Straight-line graphs Practice Questions

Free GCSE Straight-line graphs practice questions with full step-by-step worked solutions. Covers y = mx + c, gradient, y-intercept, negative gradient. Practise exam-style problems and check your method.

y = mx + cgradienty-interceptnegative gradientgradient of 1line through origin
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
A straight line has equation y=3x+2y = 3x + 2. Write down the gradient of the line.
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Worked solution

  1. Compare with y=mx+cy = mx + c

    y=3x+2y = 3x + 2

    In the form y=mx+cy = mx + c, the letter m stands for the gradient and c for the y-intercept.

  2. Identify m

    m=3m = 3

    The number multiplying x is 3, so that is the value of m.

  3. State the gradient

    gradient=3\text{gradient} = 3

    The gradient equals m, so the gradient is 3.

Answer
33
Question 2
1 markeasy
A straight line has equation y=7x−1y = 7x - 1. Write down the gradient.
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Worked solution

  1. Compare with y=mx+cy = mx + c

    y=7x−1y = 7x - 1

    The gradient is the number multiplying x.

  2. Identify m

    m=7m = 7

    x is multiplied by 7.

  3. State the gradient

    gradient=7\text{gradient} = 7

    The gradient is 77.

Answer
77
Question 3
2 marksintermediate
The graph shows the cost of hiring a bike, where xx is the number of hours and yy is the cost in pounds. Use the graph to work out the gradient (the cost per hour).
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Worked solution

  1. Pick two clear points

    (0,5) and (1,8)(0, 5) \text{ and } (1, 8)

    Use points where the line meets grid corners.

  2. Find the rise

    8−5=38 - 5 = 3

    The cost goes up by £33.

  3. Find the run

    1−0=11 - 0 = 1

    The time goes up by 11 hour.

  4. Work out the gradient

    m=31m = \frac{3}{1}

    Gradient = rise over run.

  5. Simplify

    m=3m = 3

    The gradient is 33.

  6. Interpret the rate

    £3 per hour\pounds 3 \text{ per hour}

    The cost increases by £33 for each hour of hire.

Answer
33
Question 4
4 markshard
Find the equation of the line parallel to 2y=6x+42y = 6x + 4 that passes through the point (0,−1)(0, -1).
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Worked solution

  1. Rearrange the given line

    y=3x+2y = 3x + 2

    Divide 2y=6x+42y = 6x + 4 by 22.

  2. Read the gradient

    m=3m = 3

    The given line has gradient 33.

  3. Recall the parallel rule

    m1=m2m_1 = m_2

    The parallel line also has gradient 33.

  4. Set up the new line

    y=3x+cy = 3x + c

    Use gradient 33.

  5. Substitute the point

    −1=3(0)+c-1 = 3(0) + c

    Use x=0x = 0 and y=−1y = -1.

  6. Simplify

    −1=c-1 = c

    33 times 00 is 00.

  7. State c

    c=−1c = -1

    The y-intercept is −1-1.

  8. Write the equation

    y=3x−1y = 3x - 1

    The parallel line is y=3x−1y = 3x - 1.

  9. Check the point

    3(0)−1=−13(0) - 1 = -1

    It passes through (00, −1-1).

  10. Confirm parallel

    gradient 3 in both\text{gradient } 3 \text{ in both}

    Same gradient, different intercept, so they are parallel.

Answer
y=3x−1y = 3x - 1
Question 5
6 markschallenging
A straight line passes through the points (1,4)(1, 4) and (4,13)(4, 13). Find the equation of the line and show that the point (10,31)(10, 31) lies on it.
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Worked solution

  1. Write the gradient formula

    m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}

    Change in y over change in x.

  2. Label the points

    (x1,y1)=(1,4), (x2,y2)=(4,13)(x_1, y_1) = (1, 4), \ (x_2, y_2) = (4, 13)

    Keep the order consistent.

  3. Find the rise

    13−4=913 - 4 = 9

    The y-values rise by 99.

  4. Find the run

    4−1=34 - 1 = 3

    The x-values rise by 33.

  5. Work out the gradient

    m=93=3m = \frac{9}{3} = 3

    The gradient is 33.

  6. Start y=mx+cy = mx + c

    y=3x+cy = 3x + c

    Insert the gradient.

  7. Substitute a point

    4=3(1)+c4 = 3(1) + c

    Use the point (11, 44).

  8. Simplify

    4=3+c4 = 3 + c

    33 times 11 is 33.

  9. Solve for c

    c=1c = 1

    Subtract 33 from both sides.

  10. Write the equation

    y=3x+1y = 3x + 1

    The line is y=3x+1y = 3x + 1.

  11. Check the second point

    3(4)+1=133(4) + 1 = 13

    Substituting x=4x = 4 gives 1313, which matches.

  12. Test the point (1010, 3131)

    x=10x = 10

    Substitute x=10x = 10 into the equation.

  13. Substitute

    y=3(10)+1y = 3(10) + 1

    Replace x with 10.

  14. Evaluate

    y=30+1=31y = 30 + 1 = 31

    This gives y=31y = 31, matching the point.

  15. State the conclusion

    (10,31) lies on y=3x+1(10, 31) \text{ lies on } y = 3x + 1

    The point satisfies the equation, so it is on the line.

Answer
y=3x+1y = 3x + 1

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