Hard GCSE Straight-line graphs Questions

Challenging, exam-style GCSE Straight-line graphs questions with worked solutions. Stretch yourself on the hardest equation of a line, two points, parallel lines, negative gradient problems.

equation of a linetwo pointsparallel linesnegative gradientx-interceptsolving equations
GCSE Foundation34 questionsStep-by-step solutions
Question 1
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.
Show worked solution

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
Question 2
6 markschallenging
Find the equation of the line that is parallel to 3x+y=53x + y = 5 and passes through the point (2,1)(2, 1).
Show worked solution

Worked solution

  1. Rearrange the given line

    y=5−3xy = 5 - 3x

    Subtract 3x3x from both sides of 3x+y=53x + y = 5.

  2. Write in mx + c form

    y=−3x+5y = -3x + 5

    Put the x term first.

  3. Read the gradient

    m=−3m = -3

    The given line has gradient −3-3.

  4. Apply the parallel rule

    m=−3m = -3

    The parallel line has the same gradient.

  5. Set up the new line

    y=−3x+cy = -3x + c

    Use gradient −3-3.

  6. Substitute the point

    1=−3(2)+c1 = -3(2) + c

    Use x=2x = 2 and y=1y = 1.

  7. Work out the product

    −3(2)=−6-3(2) = -6

    This gives 1=−6+c1 = -6 + c.

  8. Simplify

    1=−6+c1 = -6 + c

    Now solve for c.

  9. Solve for c

    c=7c = 7

    Add 66 to both sides.

  10. Write the equation

    y=−3x+7y = -3x + 7

    The parallel line is y=−3x+7y = -3x + 7.

  11. Check the point

    −3(2)+7=1-3(2) + 7 = 1

    It passes through (22, 11).

  12. Confirm parallel

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

    Same gradient, so the lines are parallel.

  13. Compare intercepts

    5≠75 \ne 7

    The intercepts differ, so they are distinct lines.

  14. Optional standard form

    3x+y=73x + y = 7

    The answer can also be written as 3x+y=73x + y = 7.

  15. State the answer

    y=−3x+7y = -3x + 7

    The equation of the line is y=−3x+7y = -3x + 7.

Answer
y=−3x+7y = -3x + 7
Question 3
5 markschallenging
A straight line crosses the xx-axis at (4,0)(4, 0) and the yy-axis at (0,8)(0, 8). Find the equation of the line.
Show worked solution

Worked solution

  1. Treat the intercepts as points

    (4,0) and (0,8)(4, 0) \text{ and } (0, 8)

    Both crossings are ordinary points on the line.

  2. Write the gradient formula

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

    Change in y over change in x.

  3. Find the rise

    8−0=88 - 0 = 8

    The y-values rise by 88.

  4. Find the run

    0−4=−40 - 4 = -4

    The x-values fall by 44.

  5. Work out the gradient

    m=8−4=−2m = \frac{8}{-4} = -2

    The gradient is −2-2.

  6. Read the y-intercept

    (0,8)⇒c=8(0, 8) \Rightarrow c = 8

    One point is on the y-axis, so c=8c = 8.

  7. Write the equation

    y=−2x+8y = -2x + 8

    Combine the gradient and intercept.

  8. Check the x-intercept

    −2(4)+8=0-2(4) + 8 = 0

    Substituting x=4x = 4 gives 00, matching (44, 00).

  9. Check the y-intercept

    −2(0)+8=8-2(0) + 8 = 8

    Substituting x=0x = 0 gives 88, matching (00, 88).

  10. Describe the slope

    negative gradient\text{negative gradient}

    The line slopes downwards.

  11. Note the x-intercept

    (4,0)(4, 0)

    The line crosses the x-axis at 44.

  12. Note the y-intercept

    (0,8)(0, 8)

    The line crosses the y-axis at 88.

  13. Confirm both fit

    both points satisfy the equation\text{both points satisfy the equation}

    Both intercepts lie on y=−2x+8y = -2x + 8.

  14. Alternative gradient check

    0−84−0=−2\frac{0 - 8}{4 - 0} = -2

    Using the points in the other order still gives −2-2.

  15. State the answer

    y=−2x+8y = -2x + 8

    The equation of the line is y=−2x+8y = -2x + 8.

Answer
y=−2x+8y = -2x + 8
Question 4
6 markschallenging
Plan A costs y=2x+10y = 2x + 10 pounds and Plan B costs y=4xy = 4x pounds, where xx is the number of minutes used. For how many minutes do the two plans cost the same?
Show worked solution

Worked solution

  1. Write both cost equations

    A:y=2x+10, B:y=4xA: y = 2x + 10, \ B: y = 4x

    Plan A has a fixed fee; Plan B does not.

  2. Set the costs equal

    2x+10=4x2x + 10 = 4x

    The plans cost the same when their y-values match.

  3. Collect the x terms

    10=2x10 = 2x

    Subtract 2x2x from both sides.

  4. Divide by 22

    x=5x = 5

    Solve for x.

  5. State the units

    5 minutes5 \text{ minutes}

    x is measured in minutes.

  6. Find the cost for A

    2(5)+10=202(5) + 10 = 20

    Plan A costs £2020 at 55 minutes.

  7. Find the cost for B

    4(5)=204(5) = 20

    Plan B also costs £20 at 5 minutes.

  8. Confirm they match

    20=2020 = 20

    Both plans cost £2020, so 55 minutes is correct.

  9. Read the gradient of A

    m=2m = 2

    Plan A costs £22 per minute.

  10. Read the gradient of B

    m=4m = 4

    Plan B costs £4 per minute.

  11. Compare fixed costs

    A:10, B:0A: 10, \ B: 0

    Plan A has a £10 fixed fee; Plan B has none.

  12. Check below 55 minutes

    x=0:A=10,B=0x = 0: A = 10, B = 0

    For short calls Plan B is cheaper.

  13. Check above 55 minutes

    x=10:A=30,B=40x = 10: A = 30, B = 40

    For long calls Plan A is cheaper.

  14. Interpret the crossover

    x=5 is break-evenx = 5 \text{ is break-even}

    55 minutes is where the two plans are equal.

  15. State the answer

    55

    The two plans cost the same at 55 minutes.

Answer
55
Question 5
6 markschallenging
A straight line passes through the points (2,1)(2, 1) and (6,3)(6, 3). Find the equation of the line.
Show worked solution

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)=(2,1), (x2,y2)=(6,3)(x_1, y_1) = (2, 1), \ (x_2, y_2) = (6, 3)

    Keep the order consistent.

  3. Find the rise

    3−1=23 - 1 = 2

    The y-values rise by 22.

  4. Find the run

    6−2=46 - 2 = 4

    The x-values rise by 44.

  5. Work out the gradient

    m=24=12m = \frac{2}{4} = \dfrac{1}{2}

    The gradient simplifies to one half.

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

    y=12x+cy = \dfrac{1}{2}x + c

    Insert the gradient.

  7. Substitute a point

    1=12(2)+c1 = \dfrac{1}{2}(2) + c

    Use the point (22, 11).

  8. Work out the product

    12(2)=1\dfrac{1}{2}(2) = 1

    Half of 22 is 11.

  9. Simplify

    1=1+c1 = 1 + c

    This leaves 1=1+c1 = 1 + c.

  10. Solve for c

    c=0c = 0

    Subtract 11 from both sides.

  11. Write the equation

    y=12xy = \dfrac{1}{2}x

    With c=0c = 0, the line is y=(1/2)xy = (1/2)x.

  12. Check the other point

    12(6)=3\dfrac{1}{2}(6) = 3

    Substituting x=6x = 6 gives 33, which matches.

  13. Note it passes through the origin

    (0,0)(0, 0)

    Since c=0c = 0, the line goes through the origin.

  14. Interpret the gradient

    up 1 across 2\text{up 1 across 2}

    The line rises 11 for every 22 across.

  15. State the answer

    y=12xy = \dfrac{1}{2}x

    The equation of the line is y=(1/2)xy = (1/2)x.

Answer
y=12xy = \frac{1}{2}x

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