Printable · GCSE Higher · ages 14-16
Straight-line graphs and y = mx + c worksheet — GCSE Higher
Fifteen questions on "straight-line graphs and y = mx + c" — DfE statement A9. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
part Higher
Straight-line graphs and y = mx + c worksheet — GCSE Higher
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- 1.A water butt holds 200 litres and is being drained at a steady 8 litres per minute. Write down the function for the amount of water y, in litres, left after x minutes.
- 2.Work out the equation of the straight line through the points (−2, 1) and (0, 7).
- 3.A straight line passes through the points (−1, 2) and (3, 14). Work out the equation of the line.
- 4.A straight line has gradient −2 and passes through the point (3, 1). Work out the equation of the line.
- 5.Work out the gradient of a line that is perpendicular to the line with equation 2x + 3y = 6.
- 6.A cycle route is 84 km long. Freya sets off along it at a steady 14 km/h. Write down the function for the distance y, in kilometres, that is still to be cycled after x hours.
- 7.A student is asked whether the lines y = (2/5)x − 1 and 5y = −2x + 15 are perpendicular. The student answers: "Yes, because when I rearrange the second equation I get y = −(2/5)x + 3, and 2/5 times −2/5 is negative." Which of these is a correct comment on the student's reasoning?y = -2x + 15
- 8.A straight line has gradient −3 and passes through the point (0, 1). Work out the value of y when x = 2.
- 9.A straight line has equation y = 4x + 3. A second line is parallel to the first line and passes through the point (0, −5). Work out the equation of the second line.y = 4x + 3
- 10.A garden path runs along the line 4x + y = 12. A drainage pipe must be laid perpendicular to the path, passing through the point (3, 1) where a sprinkler sits. Work out the equation of the pipe's line, giving your answer in the form y = mx + c.
- 11.A gas company charges a standing charge plus a rate per unit used. The total cost, y in pounds, for using x units is given by y = 0.15x + 20. Work out the cost of using 200 units.y = 0.15x + 20
- 12.Which of these pairs of lines are perpendicular to each other?
- 13.Points A(−1, 2) and B(5, 8) are the endpoints of a line segment. Work out the equation of the perpendicular bisector of AB.
- 14.A plumber charges a call-out fee plus an hourly rate. The total cost, y in pounds, of a job lasting x hours is given by y = 45x + 60. Work out the total cost of a job that lasts 3 hours.y = 45x + 60
- 15.Work out the equation of the straight line through the points (−3, 4) and (1, −8).
Answer key
- (c) y = 200 − 8x — Method: in a linear model the amount present at the start is the constant term and the steady rate of change is the gradient, which is negative because the amount is falling. Working: at x = 0 minutes there are 200 litres, so the constant term is 200; 8 litres are lost every minute, so after x minutes 8x litres have gone and the amount left is y = 200 − 8x. Answer: y = 200 − 8x. The distractors: y = 8x + 200 treats the draining as filling, so the butt would gain 8 litres a minute; y = 200x − 8 swaps the two numbers over, using the starting 200 litres as the rate per minute and the 8 litres per minute as the starting amount; y = −8x − 200 makes the starting amount negative as well as the rate, so the butt would begin 200 litres in deficit.
- (c) y = 3x + 7 — Method: the gradient is the change in y divided by the change in x with both taken in the same order, and a point whose x-coordinate is 0 gives the constant straight away because it lies on the y-axis. Working: m = (7 − 1) ÷ (0 − (−2)) = 6 ÷ 2 = 3; the point (0, 7) lies on the y-axis, so c = 7 and the line is y = 3x + 7. Answer: y = 3x + 7. The distractors: y = −3x + 7 comes from taking the y-difference as 1 − 7 while taking the x-difference as 0 − (−2), so the two subtractions run in opposite orders; y = 3x + 1 comes from using the y-coordinate of (−2, 1) as the constant instead of the point that actually lies on the y-axis; y = (1/3)x + 7 comes from writing the gradient upside down as the change in x over the change in y, 2 ÷ 6.
- (c) y = 3x + 5 — Method: find the gradient from the two points, then substitute one point into y = mx + c to find c. Working: gradient = (14 − 2) ÷ (3 − (−1)) = 12 ÷ 4 = 3. Using the point (3, 14): 14 = 3(3) + c, so 14 = 9 + c, giving c = 5. Answer: y = 3x + 5. y = 6x − 4 comes from mishandling the negative x-coordinate, treating 3 − (−1) as 3 − 1 = 2, so the gradient becomes 12 ÷ 2 = 6, and then c = 14 − 18 = −4. y = 3x + 23 comes from a sign error isolating c, adding 9 to 14 instead of subtracting it: c = 14 + 9 = 23. y = x/3 + 13 comes from dividing the change in x by the change in y instead of the other way round, giving a gradient of 4 ÷ 12 = 1/3, and then c = 14 − 1 = 13.
- (a) y = −2x + 7 — Using y = −2x + c, and substituting the point (3, 1): 1 = −2(3) + c, so 1 = −6 + c, and c = 7, giving y = −2x + 7. A candidate who uses the y-coordinate of the point directly as the y-intercept, instead of solving for c, would write y = −2x + 1. A candidate who drops the negative sign on the gradient would write y = 2x + 7. A candidate who makes a sign error when isolating c, writing c = 1 − 6 = −5 instead of c = 1 + 6 = 7, would write y = −2x − 5.
- (b) 3/2 — Rearrange 2x + 3y = 6 into y = mx + c: 3y = −2x + 6, so y = −(2/3)x + 2. The gradient of this line is −2/3. The perpendicular gradient is the negative reciprocal: 3/2. Distractor routes: −1/2 comes from reading the gradient straight off the x-coefficient, 2, without dividing by the y-coefficient, 3, first, then taking its negative reciprocal. −3/2 correctly finds the gradient −2/3 but only takes its reciprocal without also changing the sign, giving −3/2 instead of 3/2. 2/3 comes from negating the gradient −2/3 to 2/3, but forgetting to also take the reciprocal.
- (b) y = 84 − 14x — Method: the distance still to go is the whole route minus the distance already covered, and at a steady speed the distance covered after x hours is the speed multiplied by x. Working: after x hours Freya has cycled 14x km, so the distance remaining is the 84 km route take away 14x, giving y = 84 − 14x, a straight line with intercept 84 and gradient −14. Answer: y = 84 − 14x. The distractors: y = 84 + 14x comes from adding the distance cycled to the length of the route rather than taking it away, so the ride would grow longer the further she goes; y = 14x − 84 comes from carrying out the subtraction the wrong way round, which is negative for the whole of the ride; y = 14 − 84x comes from swapping the two numbers over, treating 84 km as the hourly speed and 14 km as the length of the route.
- (c) Wrong — the product is −4/25, not −1. — The student's rearrangement is correct: 5y = −2x + 15 gives y = −(2/5)x + 3, gradient −2/5. But the test for perpendicularity is that the product of the two gradients equals exactly −1, not merely that it is negative. Here 2/5 × (−2/5) = −4/25, which is not −1, so the lines are NOT perpendicular. Distractor routes: "a negative product always means perpendicular" states a rule that does not exist — many pairs of lines have a negative gradient product without being perpendicular, as this pair shows. "The rearrangement is incorrect" wrongly blames a correct step; 5y = −2x + 15 does rearrange to y = −(2/5)x + 3. "2/5 and −2/5 are negatives of each other" notices a true but irrelevant fact — being negatives of each other is not the perpendicularity condition; an exact product of −1 is.
- (b) −5 — Method: a point whose x-coordinate is 0 lies on the y-axis, so its y-coordinate is the constant c; once m and c are both known the equation can be written down and x substituted into it. Working: the line passes through (0, 1), so c = 1 and the equation is y = −3x + 1; substituting x = 2 gives y = −3 × 2 + 1 = −6 + 1 = −5. Answer: −5. The distractors: 7 comes from ignoring the minus sign on the gradient and working out 3 × 2 + 1; 5 comes from working out 3 × 2 = 6 and then using the minus sign to take the constant off the product, 6 − 1 = 5; −7 comes from subtracting the constant instead of adding it, −6 − 1 = −7.
- (c) y = 4x − 5 — Parallel lines have the same gradient, so the new line has gradient 4; since it passes through (0, −5), its y-intercept is −5, giving y = 4x − 5. A candidate who drops the negative sign on the y-intercept would write y = 4x + 5. A candidate who changes the sign of the gradient, instead of keeping it the same for a parallel line, would write y = −4x − 5. A candidate who confuses m and c, using the y-intercept of the first line (3) as the gradient of the second, would write y = 3x − 5.
- (a) y = (1/4)x + 1/4 — Rearrange 4x + y = 12 to y = −4x + 12, so the path has gradient −4. The perpendicular gradient is 1/4. Substituting (3, 1) into y − 1 = (1/4)(x − 3): y = (1/4)x − 3/4 + 1 = (1/4)x + 1/4. Distractor routes: y = −4x + 13 uses the path's own gradient, −4, instead of the perpendicular gradient, giving a line PARALLEL to the path through (3, 1) rather than perpendicular to it. y = (1/4)x − 1/4 makes an arithmetic slip combining −3/4 and 1, landing on −1/4 instead of the correct 1/4. y = −(1/4)x + 1/4 takes the reciprocal of −4, which is −1/4, but forgets to change its sign, so it is not the true negative reciprocal.
- (a) £50 — Substituting x = 200 into y = 0.15x + 20 gives y = 0.15 × 200 + 20 = 30 + 20 = £50. A candidate who forgets to add the standing charge would get only 0.15 × 200 = £30. A candidate who misplaces the decimal point in the rate, using 1.5 instead of 0.15, would get 1.5 × 200 + 20 = £320. A candidate who swaps the roles of the rate and the number of units would work out 0.15 × 20 + 200 = £203.
- (b) y = 2x and y = −(1/2)x + 1 — Method: two lines are perpendicular when the product of their gradients is −1, so the gradient of each line is read from the number multiplying x and the two are multiplied together. Working: for y = 2x and y = −(1/2)x + 1 the gradients are 2 and −1/2, and 2 × (−1/2) = −1, so that pair is perpendicular. Answer: y = 2x and y = −(1/2)x + 1. The distractors: the pair y = 2x and y = 2x + 1 has equal gradients, whose product is 4, so those two lines are parallel and never meet, and they are chosen by a candidate applying the parallel condition; the pair y = 2x and y = −2x + 1 comes from changing the sign of the gradient without turning it upside down, and its product is −4; the pair y = 2x and y = (1/2)x + 1 comes from turning the gradient upside down without changing its sign, and its product is 1.
- (d) x + y = 7 — The midpoint of AB is (−1 + 5)/2, (2 + 8)/2, which is (2, 5). The gradient of AB is (8 − 2)/(5 − (−1)) = 6/6 = 1. The perpendicular bisector has gradient −1 and passes through (2, 5): y − 5 = −(x − 2), which rearranges to x + y = 7. Distractor routes: x − y = −3 uses the gradient of AB itself, 1, rather than its negative reciprocal, giving a line PARALLEL to AB through the midpoint instead of perpendicular to it. x + y = 12 comes from adding the y-coordinates, 2 + 8 = 10, but forgetting to divide by 2, using the midpoint (2, 10) instead of (2, 5). x + y = 4 comes from using A's x-coordinate, −1, directly instead of averaging it with B's, giving the point (−1, 5) instead of the true midpoint (2, 5).
- (d) £195 — Substituting x = 3 into y = 45x + 60 gives y = 45 × 3 + 60 = 135 + 60 = 195. A candidate who forgets to add the call-out fee would get only 45 × 3 = £135. A candidate who adds the hours to the fee and the rate instead of multiplying would get 45 + 60 + 3 = £108. A candidate who multiplies both the hourly rate and the call-out fee by the number of hours would get 45 × 3 + 60 × 3 = £315.
- (a) y = −3x − 5 — Gradient = (−8 − 4) ÷ (1 − (−3)) = −12 ÷ 4 = −3. Using the point (1, −8): −8 = −3(1) + c, so c = −5, giving y = −3x − 5. A candidate who drops the negative sign on the gradient, using m = 3 instead, would then solve −8 = 3(1) + c to get c = −11, writing y = 3x − 11. A candidate who makes a sign error isolating c, writing c = 5 instead of −5, would write y = −3x + 5. A candidate who mixes up both mistakes — keeping the correct gradient but the wrong, positive value of c from the flipped-gradient calculation — would write y = −3x + 11.
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