Year 10 straight-line graphs worksheet answers to download as PDF

These Year 10 straight-line graphs worksheet answers give a full worked solution to all 20 questions of the straight-line graphs worksheet. Try each question yourself first, then check your method step by step.

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Answer to Question 1
1. \(2(-1)-3=-5\), \(-3\), \(-1\), \(1\), \(3\). So −5, −3, −1, 1, 3. Substitute each \(x\) value into \(2x-3\). The \(y\) values go up by 2 each time, which is the gradient.
2. Gradient 2, \(y\)-intercept −3 (the point \((0,-3)\)). In \(y=mx+c\) the number in front of \(x\) is the gradient and the constant is the \(y\)-intercept.
3. \(y=0\): \(2x=3\), so \((1.5,0)\). Every point on the \(x\)-axis has \(y=0\), so we solve \(2x-3=0\).
4. \(2\times5-3=7\ne8\), so no. Substitute \(x=5\): the line gives \(y=7\), not 8, so \((5,8)\) is not on the line.
Answer to Question 2
1. \(\dfrac{12-3}{4-1}=\dfrac93=\) 3 The \(y\) value goes up by 9 while the \(x\) value goes up by 3, so the line rises 3 for each 1 across.
2. \(\dfrac{-3-5}{2-(-2)}=\dfrac{-8}{4}=\) −2 Be careful with the signs: \(2-(-2)=4\). A negative gradient means the line slopes down from left to right.
3. \(\dfrac{4-4}{5-0}=\) 0 (a horizontal line) Both points have the same \(y\) value, so the line is horizontal and its gradient is 0.
4. The change in \(x\) is \(0\), so the gradient would involve dividing by zero: it is undefined. The equation is \(x=3\).
Answer to Question 3
1. \(y=3x-2\) The gradient is \(m=3\), and the line passes through \((0,-2)\), so the \(y\)-intercept is \(c=-2\).
2. \(y=-\tfrac12x+c\): \(1=-2+c\), \(c=3\). \(y=-\tfrac12x+3\) Substitute the point \((4,1)\) into \(y=-\tfrac12x+c\) to find \(c\). Check: \(-\tfrac12\times4+3=1\).
3. \(m=\dfrac{11-5}{3-1}=3\), then \(5=3+c\), \(c=2\). \(y=3x+2\) Find the gradient first, then substitute one of the points to find \(c\). Check with the other point: \(3\times3+2=11\).
4. \(m=\dfrac{0-6}{3}=-2\) and \(c=6\). \(y=-2x+6\) The point \((0,6)\) is on the \(y\)-axis, so \(c=6\). Check: \(-2\times3+6=0\).
Answer to Question 4
1. \(y=3x+5\) and \(2y=6x+1\) (which is \(y=3x+0.5\)) have gradient 3. Parallel lines have the same gradient. Rearrange \(2y=6x+1\) by dividing by 2 before comparing.
2. \(y=3x+c\): \(7=6+c\), \(c=1\). \(y=3x+1\) Use the same gradient 3 and substitute the point \((2,7)\) to find \(c\). Check: \(3\times2+1=7\).
3. \(y=-2x+c\): \(1=-2+c\), \(c=3\). \(y=-2x+3\) Use the gradient \(-2\) and substitute \((1,1)\). Check: \(-2+3=1\).
4. \(y=-3x+9\), gradient −3. Subtract \(3x\) from both sides to make \(y\) the subject. The gradient is the coefficient of \(x\).
Answer to Question 5
1. Midpoint \((5,7)\). Length \(\sqrt{6^2+8^2}=\sqrt{100}=\) 10 The midpoint is the mean of the \(x\) values and the mean of the \(y\) values. For the length, the changes are 6 across and 8 up, so we use Pythagoras.
2. Midpoint \((1.5,8)\). Length \(\sqrt{5^2+12^2}=\sqrt{169}=\) 13 Midpoint: \(\left(\tfrac{-1+4}{2},\tfrac{2+14}{2}\right)\). The changes are 5 across and 12 up.
3. Midpoint \((2.5,6)\). Length \(=\sqrt{25+144}=\) 13 Halve each coordinate of B for the midpoint. The changes are 5 and 12, so the length is 13.
4. Midpoint \((1,2)\). Length \(\sqrt{8^2+6^2}=\sqrt{100}=\) 10 Midpoint: \(\left(\tfrac{-3+5}{2},\tfrac{-1+5}{2}\right)\). The changes are 8 across and 6 up, so the length is 10.
Answer to Question 6
1. \(4\times3-5=7\), yes. Substitute \(x=3\) into the equation: it gives \(y=7\), so the point lies on \(L\).
2. \(4\times2-5=3\ne4\), no. Substitute \(x=2\): the line gives \(y=3\), not 4, so the point is not on \(L\).
3. \(x=-1\): \(y=-9\). \(y=11\): \(4x=16\), \(x=4\). Substitute \(x=-1\) to find \(y\). For \(y=11\), solve \(4x-5=11\).
4. \(y\)-axis: \((0,-5)\). \(x\)-axis: \(4x=5\), \(x=1.25\), so \((1.25,0)\). The \(y\)-intercept is the constant term. On the \(x\)-axis \(y=0\), so solve \(4x-5=0\).
Answer to Question 7
1. The fixed charge of £2.50 before any miles are driven. When \(d=0\), the cost is £2.50, so the intercept is the cost before the taxi moves.
2. The cost per mile, £1.20 per mile. Each extra mile adds £1.20 to the cost, so the gradient is the price per mile.
3. \(2.5+1.2\times7.5=2.5+9=\) £11.50 Substitute \(d=7.5\) into the formula. The journey costs £11.50.
4. \(20.5-2.5=18\), \(18\div1.2=\) 15 miles. Subtract the fixed charge, then divide by the cost per mile. The journey was 15 miles.
Answer to Question 8
1. 8 km in 0.5 h: 16 km/h.
2. The distance does not change, so she has stopped for 15 minutes.
3. 9 km in 45 minutes: \(9\div0.75=\) 12 km/h.
4. 17 km in 1.5 h: \(17\div1.5=\) 11.3 km/h (1 d.p.).
Answer to Question 9
1. \(\dfrac{70-30}{6-2}=\) 10 litres per minute. The gradient is the change in volume divided by the change in time: 40 litres in 4 minutes.
2. \(V=10t+c\): \(30=20+c\), \(c=10\). \(V=10t+10\). Substitute the point \((2,30)\) to find \(c\). Check: \(10\times6+10=70\).
3. At \(t=0\): 10 litres. The start is \(t=0\), so the volume is the intercept \(c\). The tank held 10 litres at the start.
4. \(150=10t+10\), \(t=\) 14 minutes. Subtract 10 from both sides, then divide by 10. The tank holds 150 litres after 14 minutes.
Answer to Question 10
1. \(m=\dfrac{58}{50}=\) 1.16. The line passes through the origin and the point \((50,58)\), so the gradient is \(58\div50\). Each pound is worth €1.16.
2. \(E=1.16P\) A straight line through the origin has \(c=0\), so the formula is \(E=1.16P\).
3. \(325\times1.16=\) €377 Multiply the pounds by 1.16. £325 is worth €377.
4. \(200\div1.16=\) £172.41 Rearrange to \(P=E\div1.16\). €200 is worth about £172.41.
Answer to Question 11
1. \(\dfrac{12}{100}=\) \(\dfrac{3}{25}\) Divide 12 and 100 by their HCF, 4. So 12% is \(\tfrac{3}{25}\).
2. \(0.12\times450=\) 54 m The road rises 12 m for every 100 m, which is 0.12 m for every metre. The road rises 54 m.
3. \(\dfrac{36}{400}=0.09=\) 9% Divide the rise by the horizontal distance and change it into a percentage. The gradient is 9%.
4. No, 9% is less than 12%, so it is less steep. A smaller percentage gradient means a gentler slope.
Answer to Question 12
1. Gradient 1.8 and intercept 32. The equation is in the form \(y=mx+c\), with \(m=1.8\) and \(c=32\).
2. \(1.8\times20+32=\) 68 °F Substitute \(C=20\). So 20 °C is 68 °F.
3. \(95-32=63\), \(63\div1.8=\) 35 °C Subtract 32, then divide by 1.8. So 95 °F is 35 °C.
4. \(C=1.8C+32\), \(-0.8C=32\), \(C=\) −40, so the point is \((-40,-40)\). Where the lines cross, \(F\) and \(C\) are equal. So \(-40\) °C is the same as \(-40\) °F.
Answer to Question 13
1. Hall A: \(y=40+5h\). Hall B: \(y=15h\).
2. \(40+5h=15h\), \(10h=40\), \(h=4\), cost \(=\) £60 at \((4,60)\).
3. 3 hours: A \(=£55\), B \(=£45\), so B. 6 hours: A \(=£70\), B \(=£90\), so A.
4. For a 4-hour hire both halls cost £60. Below 4 hours Hall B is cheaper and above 4 hours Hall A is cheaper.
Answer to Question 14
1. Yes: 0 litres costs £0. Buying no petrol costs nothing, so the point \((0,0)\) is on the line.
2. \(C=1.5L\) The cost is the price per litre times the number of litres.
3. 1.5 (pounds per litre). The gradient is the price per litre, because each extra litre adds £1.50.
4. \(63\div1.5=\) 42 litres. Divide the cost by the price per litre. Mia bought 42 litres.
Answer to Question 15
1. \(2\times m=-1\), so \(m=\) \(-\tfrac12\). The gradient of a perpendicular line is the negative reciprocal: turn 2 upside down and change the sign.
2. \(y=-\tfrac12x+c\): \(1=-2+c\), \(c=3\). \(y=-\tfrac12x+3\) Substitute the point \((4,1)\) to find \(c\). Check: \(-\tfrac12\times4+3=1\).
3. \(2x+3=-\tfrac12x+3\), \(2.5x=0\), \(x=0\), \(y=3\). \((0,3)\) At the crossing point both equations give the same \(y\).
4. The second line is \(y=-\tfrac13x+2\), gradient \(-\tfrac13\). \(3\times\left(-\tfrac13\right)=-1\) ✓. Rearrange the second equation to make \(y\) the subject. The product of the gradients is \(-1\), so the lines are perpendicular.
Answer to Question 16
1. Between \(x=2\) and \(4\): \(\dfrac{8}{2}=4\). Between \(4\) and \(6\): \(\dfrac82=4\). The gradients are equal, so it is a straight line.
2. \(y=4x+c\): \(5=8+c\), \(c=-3\). \(y=4x-3\) Substitute the point \((2,5)\) to find \(c\). Check: \(4\times6-3=21\).
3. \(4\times10-3=\) 37 Substitute \(x=10\) into \(y=4x-3\).
4. \(-3=4x-3\), so \(x=\) 0. Add 3 to both sides, so \(4x=0\).
Answer to Question 17
1. AB: \((4,2)\). BC: \((5.5,5)\). AC: \((2.5,5)\). Each midpoint is the mean of the \(x\) values and the mean of the \(y\) values of the two ends.
2. \(AC=\sqrt{3^2+6^2}=\sqrt{45}=6.7\). \(BC=\sqrt{3^2+6^2}=\sqrt{45}=6.7\). So the triangle is isosceles (AC = BC).
3. Base \(AB=6\), height \(=8-2=6\). Area \(=\tfrac12\times6\times6=\) 18 square units. AB is horizontal, so the perpendicular height is the difference in \(y\) between C and the line AB.
4. \(m=\dfrac{8-2}{4-1}=2\), \(2=2+c\), \(c=0\). \(y=2x\) Check with C: \(2\times4=8\).
Answer to Question 18
1. PQ: \(\tfrac14\). SR: \(\dfrac{5-4}{6-2}=\tfrac14\). Use \(m=\dfrac{\text{change in }y}{\text{change in }x}\) for each side.
2. QR: \(\dfrac{5-1}{6-4}=2\). PS: \(\dfrac{4-0}{2-0}=2\).
3. Both pairs of opposite sides have equal gradients, so they are parallel: PQRS is a parallelogram.
4. Midpoint of PR \(=(3,2.5)\). Midpoint of QS \(=\left(\dfrac{4+2}{2},\dfrac{1+4}{2}\right)=(3,2.5)\) ✓.
Answer to Question 19
1. \(x=6-x\), \(x=3\), \(y=3\). \((3,3)\) Where the lines cross, the two \(y\) values are equal.
2. \(y=x\) meets the \(y\)-axis at \((0,0)\). \(y=6-x\) meets it at \((0,6)\).
3. Base on the \(y\)-axis \(=6\), height \(=3\). Area \(=\tfrac12\times6\times3=\) 9 square units.
4. Sides: \(6\), \(\sqrt{18}\), \(\sqrt{18}\). Perimeter \(=6+2\times4.24=\) 14.5. Each sloping side has changes of 3 and 3, so its length is \(\sqrt{3^2+3^2}=\sqrt{18}\).
Answer to Question 20
1. First: \(y=-\dfrac k2x+4\), gradient \(-\dfrac k2\). Second: \(y=3x-1\), gradient 3.
2. \(-\dfrac k2=3\), so \(k=-6\). Parallel lines have equal gradients.
3. \(\left(-\dfrac k2\right)\times3=-1\), \(k=\dfrac23\). So \(k=\dfrac23\). For perpendicular lines the product of the gradients is \(-1\).
4. \(2x+2y=8\) and \(y=3x-1\): \(2x+6x-2=8\), \(8x=10\), \(x=1.25\), \(y=2.75\). \((1.25,2.75)\)

How to use these Year 10 straight-line graphs worksheet answers

First, try every question on the worksheet without looking. Then compare your working with the solution line by line, not only the final answer. When your answer is different, find the first step that changes, because that is where the mistake is. After that, read the matching part of the straight-line graphs lesson again. Finally, check that the method is secure with the quiz or the test.

Also, the Year 10 straight-line graphs worksheet answers are available as a PDF, so you can mark printed work quickly. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, these Year 10 straight-line graphs worksheet answers show every step, so nothing is left unexplained.

Straight-line graphs – Year 10 GCSE maths worksheet answers to download as PDF

Answer to Question 1

Answer to Question 2

Answer to Question 3

Answer to Question 4

Answer to Question 5

Answer to Question 6

Answer to Question 7

Answer to Question 8

Answer to Question 9

Answer to Question 10

Answer to Question 11

Answer to Question 12

Answer to Question 13

Answer to Question 14

Answer to Question 15

Answer to Question 16

Answer to Question 17

Answer to Question 18

Answer to Question 19

Answer to Question 20

How to use these Year 10 straight-line graphs worksheet answers

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