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

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1. Answer to Question 1 (4 marks)

   1. \(\dfrac{16-4}{5-1}=\dfrac{12}{4}=\) 3 (1 for the method, 1 for the answer) Divide the change in \(y\) by the change in \(x\). The gradient is 3.
   2. \(y=3x+c\), \(4=3+c\), \(c=1\): \(y=3x+1\) (1, 1) Substitute the point \((1,4)\) to find \(c\). Check with \((5,16)\): \(3\times5+1=16\). The equation is \(y=3x+1\).

   A common error in (a) is to divide the change in \(x\) by the change in \(y\), giving \(\tfrac13\). Another is to subtract the coordinates in a different order on the top and the bottom. Here both changes are positive, so the line slopes upwards.

2. Answer to Question 2 (4 marks)

   1. \(3\times4-2=10\), yes (1) Substitute \(x=4\): the equation gives \(y=10\), so the point lies on the line.
   2. \(y=3x+5\) (1) A parallel line has the same gradient, 3, and the point \((0,5)\) gives the intercept 5.
   3. \(0=3x-2\) (1), \(x=\dfrac23\): \(\left(\tfrac23,0\right)\) (1) Every point on the \(x\)-axis has \(y=0\). Solve \(3x-2=0\).

   In (b), any line \(y=3x+c\) is parallel, but only \(c=5\) passes through \((0,5)\). In (c), the answer as a decimal, \(0.67\), is not exact, so the fraction is better.

3. Answer to Question 3 (4 marks)

   1. \(\left(\dfrac{-2+6}2,\dfrac{1+7}2\right)=\) \((2,4)\) (1, 1) The midpoint is the mean of the \(x\) values and the mean of the \(y\) values. The midpoint is \((2,4)\).
   2. \(\sqrt{8^2+6^2}=\sqrt{100}=\) 10 (1, 1) The changes are 8 across and 6 up, so we use Pythagoras. The length of AB is 10 units.

   In (a), take care with the negative: \(-2+6=4\). In (b), the horizontal change is \(6-(-2)=8\), not 4, and the vertical change is \(7-1=6\).

4. Answer to Question 4 (4 marks)

   1. \(\dfrac{93-48}{500-200}=\dfrac{45}{300}=\) £0.15 per unit (1, 1) The gradient is the change in the bill divided by the change in units. Each unit costs 15p.
   2. \(48-0.15\times200=48-30=\) £18 (1) Subtract the cost of 200 units from the £48 bill. The standing charge is the intercept, £18.
   3. \(18+0.15\times350=18+52.50=\) £70.50 (1) Use \(C=18+0.15u\) with \(u=350\). The bill is £70.50.

   Check (c): 350 units is halfway between 200 and 500, and the bill £70.50 is halfway between £48 and £93.

5. Answer to Question 5 (4 marks)

   1. −2 (1) The gradient of L is \(\tfrac12\). The perpendicular gradient is the negative reciprocal, \(-2\), because \(\tfrac12\times(-2)=-1\).
   2. \(y=-2x+c\) (1), \(7=-4+c\) (1), \(c=11\): \(y=-2x+11\) (1) Substitute the point \((2,7)\) to find \(c\). Check: \(-2\times2+11=7\). The equation is \(y=-2x+11\).

   In (b), a common error is to use the gradient \(-\tfrac12\) or \(2\) instead of the negative reciprocal. The new line crosses the \(y\)-axis at \((0,11)\). Check that the gradients multiply to \(-1\): \(\tfrac12\times(-2)=-1\), so the two lines meet at right angles.

Total : 20 marks