Gradient of a Line: Practise Online

Either paper 2 marks

The grid shows the points \(C(-5, 1)\) and \(D(-3, 3)\).

Find the gradient of the line \(CD\).
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\(1\)
Gradient is the change in \(y\) divided by the change in \(x\):

\(\text{gradient} = \frac{ 3 - (1) }{ -3 - (-5) } = \frac{ 2 }{ 2 } = 1\)
Either paper 2 marks

The line through \(A(-3, 2)\) and \(B(-2, -1)\) is drawn on the grid.

Calculate the gradient of the line.
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\(-3\)
Going from \(A\) to \(B\) the line moves \(1\) across and falls \(3\).

Falling counts as a negative change in \(y\):

\(\text{gradient} = \frac{ -3 }{ 1 } = -3\)

A line sloping downwards always has a negative gradient.
Either paper 2 marks

A straight line has equation \(y = -x - 7\)

(a)
Write down the gradient of the line.

(b)
Write down the coordinates of the point where the line crosses the \(y\)-axis.
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(a) \(-1\)
(b) \((0, -7)\)
In the form \(y = mx + c\) the number in front of \(x\) is the gradient.

The gradient is \(-1\).
The line crosses the \(y\)-axis where \(x = 0\), and in \(y = mx + c\) the constant \(c\) is that \(y\)-value.

The point is \((0, -7)\).
Either paper 4 marks

A taxi firm charges a fixed amount for every journey, plus an extra amount for each mile travelled.

The graph shows the cost of a journey of up to \(14\) miles.

(a)
State the amount the taxi firm charges before it has travelled any distance at all.

(b)
Work out the gradient of the line.

(c)
Write down what the gradient tells you about the cost of a journey.
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(a) \(\text{£}2\)
(b) \(2\)
(c) The taxi firm charges £2 for each extra mile travelled.
A journey of \(0\) miles is where the line crosses the cost axis.

It crosses at \(2\), so the fixed charge is £\(2\).
Take the two ends of the line: \((0, 2)\) and \((14, 30)\).

\(\text{gradient} = \frac{ 30 - 2 }{ 14 - 0 } = \frac{ 28 }{ 14 } = 2\)
The gradient is how much the cost goes up for each extra mile, so it is measured in pounds per mile.

A gradient of \(2\) means the firm charges £\(2\) for every extra mile travelled — on top of the £\(2\) fixed charge.
Either paper 3 marks

A straight line passes through \((-2, 5)\) and \((2, k)\).

The gradient of the line is \(-3\).

Find the value of \(k\).
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\(-7\)
Gradient is the change in \(y\) divided by the change in \(x\).

The change in \(x\) is \(2 - -2 = 4\).

So the change in \(y\) is \(-3 \times 4 = -12\).

\(k = 5 - 12 = -7\)
Either paper 4 marks

A cafe makes a drink from orange squash and water.

The number of cartons of water \((y)\) : the number of cartons of squash \((x)\) = \(3 : 2\)

(a)
Use this information to draw a graph on the grid to show the relationship between the number of cartons of squash and the number of cartons of water.

(b)
Find the gradient of the line drawn in part (a).

(c)
Explain what this gradient represents.
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You cannot draw on this page. Draw (a) on paper, then show the answer and check your drawing against the model.
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(a) A straight line through \((0, 0)\), \((2, 3)\) and \((4, 6)\)
(b) \(\frac{3}{2}\)
(c) The number of cartons of water needed for 1 carton of squash
(a)

Written with x first, the ratio is \(x : y = 2 : 3\), so \(y = \frac{ 3 }{ 2 }x\) and the line goes through the origin.

When \(x = 2\), \(y = 3\). When \(x = 4\), \(y = 6\).

Plot \((0, 0)\), \((2, 3)\) and \((4, 6)\) and join them with a straight line.
The line passes through \((0, 0)\) and \((2, 3)\).

gradient \(= \frac{ \text{change in } y }{ \text{change in } x } = \frac{ 3 }{ 2 }\)
The gradient is \(y\) divided by \(x\), so it is the number of cartons of water for each 1 carton of squash.

It is the amount of water needed for each 1 carton of squash.
- (a) B1a straight line through \((0, 0)\)
- (a) B1the correct line, through \((0, 0)\), \((2, 3)\) and \((4, 6)\)
Either paper 3 marks

The line \(L\) has equation \(y = -x - 1\)

Work out the equation of the line that is parallel to \(L\) and passes through the point \((4, 6)\).
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\(y = -x + 10\)
Parallel lines have the same gradient, so the new line is \(y = -x + k\) for some number \(k\).

Substitute \((4, 6)\):

\(6 = -1 \times (4) + k\), so \(k = 10\).

The equation is \(y = -x + 10\)
Either paper 3 marks

The points \(R\) and \(S\) are marked on the grid.

Work out the gradient of the line \(RS\).

Give your answer as a fraction in its simplest form.
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\(-\frac{3}{4}\)
Read the points off the grid: \(R\) is \((-6, 4)\) and \(S\) is \((-2, 1)\).

From \(R\) to \(S\) the line goes \(4\) across and \(3\) down.

\(\text{gradient} = \frac{ -3 }{ 4 } = -\frac{3}{4}\)
- M1reads both points correctly, \(R(-6, 4)\) and \(S(-2, 1)\)
- M1change in y \(-3\) over change in x \(4\)
- A1gradient \(-\frac{3}{4}\) in simplest form
Either paper 4 marks

\(A\) is the point \((5, -9)\).

\(B\) is the point \((8, -15)\).

(a)
Show that the gradient of the straight line passing through \(A\) and \(B\) is \(-2\).

(b)
\(C\) is the point \((264, -528)\).

Does \(C\) lie on the straight line passing through \(A\) and \(B\)?

You must show your working.
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(a) \(\frac{ -15 - (-9) }{ 8 - 5 } = \frac{ -6 }{ 3 } = -2\)
(b) No: the line gives \(y = -527\) at \(x = 264\), and \(C\) has \(y = -528\)
The gradient is the change in \(y\) divided by the change in \(x\):

\(\frac{ -15 - (-9) }{ 8 - 5 } = \frac{ -6 }{ 3 } = -2\)
Find the equation of \(AB\). The gradient is \(-2\), so \(y = -2x + c\).

Substitute \(A\) to find \(c\):

\(-9 = -2 \times 5 + c\)

\(-9 = -10 + c\)

\(c = 1\)

So the line is \(y = -2x + 1\).

Put in the \(x\)-coordinate of \(C\):

\(y = -2 \times 264 + 1 = -528 + 1 = -527\)

The line is at \(y = -527\) when \(x = 264\), and \(C\) is at \(y = -528\). These are different, so \(C\) does not lie on the line.

Alternative: the gradient of \(AC\) is \(\frac{ -519 }{ 259 }\), which is not equal to \(-2\), so \(C\) does not lie on the line.

Answer: No
- (b) M1equation of \(AB\): \(y = -2x + 1\), with \(c = 1\) (or the gradient of \(AC\): \(\frac{ -519 }{ 259 }\))
- (b) M1substitute \(x = 264\): \(y = -528 + 1 = -527\) (or compare the gradient of \(AC\) with \(-2\))
- (b) A1No, comparing \(-527\) with \(-528\)
Either paper 3 marks

The graph gives information about the volume, \(v\) litres, of diesel in the tank of Anya's car after it has travelled a distance of \(d\) kilometres.

(a)
Calculate the gradient of the graph.

(b)
What does the gradient of the graph represent?
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(a) \(-0.075\) litres per km
(b) The diesel used per kilometre travelled (the rate of decrease of the volume): the tank loses 0.075 litres of diesel for every 1 km driven.
Read two points on the line from the scaled axes: \((0, 45)\) and \((200, 30)\).

The volume changes by \(30 - 45 = -15\) litres while the distance changes by \(200 - 0 = 200\) km:

\(\text{gradient} = \frac{ -15 }{ 200 } = -0.075\)
The gradient is the change in volume for each 1 km of distance.

It is negative because the volume of diesel is falling, so it represents the diesel used per kilometre: the tank loses \(0.075\) litres for every 1 km travelled.
- (a) M1reads two points on the line, \((0, 45)\) and \((200, 30)\), and divides the change in \(v\) by the change in \(d\)
- (a) A1gradient \(-0.075\)