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Either paper 2 marks
\(A\) is the point \((3, 7)\) and \(B\) is the point \((5, 11)\).
Find the coordinates of the midpoint of \(AB\).
Show answer
\((4, 9)\)The midpoint is halfway between the two points, so take the mean of each coordinate:
\(\frac{ 3 + 5 }{2} = 4\) and \(\frac{ 7 + 11 }{2} = 9\)
The midpoint is \((4, 9)\).
- M1mean of the coordinates, \(\frac{ 3 + 5 }{2}\) or \(\frac{ 7 + 11 }{2}\)
- A1midpoint \((4, 9)\)
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Either paper 1 mark
The line segment \(AB\) is drawn on the grid.
Plot the midpoint of \(AB\) on the grid.
You cannot draw on this page. Draw this on paper, then show the answer and check your drawing against the model.
Show answer
The midpoint is at \((0, 2)\)The midpoint is halfway along in both directions, so average the two \(x\)-coordinates and the two \(y\)-coordinates.
\(A\) is \((-3, 1)\) and \(B\) is \((3, 3)\).
\(x\): \(\dfrac{ -3 + 3 }{2} = \dfrac{ 0 }{2} = 0\)
\(y\): \(\dfrac{ 1 + 3 }{2} = \dfrac{ 4 }{2} = 2\)
The midpoint is \((0, 2)\).
- B1the midpoint plotted at \((0, 2)\)
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Either paper 2 marks
\(P\) has coordinates \((-15, 10)\) and \(Q\) has coordinates \((-1, -8)\).
Work out the coordinates of the midpoint of \(PQ\).
Show answer
\((-8, 1)\)Take the mean of the \(x\)-coordinates and the mean of the \(y\)-coordinates:
\(\frac{ -15 - 1 }{2} = -8\)
\(\frac{ 10 - 8 }{2} = 1\)
The midpoint of \(PQ\) is \((-8, 1)\).
- M1mean of the \(x\)-coordinates \(-8\) or of the \(y\)-coordinates \(1\)
- A1midpoint \((-8, 1)\)
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Either paper 2 marks
The line segment \(ST\) is drawn on the grid.
Work out the coordinates of the midpoint of \(ST\).
Show answer
\((-1, 3)\)Read the ends off the grid: \(S\) is \((-3, 5)\) and \(T\) is \((1, 1)\).
The midpoint is the mean of each coordinate:
\(\frac{ -3 + 1 }{2} = \frac{ -2 }{2} = -1\)
\(\frac{ 5 + 1 }{2} = \frac{ 6 }{2} = 3\)
The midpoint of \(ST\) is \((-1, 3)\).
- M1mean of the coordinates of \(S\) \((-3, 5)\) and \(T\) \((1, 1)\)
- A1midpoint \((-1, 3)\)
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Either paper 2 marks
\(A\) has coordinates \((-2, -6)\) and \(B\) has coordinates \((-3, -1)\).
Calculate the coordinates of the midpoint of \(AB\).
Give each coordinate as a decimal.
Show answer
\((-2.5, -3.5)\)\(\frac{ -2 - 3 }{2} = \frac{ -5 }{2} = -2.5\)
\(\frac{ -6 - 1 }{2} = \frac{ -7 }{2} = -3.5\)
The midpoint is \((-2.5, -3.5)\).
- M1mean of the \(x\)-coordinates \(-2.5\) or of the \(y\)-coordinates \(-3.5\)
- A1midpoint \((-2.5, -3.5)\)
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Either paper 4 marks
\(AB\) is a diameter of a circle.
\(A\) has coordinates \((-4, 2)\) and \(B\) has coordinates \((12, -14)\).
(a)Write down the coordinates of the centre of the circle.
(b)\(PQ\) is a different diameter of the same circle.
\(P\) has coordinates \((12, 2)\).
Find the coordinates of \(Q\).
Show answer
(a) \((4, -6)\)
(b) \((-4, -14)\)The centre is the midpoint of the diameter:
\(\frac{ -4 + 12 }{2} = 4\) and \(\frac{ 2 - 14 }{2} = -6\)
The centre is \((4, -6)\).
The centre is the midpoint of every diameter, so it is the midpoint of \(PQ\).
The step from \(P\) to the centre happens again from the centre to \(Q\).
\(x\)-coordinate: the step is \(-8\), so \(4 - 8 = -4\)
\(y\)-coordinate: the step is \(-8\), so \(-6 - 8 = -14\)
\(Q\) is \((-4, -14)\).
- (a) M1midpoint of \(AB\): half the sum of the coordinates
- (a) A1centre \((4, -6)\)
- (b) M1centre is the midpoint of \(PQ\), so the step from \(P\) to the centre is repeated
- (b) A1coordinates of \(Q\) \((-4, -14)\)
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Either paper 4 marks
The triangle \(LMN\) is drawn on the grid.
(a)Work out the coordinates of the midpoint of \(LM\).
(b)Work out the coordinates of the midpoint of \(MN\).
Show answer
(a) \((0, 5)\)
(b) \((-2, -1)\)\(L\) is \((4, 6)\) and \(M\) is \((-4, 4)\).
\(\frac{ 4 - 4 }{2} = \frac{ 0 }{2} = 0\) and \(\frac{ 6 + 4 }{2} = \frac{ 10 }{2} = 5\)
The midpoint of \(LM\) is \((0, 5)\).
\(M\) is \((-4, 4)\) and \(N\) is \((0, -6)\).
\(\frac{ -4 }{2} = \frac{ -4 }{2} = -2\) and \(\frac{ 4 - 6 }{2} = \frac{ -2 }{2} = -1\)
The midpoint of \(MN\) is \((-2, -1)\).
- (a) M1adding the coordinates of \((4, 6)\) and \((-4, 4)\) and halving
- (a) A1\((0, 5)\)
- (b) M1adding the coordinates of \((-4, 4)\) and \((0, -6)\) and halving
- (b) A1\((-2, -1)\)
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Either paper 3 marks
The grid is a map of a village. Each unit stands for 1 km.
(a)State the coordinates of the Pool.
(b)Reuben lives exactly halfway along the straight line from the Pool to the Market.
Work out the coordinates of he house.
Show answer
(a) \((3, 5)\)
(b) \((4, 4)\)The Pool is \(3\) km east and \(5\) km north of the origin.
Its coordinates are \((3, 5)\).
Halfway means the midpoint of \((3, 5)\) and \((5, 3)\).
\(\frac{ 3 + 5 }{2} = \frac{ 8 }{2} = 4\) and \(\frac{ 5 + 3 }{2} = \frac{ 8 }{2} = 4\)
The house is at \((4, 4)\).
- (b) M1midpoint of \((3, 5)\) and \((5, 3)\): mean of the \(x\) values and of the \(y\) values
- (b) A1house at \((4, 4)\)
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Either paper 4 marks
The line \(AB\) is shown on the grid. \(A\) is the point \((2, -2)\) and \(B\) is the point \((-7, 4)\).
(a)Find the coordinates of the midpoint of \(AB\).
(b)\(C\) is a point on \(AB\), not at \(A\) or \(B\). \(C\) is closer to \(A\) than to \(B\). The coordinates of \(C\) are whole numbers.
Find the coordinates of \(C\).
(c)On the grid, draw a line from the point \((-1, -3)\) that is parallel to \(AB\), runs in the same direction as \(A\) to \(B\), and is two thirds as long as \(AB\).
Give the coordinates of the other end of your line.
Show answer
(a) \((-2.5, 1)\)
(b) \((-1, 0)\)
(c) \((-7, 1)\)The midpoint is the mean of each coordinate:
\(\frac{ 2 - 7 }{2} = -2.5\)
\(\frac{ -2 + 4 }{2} = 1\)
The midpoint is \((-2.5, 1)\).
Follow \(AB\) from \(A\). It passes through a grid point every \(3\) left and \(2\) up, so the grid points on \(AB\) are:
\((2, -2)\), \((-1, 0)\), and so on in steps of \((-3, 2)\), up to \((-7, 4)\).
The grid point closest to \(A\) (other than \(A\) itself) is one step from \(A\) back along the line, so \(C = (-1, 0)\).
The move from \(A\) to \(B\) is \((-9, 6)\), and your line runs in the same direction, so its move is two thirds of that:
\(\frac{ 2 }{ 3 } \times (-9, 6) = (-6, 4)\)
Starting at \((-1, -3)\), add this move:
\((-1 - 6, -3 + 4) = (-7, 1)\)
The line ends at \((-7, 1)\).
- (c) M1scaled vector \((-6, 4)\)
- (c) A1end of the line \((-7, 1)\)
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Either paper 3 marks
Points \(E\), \(F\) and \(G\) are plotted on the grid.
(a)State the coordinates of \(G\).
(b)State the coordinates of the midpoint of \(EG\).
(c)The midpoint of \(EG\) is also the midpoint of \(FH\).
Work out the coordinates of \(H\).
Show answer
(a) \((-4, -5)\)
(b) \((-3, -3)\)
(c) \((-7, 1)\)Reading across and then up from the origin, \(G\) is \((-4, -5)\).
\(E\) is \((-2, -1)\) and \(G\) is \((-4, -5)\).
\(\frac{ -2 - 4 }{2} = \frac{ -6 }{2} = -3\) and \(\frac{ -1 - 5 }{2} = \frac{ -6 }{2} = -3\)
The midpoint of \(EG\) is \((-3, -3)\).
\((-3, -3)\) is halfway from \(F\ (1, -7)\) to \(H\), so the \(x\)-coordinates of \(F\) and \(H\) add to \(-6\) and the \(y\)-coordinates add to \(-6\).
\(x = -6 - 1 = -7\) and \(y = -6 + 7 = 1\)
\(H\) is \((-7, 1)\).