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Either paper 3 marks
The points \(D\), \(E\) and \(F\) are marked on the grid.
(a)State the coordinates of \(D\).
(b)State the coordinates of \(E\).
(c)State the coordinates of \(F\).
Show answer
(a) \((5, 4)\)
(b) \((-3, 3)\)
(c) \((-4, -2)\)Read down to the \(x\)-axis, then across to the \(y\)-axis.
\(D\) is \(5\) across and \(4\) up, so \(D\) is \((5, 4)\).
\(E\) is to the left of the \(y\)-axis, so its \(x\)-coordinate is negative.
\(E\) is \((-3, 3)\).
\(F\) is below the \(x\)-axis, so its \(y\)-coordinate is negative.
\(F\) is \((-4, -2)\).
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Either paper 2 marks
Plot the points \(L\,(5, 2)\) and \(M\,(-3, -4)\) on the grid.
You cannot draw on this page. Draw this on paper, then show the answer and check your drawing against the model.
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\(L\) at \((5, 2)\) and \(M\) at \((-3, -4)\)The first number is how far across, the second is how far up.
For \(L\,(5, 2)\): start at the origin, go \(5\) along the \(x\)-axis, then \(2\) in the \(y\) direction. A negative number means left, or down.
For \(M\,(-3, -4)\): go \(-3\) along the \(x\)-axis, which is to the left, then \(-4\) in the \(y\) direction.
- B1one of the two points in the right place
- B1both points in the right place
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Either paper 4 marks
Every point of a shape is moved 3 units to the right and 2 units down.
(a)The point \(A\) has coordinates \((-6, -2)\).
Write down the coordinates of the image of \(A\).
(b)The image of the point \(B\) has coordinates \((-3, -5)\).
Write down the coordinates of \(B\).
Show answer
(a) \((-3, -4)\)
(b) \((-6, -3)\)Moving right changes the \(x\)-coordinate: \(-6 + 3 = -3\)
Moving down changes the \(y\)-coordinate: \(-2 - 2 = -4\)
The image of \(A\) is \((-3, -4)\).
Undo the translation, so take the moves off instead of adding them:
\(-3 - 3 = -6\) and \(-5 + 2 = -3\)
\(B\) is \((-6, -3)\).
- (a) M1one coordinate of the image correct, \(x = -3\) or \(y = -4\)
- (a) A1image \((-3, -4)\)
- (b) M1subtracting the move from a coordinate, \(x = -6\) or \(y = -3\)
- (b) A1\(B\) is \((-6, -3)\)
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Either paper 2 marks
\(A\) is the point \((6, 7)\) and \(B\) is the point \((10, 11)\).
Calculate the coordinates of the midpoint of \(AB\).
Show answer
\((8, 9)\)The midpoint is halfway between the two points, so take the mean of each coordinate:
\(\frac{ 6 + 10 }{2} = 8\) and \(\frac{ 7 + 11 }{2} = 9\)
The midpoint is \((8, 9)\).
- M1mean of the coordinates, \(\frac{ 6 + 10 }{2}\) or \(\frac{ 7 + 11 }{2}\)
- A1midpoint \((8, 9)\)
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Either paper 2 marks
Four points are marked on the grid.
(a)Write down the coordinates of \(P\).
(b)One of the points has coordinates \((-3, -5)\).
Write down the letter of that point.
Show answer
(a) \((5, 2)\)
(b) RCount across from the origin first, then up or down.
\(P\) is \((5, 2)\).
Go \(-3\) along the \(x\)-axis and \(-5\) along the \(y\)-axis.
That is the point \(R\).
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Either paper 1 mark
\(PQRS\) is a rectangle. Three of its corners are marked on the grid.
Plot the point \(S\) on the grid.
You cannot draw on this page. Draw this on paper, then show the answer and check your drawing against the model.
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\(S\) at \((-3, 1)\)The sides of the rectangle follow the grid lines, so \(S\) is straight above or below \(P\) and level with \(R\).
It has the same \(x\)-coordinate as \(P\), which is \(-3\), and the same \(y\)-coordinate as \(R\), which is \(1\).
\(S\) is at \((-3, 1)\).
- B1\(S\) plotted at \((-3, 1)\)
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Either paper 2 marks
\(P\) is the point \((-5, -3)\).
\(P\) is reflected in the \(x\)-axis to give the point \(P'\).
Write down the coordinates of \(P'\).
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\((-5, 3)\)A reflection in the \(x\)-axis keeps the \(x\)-coordinate the same and reverses the sign of the \(y\)-coordinate.
So \((-5, -3)\) maps to \((-5, 3)\).
- M1\(x\)-coordinate unchanged, sign of \(y\)-coordinate reversed
- A1image \((-5, 3)\)
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Either paper 4 marks
\(ABCD\) is a rectangle. Its sides are parallel to the axes.
\(A\) is the point \((-3, 3)\) and \(C\) is the point \((5, 11)\).
(a)\(B\) is the point \((5, 3)\).
Write down the coordinates of \(D\).
(b)Calculate the coordinates of the centre of the rectangle.
Show answer
(a) \((-3, 11)\)
(b) \((1, 7)\)\(D\) is opposite \(B\), so it shares the \(x\)-coordinate of \(A\) and the \(y\)-coordinate of \(C\).
\(D\) is \((-3, 11)\).
The centre is the midpoint of the diagonal \(AC\):
\(\frac{ -3 + 5 }{2} = 1\) and \(\frac{ 3 + 11 }{2} = 7\)
The centre is \((1, 7)\).
- (a) M1\(D\) has \(x = -3\) or \(y = 11\)
- (a) A1\(D\) is \((-3, 11)\)
- (b) M1mean of the coordinates of \(A\) and \(C\), \(\frac{ -3 + 5 }{2}\) or \(\frac{ 3 + 11 }{2}\)
- (b) A1centre \((1, 7)\)
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Either paper 2 marks
\(JKLM\) is a parallelogram.
Three of its vertices are marked on the grid.
Calculate the coordinates of \(M\).
Show answer
\((5, -1)\)In a parallelogram \(KJ\) and \(LM\) are parallel and the same length.
From \(K\ (1, -3)\) to \(J\ (4, -5)\) the move is \(3\) across and \(-2\) up.
Make the same move from \(L\ (2, 1)\):
\(2 + 3 = 5\) and \(1 - 2 = -1\)
\(M\) is \((5, -1)\).
- M1uses the move from \(K\) to \(J\): \(3\) across and \(-2\) up
- A1coordinates \((5, -1)\)
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Either paper 3 marks
Three points are marked on the grid.
(a)State the coordinates of \(S\).
(b)State the coordinates of \(T\).
(c)State the coordinates of \(U\).
Show answer
(a) \((6, 16)\)
(b) \((1, 14)\)
(c) \((9, 19)\)The axes go up in twos, so every square is \(2\) units wide.
\(S\) is \((6, 16)\).
\(T\) sits halfway between \(0\) and \(2\) on the \(x\)-axis, so its \(x\)-coordinate is \(1\).
\(T\) is \((1, 14)\).
\(U\) is in the middle of a square, so both coordinates are odd numbers.
\(U\) is \((9, 19)\).