Exam questions · Maths
Vectors, Constructions and Loci
- 30 exam questions
- 89 marks
- 45 quick checks
Column Vectors and Vector Arithmetic
Just this lesson-
1 Write down [2 marks]
Write down the column vector for a move of 2 to the left and 4 up.
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Model answer
Left is negative and up is positive, so the vector is \(\begin{pmatrix} -2 \\ 4 \end{pmatrix}\).
Mark scheme
- \(-2\) as the top number — B1
- \(\begin{pmatrix} -2 \\ 4 \end{pmatrix}\) — B1
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2 Work out [4 marks]
The vectors \(\mathbf{p}\) and \(\mathbf{q}\) are drawn on the grid. (a) Write \(\mathbf{p}\) and \(\mathbf{q}\) as column vectors. (2 marks) (b) Work out \(2\mathbf{p} + \mathbf{q}\). (2 marks)
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Model answer
(a) \(\mathbf{p}\) goes 3 right and 2 up, so \(\mathbf{p} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}\). \(\mathbf{q}\) goes 2 left and 3 up, so \(\mathbf{q} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}\). (b) \(2\mathbf{p} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}\), so \(2\mathbf{p} + \mathbf{q} = \begin{pmatrix} 4 \\ 7 \end{pmatrix}\).
Mark scheme
- (a) \(\mathbf{p} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}\) — B1
- (a) \(\mathbf{q} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}\) — B1
- (b) \(2\mathbf{p} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}\) or a correct method — M1
- (b) \(\begin{pmatrix} 4 \\ 7 \end{pmatrix}\) — A1 (follow through from (a))
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3 Work out [3 marks]
\(A\) is the point \((-1, 2)\) and \(B\) is the point \((4, -3)\). (a) Write \(\overrightarrow{AB}\) as a column vector. (2 marks) (b) Write \(\overrightarrow{BA}\) as a column vector. (1 mark)
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Model answer
(a) \((4 - (-1), -3 - 2) = (5, -5)\), so \(\overrightarrow{AB} = \begin{pmatrix} 5 \\ -5 \end{pmatrix}\). (b) \(\overrightarrow{BA} = -\overrightarrow{AB} = \begin{pmatrix} -5 \\ 5 \end{pmatrix}\).
Mark scheme
- (a) \(4 - (-1)\) or \(-3 - 2\) — M1
- (a) \(\begin{pmatrix} 5 \\ -5 \end{pmatrix}\) — A1
- (b) \(\begin{pmatrix} -5 \\ 5 \end{pmatrix}\) — B1 (follow through from (a))
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4 Work out [3 marks]
\(\mathbf{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}\). (a) Work out \(\mathbf{a} + 2\mathbf{b}\). (2 marks) (b) Write down a vector that is parallel to \(\mathbf{a}\). (1 mark)
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Model answer
(a) \(2\mathbf{b} = \begin{pmatrix} 2 \\ 8 \end{pmatrix}\), so \(\mathbf{a} + 2\mathbf{b} = \begin{pmatrix} 5 \\ 6 \end{pmatrix}\). (b) Any multiple of \(\mathbf{a}\), such as \(\begin{pmatrix} 6 \\ -4 \end{pmatrix}\).
Mark scheme
- (a) \(2\mathbf{b} = \begin{pmatrix} 2 \\ 8 \end{pmatrix}\) or a correct method — M1
- (a) \(\begin{pmatrix} 5 \\ 6 \end{pmatrix}\) — A1
- (b) A multiple of \(\begin{pmatrix} 3 \\ -2 \end{pmatrix}\), such as \(\begin{pmatrix} 6 \\ -4 \end{pmatrix}\) — B1
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5 Show that [2 marks]
Show that the vectors \(\begin{pmatrix} 2 \\ -3 \end{pmatrix}\) and \(\begin{pmatrix} -6 \\ 9 \end{pmatrix}\) are parallel.
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Model answer
\(\begin{pmatrix} -6 \\ 9 \end{pmatrix} = -3 \times \begin{pmatrix} 2 \\ -3 \end{pmatrix}\). One vector is a multiple of the other, so they are parallel.
Mark scheme
- \(-3 \times \begin{pmatrix} 2 \\ -3 \end{pmatrix}\) or \(\begin{pmatrix} -6 \\ 9 \end{pmatrix} = -3 \times\) the first — M1
- One is a multiple of the other, so they are parallel — C1
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6 Work out [3 marks]
(a) Work out the length of the vector \(\begin{pmatrix} 6 \\ 8 \end{pmatrix}\). (2 marks) (b) \(A\) is the point \((1, 2)\) and \(B\) is the point \((5, 5)\). Work out the length of \(AB\). (1 mark)
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Model answer
(a) \(\sqrt{6^2 + 8^2} = \sqrt{100} = 10\). (b) \(\overrightarrow{AB} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}\), and \(\sqrt{4^2 + 3^2} = 5\).
Mark scheme
- (a) \(\sqrt{6^2 + 8^2}\) or \(\sqrt{100}\) — M1
- (a) 10 — A1
- (b) 5 — B1
Quick check
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1
What does the column vector \(\begin{pmatrix} -2 \\ 5 \end{pmatrix}\) mean?
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B: 2 left and 5 up
The top number is the horizontal move, and the bottom number is the vertical move.
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2
\(A\) is \((2, 5)\) and \(B\) is \((6, 2)\). What is \(\overrightarrow{AB}\)?
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A: \(\begin{pmatrix} 4 \\ -3 \end{pmatrix}\)
Subtract the start from the end: \((6 - 2, 2 - 5) = (4, -3)\).
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3
What is \(\begin{pmatrix} 3 \\ 2 \end{pmatrix} + \begin{pmatrix} 1 \\ 4 \end{pmatrix}\)?
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D: \(\begin{pmatrix} 4 \\ 6 \end{pmatrix}\)
Add the top numbers and add the bottom numbers.
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4
What is \(\begin{pmatrix} 3 \\ 2 \end{pmatrix} - \begin{pmatrix} 1 \\ 4 \end{pmatrix}\)?
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C: \(\begin{pmatrix} 2 \\ -2 \end{pmatrix}\)
\((3 - 1, 2 - 4) = (2, -2)\).
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5
What is \(3\begin{pmatrix} 2 \\ -1 \end{pmatrix}\)?
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B: \(\begin{pmatrix} 6 \\ -3 \end{pmatrix}\)
Multiply both numbers by 3.
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6
\(\mathbf{p} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\) and \(\mathbf{q} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}\). What is \(2\mathbf{p} - \mathbf{q}\)?
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A: \(\begin{pmatrix} 5 \\ 2 \end{pmatrix}\)
\(2\mathbf{p} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}\), then \((4 - (-1), 6 - 4) = (5, 2)\).
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7
Which vector is parallel to \(\begin{pmatrix} 2 \\ 3 \end{pmatrix}\)?
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D: \(\begin{pmatrix} 6 \\ 9 \end{pmatrix}\)
\(\begin{pmatrix} 6 \\ 9 \end{pmatrix} = 3\begin{pmatrix} 2 \\ 3 \end{pmatrix}\), so it is parallel.
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8
\(\overrightarrow{AB} = \begin{pmatrix} 4 \\ -3 \end{pmatrix}\). What is \(\overrightarrow{BA}\)?
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C: \(\begin{pmatrix} -4 \\ 3 \end{pmatrix}\)
\(\overrightarrow{BA} = -\overrightarrow{AB}\), so both signs change.
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9
What is the length of the vector \(\begin{pmatrix} 3 \\ 4 \end{pmatrix}\)?
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B: \(5\)
The length is \(\sqrt{3^2 + 4^2} = \sqrt{25} = 5\).
Vector Geometry and Proof
Just this lesson-
1 Write down [2 marks]
In the triangle \(OAB\), \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). Write down (a) \(\overrightarrow{AO}\), (b) \(\overrightarrow{AB}\).
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Model answer
(a) \(\overrightarrow{AO} = -\mathbf{a}\). (b) \(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\).
Mark scheme
- (a) \(-\mathbf{a}\) — B1
- (b) \(\mathbf{b} - \mathbf{a}\) — B1
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2 Find [5 marks]
\(OAB\) is a triangle with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(M\) is the midpoint of \(AB\), and \(N\) is the point on \(OB\) with \(ON = \dfrac{2}{3}OB\). (a) Find \(\overrightarrow{AB}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\). (1 mark) (b) Find \(\overrightarrow{OM}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\). (2 marks) (c) Find \(\overrightarrow{MN}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\). (2 marks)
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Model answer
(a) \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\). (b) \(\overrightarrow{OM} = \mathbf{a} + \dfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \dfrac{1}{2}\mathbf{a} + \dfrac{1}{2}\mathbf{b}\). (c) \(\overrightarrow{ON} = \dfrac{2}{3}\mathbf{b}\), so \(\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} = \dfrac{2}{3}\mathbf{b} - \dfrac{1}{2}\mathbf{a} - \dfrac{1}{2}\mathbf{b} = \dfrac{1}{6}\mathbf{b} - \dfrac{1}{2}\mathbf{a}\).
Mark scheme
- (a) \(\mathbf{b} - \mathbf{a}\) — B1
- (b) \(\mathbf{a} + \dfrac{1}{2}(\mathbf{b} - \mathbf{a})\) or \(\overrightarrow{OA} + \dfrac{1}{2}\overrightarrow{AB}\) — M1
- (b) \(\dfrac{1}{2}\mathbf{a} + \dfrac{1}{2}\mathbf{b}\) — A1
- (c) \(\overrightarrow{ON} - \overrightarrow{OM}\) or \(\overrightarrow{MO} + \overrightarrow{ON}\) — M1
- (c) \(\dfrac{1}{6}\mathbf{b} - \dfrac{1}{2}\mathbf{a}\) — A1
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3 Prove [3 marks]
\(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\) and \(\overrightarrow{OC} = 3\mathbf{b} - 2\mathbf{a}\). Prove that \(A\), \(B\) and \(C\) lie on a straight line.
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Model answer
\(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) and \(\overrightarrow{BC} = -\mathbf{b} + 3\mathbf{b} - 2\mathbf{a} = 2\mathbf{b} - 2\mathbf{a} = 2\overrightarrow{AB}\). The vectors are parallel and share the point \(B\), so \(A\), \(B\) and \(C\) are on a straight line.
Mark scheme
- \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) — M1
- \(\overrightarrow{BC} = 2\mathbf{b} - 2\mathbf{a}\) or \(2\overrightarrow{AB}\) — M1
- Parallel and a common point, so collinear — C1
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4 Find [3 marks]
\(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(P\) is the point on \(AB\) such that \(AP : PB = 2 : 3\). Find \(\overrightarrow{OP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
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Model answer
\(AP = \dfrac{2}{5}AB\), so \(\overrightarrow{AP} = \dfrac{2}{5}(\mathbf{b} - \mathbf{a})\). Then \(\overrightarrow{OP} = \mathbf{a} + \dfrac{2}{5}(\mathbf{b} - \mathbf{a}) = \dfrac{3}{5}\mathbf{a} + \dfrac{2}{5}\mathbf{b}\).
Mark scheme
- \(\overrightarrow{AP} = \dfrac{2}{5}\overrightarrow{AB}\) — M1
- \(\mathbf{a} + \dfrac{2}{5}(\mathbf{b} - \mathbf{a})\) — M1
- \(\dfrac{3}{5}\mathbf{a} + \dfrac{2}{5}\mathbf{b}\) — A1
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5 Show that [3 marks]
\(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(X\) is the point such that \(\overrightarrow{OX} = 3\mathbf{a}\), and \(Y\) is the point such that \(\overrightarrow{OY} = 3\mathbf{b}\). Show that \(XY\) is parallel to \(AB\).
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Model answer
\(\overrightarrow{XY} = -3\mathbf{a} + 3\mathbf{b} = 3(\mathbf{b} - \mathbf{a})\). Since \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\), \(\overrightarrow{XY} = 3\overrightarrow{AB}\), so the lines are parallel.
Mark scheme
- \(\overrightarrow{XY} = 3\mathbf{b} - 3\mathbf{a}\) — M1
- \(\overrightarrow{XY} = 3\overrightarrow{AB}\) or \(3(\mathbf{b} - \mathbf{a})\) — M1
- A multiple, so parallel — C1
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6 Show that [3 marks]
\(\overrightarrow{AB} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\) and \(\overrightarrow{BC} = \begin{pmatrix} 6 \\ 9 \end{pmatrix}\). Show that \(A\), \(B\) and \(C\) are on a straight line.
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Model answer
\(\begin{pmatrix} 6 \\ 9 \end{pmatrix} = 3 \times \begin{pmatrix} 2 \\ 3 \end{pmatrix}\), so \(\overrightarrow{BC} = 3\overrightarrow{AB}\). The vectors are parallel and share the point \(B\), so \(A\), \(B\) and \(C\) are on a straight line.
Mark scheme
- \(\overrightarrow{BC} = 3\overrightarrow{AB}\) — M1
- Parallel — A1
- Common point \(B\), so collinear — C1
Quick check
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1
\(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). What is \(\overrightarrow{AB}\)?
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C: \(\mathbf{b} - \mathbf{a}\)
Go from \(A\) to \(O\) (\(-\mathbf{a}\)) and then to \(B\) (\(\mathbf{b}\)).
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2
\(\overrightarrow{OA} = \mathbf{a}\). What is \(\overrightarrow{AO}\)?
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B: \(-\mathbf{a}\)
Going backwards reverses the vector.
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3
\(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(M\) is the midpoint of \(AB\). What is \(\overrightarrow{OM}\)?
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A: \(\dfrac{1}{2}(\mathbf{a} + \mathbf{b})\)
\(\overrightarrow{OM} = \mathbf{a} + \dfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})\).
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4
\(P\) is on \(AB\) with \(AP : PB = 1 : 2\). What fraction of \(AB\) is \(AP\)?
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D: \(\dfrac{1}{3}\)
There are \(1 + 2 = 3\) parts, and \(AP\) is 1 of them.
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5
\(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\), and \(P\) is on \(AB\) with \(AP : PB = 1 : 2\). What is \(\overrightarrow{OP}\)?
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C: \(\dfrac{2}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b}\)
\(\overrightarrow{OP} = \mathbf{a} + \dfrac{1}{3}(\mathbf{b} - \mathbf{a}) = \dfrac{2}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b}\).
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6
How do you show that two lines are parallel using vectors?
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B: Show that one vector is a multiple of the other
Parallel vectors are scalar multiples of each other.
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7
\(\overrightarrow{AB} = 2\overrightarrow{BC}\). What does this show?
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A: \(A\), \(B\) and \(C\) are on a straight line
The vectors are parallel and share the point \(B\), so the three points are collinear.
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8
\(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\) and \(\overrightarrow{OC} = 3\mathbf{b} - 2\mathbf{a}\). What is \(\overrightarrow{BC}\)?
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D: \(2\mathbf{b} - 2\mathbf{a}\)
\(\overrightarrow{BC} = -\mathbf{b} + 3\mathbf{b} - 2\mathbf{a} = 2\mathbf{b} - 2\mathbf{a}\).
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9
What is the length of the vector \(\begin{pmatrix} 5 \\ 12 \end{pmatrix}\)?
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C: \(13\)
\(\sqrt{5^2 + 12^2} = \sqrt{169} = 13\).
Ruler-and-Compass Constructions
Just this lesson-
1 Construct [2 marks]
Use ruler and compasses to construct an angle of \(60^\circ\) at the point \(A\) on a line. You must show all your construction lines.
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Model answer
Draw an arc with centre \(A\) that crosses the line at \(P\). With the same radius and centre \(P\), draw an arc that crosses the first arc at \(Q\). Join \(A\) to \(Q\). The angle \(PAQ\) is \(60^\circ\), because \(APQ\) is an equilateral triangle.
Mark scheme
- Arc from A and an arc of the same radius from the crossing point — M1
- Line through \(A\) and the point where the arcs cross — A1
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2 Construct [3 marks]
The diagram shows a line \(AB\) of length 6 cm. (a) Use ruler and compasses to construct the perpendicular bisector of \(AB\). You must show all your construction lines. (2 marks) (b) Write down the distance from \(A\) to the point where the bisector crosses \(AB\). (1 mark)
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Model answer
(a) Open the compasses to more than 3 cm. Draw arcs of the same radius from \(A\) and from \(B\), crossing above and below the line. Join the two crossing points with a straight line. (b) The bisector crosses \(AB\) at its midpoint, 3 cm from \(A\).
Mark scheme
- (a) Arcs of equal radius from A and B, crossing above and below — M1
- (a) A straight line through the crossing points, with the arcs left on — A1
- (b) 3 cm — B1
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3 Construct [3 marks]
Use ruler and compasses to construct a triangle \(ABC\) with \(AB = 6\) cm, \(BC = 5\) cm and \(AC = 4\) cm. You must show all your construction lines.
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Model answer
Draw \(AB\) 6 cm long. With the compasses set to 4 cm and the point on \(A\), draw an arc. With the compasses set to 5 cm and the point on \(B\), draw an arc that crosses the first. Join the crossing point \(C\) to \(A\) and \(B\).
Mark scheme
- \(AB = 6\) cm drawn accurately — B1
- Arc of radius 4 cm from \(A\) and arc of radius 5 cm from \(B\) — M1
- Triangle completed, with \(C\) at the crossing — A1
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4 Explain [2 marks]
\(M\) is a point on the perpendicular bisector of \(AB\). Explain why \(MA = MB\).
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Model answer
The perpendicular bisector cuts \(AB\) in half at right angles. The two triangles formed by \(M\), \(A\), \(B\) and the midpoint have equal sides next to the right angle and a common side, so they are congruent and \(MA = MB\). More simply, every point on the perpendicular bisector is the same distance from \(A\) and \(B\).
Mark scheme
- States that the bisector cuts AB in half at a right angle — B1
- Uses congruent triangles, or states that points on the bisector are equidistant — B1
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5 Construct [3 marks]
Use ruler and compasses to construct an angle of \(30^\circ\). You must show all your construction lines.
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Model answer
Construct an angle of \(60^\circ\) with two arcs of the same radius. Then bisect the \(60^\circ\) angle by drawing an arc on both arms, then matching arcs from those points, and a line through the crossing. This gives two angles of \(30^\circ\).
Mark scheme
- A construction of \(60^\circ\) — M1
- A bisector construction with arcs of equal radius — M1
- A \(30^\circ\) angle completed — A1
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6 Construct [3 marks]
Use ruler and compasses to construct an equilateral triangle with sides of 5 cm. You must show all your construction lines.
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Model answer
Draw a line \(AB\) of 5 cm. Set the compasses to 5 cm. Draw an arc from \(A\) and an arc from \(B\), crossing at \(C\). Join \(C\) to \(A\) and \(B\). All three sides are 5 cm.
Mark scheme
- A line of 5 cm drawn — B1
- Two arcs of radius 5 cm from the ends — M1
- Triangle completed — A1
Quick check
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1
What does the perpendicular bisector of a line do?
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D: Cuts it in half at right angles
Perpendicular means at right angles, and bisector means cuts in half.
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2
What must you leave on your drawing in a construction?
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C: The construction arcs
The arcs show the method, and earn the marks.
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3
Which instruments do you use for a construction?
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B: A ruler and compasses
Constructions use a ruler and a pair of compasses.
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4
Which angle is constructed using two arcs of the same radius, as in an equilateral triangle?
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A: \(60^\circ\)
The triangle with three equal sides has three angles of \(60^\circ\).
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5
A \(60^\circ\) angle is bisected. What is the size of each part?
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D: \(30^\circ\)
\(60 \div 2 = 30\).
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6
Why must the compasses be opened to more than half the length of the line when constructing a perpendicular bisector?
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C: So that the arcs from both ends cross
If the radius is too small the arcs do not meet.
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7
What is true of every point on an angle bisector?
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B: It is the same distance from both arms
The bisector is equidistant from the two arms.
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8
A triangle has sides 6 cm, 5 cm and 4 cm. After drawing the 6 cm side \(AB\), how do you find \(C\) if \(AC = 4\) cm and \(BC = 5\) cm?
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A: Arc of radius 4 cm from \(A\) and arc of radius 5 cm from \(B\), where they cross
The third corner is the point 4 cm from \(A\) and 5 cm from \(B\).
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9
What is the shortest distance from a point to a line?
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D: The perpendicular distance
The shortest path to a line meets it at a right angle.
Loci
Just this lesson-
1 Draw [2 marks]
Draw the locus of all the points that are 3 cm from a point \(P\).
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Model answer
The locus is a circle with centre \(P\) and radius 3 cm.
Mark scheme
- A circle — M1
- With centre \(P\) and radius 3 cm — A1
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2 Shade [4 marks]
The diagram shows a rectangular garden \(ABCD\), drawn to a scale of 1 cm to 1 m. A bush must be planted so that it is closer to \(AB\) than to \(AD\), and less than 6 m from \(A\). Show how to find the region where the bush could be planted. (4 marks)
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Model answer
Bisect the angle at \(A\), which makes a \(45^\circ\) line from \(A\). The region closer to \(AB\) is below this line. Draw an arc of radius 6 cm with centre \(A\). The region is inside the arc, below the bisector, and inside the rectangle.
Mark scheme
- Angle bisector at \(A\), with arcs — M1
- Side nearer \(AB\) chosen — A1
- Arc of radius 6 cm centred on \(A\) — M1
- The correct region shaded, inside the rectangle — A1
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3 Draw [3 marks]
\(A\) and \(B\) are two points 6 cm apart. Describe the locus of the points that are the same distance from \(A\) and from \(B\), and say how far from \(A\) it crosses the line \(AB\).
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Model answer
The locus is the perpendicular bisector of \(AB\). It crosses \(AB\) at its midpoint, which is 3 cm from \(A\).
Mark scheme
- The perpendicular bisector — M1
- Of the line AB — A1
- 3 cm from A — B1
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4 Draw [3 marks]
\(PQ\) is a line of length 5 cm. Describe the locus of the points that are 2 cm from the line \(PQ\), and work out the length of the straight parts of it.
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Model answer
The locus is two straight lines, 2 cm from \(PQ\) on each side and parallel to it, joined by semicircles of radius 2 cm round \(P\) and \(Q\). Each straight part is 5 cm long, so the straight parts total \(2 \times 5 = 10\) cm.
Mark scheme
- Two parallel lines 2 cm from PQ — M1
- Semicircles at both ends — A1
- 10 cm — B1
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5 Shade [4 marks]
\(P\) and \(Q\) are two points 6 cm apart. Shade the region of points that are closer to \(P\) than to \(Q\) and are less than 4 cm from \(Q\). (4 marks)
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Model answer
Construct the perpendicular bisector of \(PQ\), which is 3 cm from each point, and draw a circle of radius 4 cm around \(Q\). The region is inside the circle and on the \(P\) side of the bisector.
Mark scheme
- Perpendicular bisector of PQ with arcs — M1
- The \(P\) side chosen — A1
- Circle of radius 4 cm centred on \(Q\) — M1
- The correct region shaded — A1
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6 Work out [3 marks]
\(A\) and \(B\) are 8 cm apart. Is there a point that is closer to \(B\) than to \(A\) and less than 3 cm from \(A\)? Explain your answer.
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Model answer
No. The points that are closer to \(B\) than \(A\) are more than 4 cm from \(A\), because the perpendicular bisector is 4 cm from \(A\). A point less than 3 cm from \(A\) is inside a circle that does not reach the bisector.
Mark scheme
- No — B1
- The bisector is 4 cm from \(A\) — M1
- The circle of radius 3 cm does not reach it — A1
Quick check
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1
What is the locus of points 3 cm from a point \(P\)?
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A: A circle with centre \(P\) and radius 3 cm
All points the same distance from \(P\) form a circle.
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2
What is the locus of points that are the same distance from two points \(A\) and \(B\)?
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D: The perpendicular bisector of \(AB\)
Every point on the perpendicular bisector is equidistant from \(A\) and \(B\).
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3
What is the locus of points that are the same distance from two lines that meet at a point?
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C: The bisector of the angle between the lines
The angle bisector is equidistant from both arms.
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4
On a drawing, which region is “less than 3 cm from \(P\)”?
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B: Inside the circle of radius 3 cm around \(P\)
Less than 3 cm means closer than the circle, so inside it.
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5
What is the locus of points 2 cm from a line segment?
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A: Parallel lines 2 cm on each side with semicircular ends
Near the ends, the points 2 cm away form semicircles.
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6
\(A\) and \(B\) are two points. Which side of the perpendicular bisector is “closer to \(A\) than \(B\)”?
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D: The side containing \(A\)
Points nearer to \(A\) are on \(A\)'s side of the bisector.
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7
On a scale drawing with 1 cm for 1 m, a tree must be within 4 m of a post. What do you draw?
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C: A circle of radius 4 cm around the post
4 m is 4 cm on the drawing.
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8
A rectangle \(ABCD\) has \(A\) at the bottom left and \(B\) to its right. Which line separates the points closer to \(AB\) from those closer to \(AD\)?
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B: The bisector of angle \(A\)
The angle bisector at \(A\) is the same distance from \(AB\) and \(AD\).
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9
\(A\) and \(B\) are 8 cm apart. Is there a point closer to \(B\) than \(A\) that is less than 3 cm from \(A\)?
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A: No
The bisector is 4 cm from \(A\), and a circle of radius 3 cm around \(A\) does not reach it.
Bearings, Scale Drawings, Plans and Elevations
Just this lesson-
1 Write down [2 marks]
Write down the three-figure bearing of (a) east, (b) south-west.
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Model answer
(a) \(090^\circ\). (b) South is \(180^\circ\) and west is \(270^\circ\), so south-west is halfway: \(225^\circ\).
Mark scheme
- (a) \(090^\circ\) — B1
- (b) \(225^\circ\) — B1
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2 Work out [4 marks]
The diagram shows three towns \(A\), \(B\) and \(C\). The bearing of \(B\) from \(A\) is \(070^\circ\). The bearing of \(C\) from \(B\) is \(150^\circ\). (a) Work out the bearing of \(A\) from \(B\). (2 marks) (b) Work out the size of angle \(ABC\). (2 marks)
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Model answer
(a) \(070 + 180 = 250\), so the bearing is \(250^\circ\). (b) The bearing of \(A\) from \(B\) is \(250^\circ\) and the bearing of \(C\) from \(B\) is \(150^\circ\), so angle \(ABC = 250 - 150 = 100^\circ\).
Mark scheme
- (a) \(070 + 180\) or parallel north lines — M1
- (a) \(250^\circ\) — A1
- (b) \(250 - 150\) — M1
- (b) \(100^\circ\) — A1
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3 Work out [2 marks]
The bearing of \(B\) from \(A\) is \(130^\circ\). Work out the bearing of \(A\) from \(B\).
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Model answer
\(130 + 180 = 310\), so the bearing is \(310^\circ\).
Mark scheme
- \(130 + 180\) — M1
- \(310^\circ\) — A1
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4 Work out [3 marks]
A map has a scale of 1 : 50 000. (a) Two villages are 4 cm apart on the map. Work out the real distance in kilometres. (2 marks) (b) Two other villages are 7.5 km apart. How far apart are they on the map? (1 mark)
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Model answer
(a) \(4 \times 50\,000 = 200\,000\) cm \(= 2\) km. (b) \(7.5\) km \(= 750\,000\) cm, and \(750\,000 \div 50\,000 = 15\) cm.
Mark scheme
- (a) \(4 \times 50\,000\) — M1
- (a) 2 km — A1
- (b) 15 cm — B1
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5 Write down [3 marks]
A cuboid is 4 cm long, 3 cm wide and 2 cm high. Write down the dimensions of (a) its plan, (b) its front elevation, looking along the 3 cm width, (c) its side elevation, looking along the 4 cm length.
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Model answer
(a) The plan is a rectangle 4 cm by 3 cm. (b) The front elevation shows the length and the height, a rectangle 4 cm by 2 cm. (c) The side elevation shows the width and the height, a rectangle 3 cm by 2 cm.
Mark scheme
- (a) 4 cm by 3 cm — B1
- (b) 4 cm by 2 cm — B1
- (c) 3 cm by 2 cm — B1
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6 Work out [4 marks]
\(B\) is 5 km from \(A\) on a bearing of \(030^\circ\). \(C\) is 5 km from \(B\) on a bearing of \(150^\circ\). (a) Work out the size of angle \(ABC\). (2 marks) (b) Work out the bearing of \(C\) from \(A\). (2 marks)
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Model answer
(a) The bearing of \(A\) from \(B\) is \(030 + 180 = 210^\circ\), and the bearing of \(C\) from \(B\) is \(150^\circ\), so angle \(ABC = 210 - 150 = 60^\circ\). (b) Triangle \(ABC\) is isosceles with a \(60^\circ\) angle, so it is equilateral, and angle \(BAC = 60^\circ\). The bearing of \(C\) from \(A\) is \(030 + 60 = 090^\circ\).
Mark scheme
- (a) \(210\) seen — M1
- (a) \(60^\circ\) — A1
- (b) Triangle is equilateral, or angle \(BAC = 60^\circ\) — M1
- (b) \(090^\circ\) — A1
Quick check
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1
What is the bearing of east?
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B: \(090^\circ\)
North is 000, east is 090, south is 180 and west is 270.
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2
What is the bearing of south?
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A: \(180^\circ\)
South is half a turn from north.
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3
The bearing of \(B\) from \(A\) is \(130^\circ\). What is the bearing of \(A\) from \(B\)?
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D: \(310^\circ\)
Add \(180^\circ\): \(130 + 180 = 310\).
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4
The bearing of \(B\) from \(A\) is \(072^\circ\). What is the bearing of \(A\) from \(B\)?
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C: \(252^\circ\)
Add \(180^\circ\): \(072 + 180 = 252\).
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5
The bearing of \(B\) from \(A\) is \(250^\circ\). What is the bearing of \(A\) from \(B\)?
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B: \(070^\circ\)
Subtract \(180^\circ\): \(250 - 180 = 70\), written as 070.
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6
From where, and in which direction, is a bearing measured?
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A: Clockwise from the north line
Bearings are measured clockwise from north.
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7
A map has a scale of \(1 : 25\,000\). Two towns are 6 cm apart on the map. What is the real distance?
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D: 1.5 km
\(6 \times 25\,000 = 150\,000\) cm \(= 1.5\) km.
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8
A plan has a scale of \(1 : 1000\). A length of 5 cm on the plan is how long in real life?
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C: 50 m
\(5 \times 1000 = 5000\) cm \(= 50\) m.
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9
What is the plan of a solid?
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B: The view from above
A plan is a view looking straight down.