Exam questions · Maths
Vectors, Constructions and Loci
- 30 exam questions
- 90 marks
- 45 quick checks
Column Vectors and Vector Arithmetic
Just this lesson-
1 Calculate [2 marks]
Calculate \(\begin{pmatrix} 4 \\ -2 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \end{pmatrix}\). [2 marks]
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Model answer
\(\begin{pmatrix} 3 \\ -5 \end{pmatrix} = \begin{pmatrix} 3 \\ -5 \end{pmatrix}\).
Mark scheme
- Subtracts the top numbers or the bottom numbers — M1
- \(\begin{pmatrix} 3 \\ -5 \end{pmatrix}\) — A1
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2 Calculate [4 marks]
The vectors \(\mathbf{m}\) and \(\mathbf{n}\) are drawn on the grid. (a) Write \(\mathbf{m}\) and \(\mathbf{n}\) as column vectors. [2 marks] (b) Calculate \(\mathbf{m} + 2\mathbf{n}\). [2 marks]
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Model answer
(a) \(\mathbf{m}\) goes 3 right and 1 up, so \(\mathbf{m} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}\). \(\mathbf{n}\) goes 2 left and 3 down, so \(\mathbf{n} = \begin{pmatrix} -2 \\ -3 \end{pmatrix}\). (b) \(2\mathbf{n} = \begin{pmatrix} -4 \\ -6 \end{pmatrix}\), so \(\mathbf{m} + 2\mathbf{n} = \begin{pmatrix} -1 \\ -5 \end{pmatrix}\).
Mark scheme
- (a) \(\mathbf{m} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}\) — B1
- (a) \(\mathbf{n} = \begin{pmatrix} -2 \\ -3 \end{pmatrix}\) — B1
- (b) \(2\mathbf{n} = \begin{pmatrix} -4 \\ -6 \end{pmatrix}\) or a correct method — M1
- (b) \(\begin{pmatrix} -1 \\ -5 \end{pmatrix}\) — A1 (follow through from (a))
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3 Calculate [3 marks]
\(A\) is the point \((5, -2)\) and \(B\) is the point \((-1, 4)\). (a) Calculate \(\overrightarrow{AB}\) as a column vector. [2 marks] (b) Write down \(\overrightarrow{BA}\) as a column vector. [1 mark]
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Model answer
(a) \((-1 - 5, 4 - (-2)) = (-6, 6)\), so \(\overrightarrow{AB} = \begin{pmatrix} -6 \\ 6 \end{pmatrix}\). (b) \(\overrightarrow{BA} = \begin{pmatrix} 6 \\ -6 \end{pmatrix}\).
Mark scheme
- (a) \(-1 - 5\) or \(4 - (-2)\) — M1
- (a) \(\begin{pmatrix} -6 \\ 6 \end{pmatrix}\) — A1
- (b) \(\begin{pmatrix} 6 \\ -6 \end{pmatrix}\) — B1 (follow through from (a))
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4 Calculate [3 marks]
\(\mathbf{u} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}\) and \(\mathbf{v} = \begin{pmatrix} 3 \\ 5 \end{pmatrix}\). (a) Calculate \(3\mathbf{u} - \mathbf{v}\). [2 marks] (b) Calculate \(2\mathbf{u} + \mathbf{v}\). [1 mark]
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Model answer
(a) \(3\mathbf{u} = \begin{pmatrix} 3 \\ -6 \end{pmatrix}\), so \(3\mathbf{u} - \mathbf{v} = \begin{pmatrix} 0 \\ -11 \end{pmatrix}\). (b) \(2\mathbf{u} = \begin{pmatrix} 2 \\ -4 \end{pmatrix}\), so \(2\mathbf{u} + \mathbf{v} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}\).
Mark scheme
- (a) \(3\mathbf{u} = \begin{pmatrix} 3 \\ -6 \end{pmatrix}\) or a correct method — M1
- (a) \(\begin{pmatrix} 0 \\ -11 \end{pmatrix}\) — A1
- (b) \(\begin{pmatrix} 5 \\ 1 \end{pmatrix}\) — B1
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5 Show that [2 marks]
Show that the vectors \(\begin{pmatrix} -4 \\ 6 \end{pmatrix}\) and \(\begin{pmatrix} 2 \\ -3 \end{pmatrix}\) are parallel. [2 marks]
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Model answer
\(\begin{pmatrix} -4 \\ 6 \end{pmatrix} = -2 \times \begin{pmatrix} 2 \\ -3 \end{pmatrix}\). One vector is a multiple of the other, so they are parallel.
Mark scheme
- \(-2 \times \begin{pmatrix} 2 \\ -3 \end{pmatrix}\) — M1
- A multiple, so parallel — A1
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6 Calculate [3 marks]
(a) Calculate the length of the vector \(\begin{pmatrix} 5 \\ 12 \end{pmatrix}\). [2 marks] (b) Calculate the length of the vector \(\begin{pmatrix} 8 \\ 6 \end{pmatrix}\). [1 mark]
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Model answer
(a) \(\sqrt{5^2 + 12^2} = \sqrt{169} = 13\). (b) \(\sqrt{8^2 + 6^2} = \sqrt{100} = 10\).
Mark scheme
- (a) \(\sqrt{5^2 + 12^2}\) or \(\sqrt{169}\) — M1
- (a) 13 — A1
- (b) 10 — 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) \(\overrightarrow{BO}\), (b) \(\overrightarrow{BA}\). [2 marks]
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Model answer
(a) \(\overrightarrow{BO} = -\mathbf{b}\). (b) \(\overrightarrow{BA} = \overrightarrow{BO} + \overrightarrow{OA} = -\mathbf{b} + \mathbf{a} = \mathbf{a} - \mathbf{b}\).
Mark scheme
- (a) \(-\mathbf{b}\) — B1
- (b) \(\mathbf{a} - \mathbf{b}\) — B1
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2 Calculate [4 marks]
\(OAB\) is a triangle with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(P\) is the point on \(AB\) such that \(AP : PB = 2 : 1\). (a) Write down \(\overrightarrow{AB}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\). [1 mark] (b) Find \(\overrightarrow{OP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\). Give your answer in its simplest form. [3 marks]
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Model answer
(a) \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\). (b) \(AP = \dfrac{2}{3}AB\), so \(\overrightarrow{AP} = \dfrac{2}{3}(\mathbf{b} - \mathbf{a})\). Then \(\overrightarrow{OP} = \mathbf{a} + \dfrac{2}{3}(\mathbf{b} - \mathbf{a}) = \dfrac{1}{3}\mathbf{a} + \dfrac{2}{3}\mathbf{b}\).
Mark scheme
- (a) \(\mathbf{b} - \mathbf{a}\) — B1
- (b) \(\overrightarrow{AP} = \dfrac{2}{3}\overrightarrow{AB}\) — M1
- (b) \(\mathbf{a} + \dfrac{2}{3}(\mathbf{b} - \mathbf{a})\) — M1
- (b) \(\dfrac{1}{3}\mathbf{a} + \dfrac{2}{3}\mathbf{b}\) — A1
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3 Prove [3 marks]
\(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\) and \(\overrightarrow{OC} = 4\mathbf{b} - 3\mathbf{a}\). Prove that \(A\), \(B\) and \(C\) lie on a straight line. [3 marks]
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Model answer
\(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) and \(\overrightarrow{BC} = -\mathbf{b} + 4\mathbf{b} - 3\mathbf{a} = 3\mathbf{b} - 3\mathbf{a} = 3\overrightarrow{AB}\). The vectors are parallel and share the point \(B\), so \(A\), \(B\) and \(C\) are on a straight line.
Mark scheme
- \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) — M1
- \(\overrightarrow{BC} = 3\mathbf{b} - 3\mathbf{a}\) or \(3\overrightarrow{AB}\) — M1
- Parallel with a common point, so collinear — A1
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4 Find [3 marks]
\(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(Q\) is the point on \(OA\) such that \(OQ : QA = 1 : 3\). Find \(\overrightarrow{BQ}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\). [3 marks]
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Model answer
\(OQ = \dfrac{1}{4}OA\), so \(\overrightarrow{OQ} = \dfrac{1}{4}\mathbf{a}\). Then \(\overrightarrow{BQ} = \overrightarrow{BO} + \overrightarrow{OQ} = -\mathbf{b} + \dfrac{1}{4}\mathbf{a} = \dfrac{1}{4}\mathbf{a} - \mathbf{b}\).
Mark scheme
- \(\overrightarrow{OQ} = \dfrac{1}{4}\mathbf{a}\) — M1
- \(\overrightarrow{BQ} = \overrightarrow{BO} + \overrightarrow{OQ}\) or \(-\mathbf{b} + \dfrac{1}{4}\mathbf{a}\) — M1
- \(\dfrac{1}{4}\mathbf{a} - \mathbf{b}\) — 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 \(\overrightarrow{OX} = 2\mathbf{a}\), and \(Y\) is the point such that \(\overrightarrow{OY} = 2\mathbf{b}\). Show that \(XY\) is parallel to \(AB\). [3 marks]
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Model answer
\(\overrightarrow{XY} = -2\mathbf{a} + 2\mathbf{b} = 2(\mathbf{b} - \mathbf{a}) = 2\overrightarrow{AB}\). It is a multiple of \(\overrightarrow{AB}\), so the lines are parallel.
Mark scheme
- \(\overrightarrow{XY} = 2\mathbf{b} - 2\mathbf{a}\) — M1
- \(2\overrightarrow{AB}\) or \(2(\mathbf{b} - \mathbf{a})\) — M1
- A multiple, so parallel — A1
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6 Calculate [3 marks]
\(\overrightarrow{PQ} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}\) and \(\overrightarrow{QR} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}\). (a) Show that \(P\), \(Q\) and \(R\) lie on a straight line. [2 marks] (b) Calculate the length of \(PR\). [1 mark]
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Model answer
(a) \(\begin{pmatrix} 6 \\ 8 \end{pmatrix} = 2 \times \begin{pmatrix} 3 \\ 4 \end{pmatrix}\), so \(\overrightarrow{PQ} = 2\overrightarrow{QR}\). The vectors are parallel and share \(Q\), so the points are collinear. (b) \(\overrightarrow{PR} = \begin{pmatrix} 9 \\ 12 \end{pmatrix}\), and the length is \(\sqrt{81 + 144} = \sqrt{225} = 15\).
Mark scheme
- (a) \(\overrightarrow{PQ} = 2\overrightarrow{QR}\) — M1
- (a) Parallel with a common point, so collinear — A1
- (b) 15 — B1
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 [3 marks]
Use ruler and compasses to construct a triangle \(ABC\) with \(AB = 7\) cm, \(BC = 6\) cm and \(AC = 5\) cm. You must show all your construction lines. [3 marks]
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Model answer
Draw \(AB\) 7 cm long. With the compasses set to 5 cm and the point on \(A\), draw an arc. With the compasses set to 6 cm and the point on \(B\), draw an arc that crosses the first. Join the crossing point \(C\) to \(A\) and \(B\).
Mark scheme
- \(AB = 7\) cm drawn accurately — B1
- Arc of radius 5 cm from \(A\) and arc of radius 6 cm from \(B\) — M1
- Triangle completed — A1
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2 Construct [3 marks]
The diagram shows a straight line \(l\) and a point \(P\) that is not on the line. (a) Use ruler and compasses to construct the perpendicular from \(P\) to the line \(l\). You must show all your construction lines. [2 marks] (b) The point \(P\) is 3 cm above the line. Write down the shortest distance from \(P\) to \(l\). [1 mark]
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Model answer
(a) With the point on \(P\), draw an arc that cuts \(l\) twice, at \(X\) and \(Y\). From \(X\) and \(Y\) draw arcs of equal radius that cross below the line. Join \(P\) to the crossing point. (b) The shortest distance is the perpendicular distance, 3 cm.
Mark scheme
- (a) An arc from P that cuts the line twice, and matching arcs that cross — M1
- (a) A straight line from \(P\) through the crossing — A1
- (b) 3 cm — B1
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3 Construct [3 marks]
Use ruler and compasses to construct the perpendicular bisector of a line \(XY\) of length 8 cm. You must show all your construction lines. [3 marks]
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Model answer
Open the compasses to more than 4 cm. Draw arcs of equal radius from \(X\) and \(Y\) that cross above and below the line. Join the crossing points with a straight line.
Mark scheme
- Line \(XY\) of 8 cm drawn — B1
- Arcs of equal radius from both ends — M1
- A straight line through the crossing points — A1
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4 Explain [2 marks]
Explain why the compasses must be set to more than half the length of the line when constructing a perpendicular bisector. [2 marks]
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Model answer
The arcs from the two ends only meet if the radius is more than half the length of the line. If the radius is half or less, the arcs touch at the midpoint or do not meet, so there are no two crossing points to join.
Mark scheme
- The arcs must cross — B1
- If the radius is half or less the arcs do not cross twice — B1
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5 Construct [3 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. [3 marks]
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Model answer
With the point of the compasses on \(A\), draw an arc that crosses the line at \(P\). With the same radius and the point on \(P\), draw an arc that crosses the first arc at \(Q\). Join \(A\) to \(Q\). Triangle \(APQ\) is equilateral, so the angle \(PAQ\) is \(60^\circ\).
Mark scheme
- An arc from A that crosses the line — M1
- An arc of the same radius from the crossing point — M1
- A line through \(A\) and the crossing point — A1
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6 Construct [4 marks]
Use ruler and compasses to construct an angle of \(30^\circ\) at the point \(A\) on a line. You must show all your construction lines. [4 marks]
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Model answer
Construct \(60^\circ\) at \(A\) with two arcs of the same radius. Then bisect the \(60^\circ\) angle: draw an arc from \(A\) that cuts both arms, draw matching arcs from those points that cross, and join \(A\) to the crossing.
Mark scheme
- A construction of \(60^\circ\) — M1
- An arc on both arms of the \(60^\circ\) angle — M1
- Matching arcs that cross — M1
- A line from \(A\) through the crossing — 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 2 cm from a point \(T\). [2 marks]
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Model answer
The locus is a circle with centre \(T\) and radius 2 cm.
Mark scheme
- A circle — M1
- With centre \(T\) and radius 2 cm — A1
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2 Shade [4 marks]
The diagram shows the line \(XY\), which is 6 cm long. Shade the region of points that are less than 2 cm from the line \(XY\) and closer to \(X\) than to \(Y\). [4 marks]
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Model answer
Draw the shape that is 2 cm from \(XY\): parallel lines 2 cm on each side with semicircular ends. Then construct the perpendicular bisector of \(XY\), which crosses at 3 cm. The region is the part of the shape on the \(X\) side of the bisector.
Mark scheme
- Parallel lines 2 cm from XY with semicircular ends — M1
- Perpendicular bisector of XY with arcs — M1
- The \(X\) side chosen — A1
- The correct region shaded — A1
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3 Draw [3 marks]
A square garden has sides of 10 m. A path must be the same distance from two adjacent sides, \(AB\) and \(AD\), starting at the corner \(A\). Describe the locus of points inside the garden that are the same distance from \(AB\) and \(AD\). [3 marks]
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Model answer
The locus is the bisector of angle \(A\), a straight line from \(A\) at \(45^\circ\) to both sides, which is the diagonal \(AC\) of the square.
Mark scheme
- The bisector of the angle at A — M1
- At \(45^\circ\) to the sides — A1
- The diagonal AC — B1
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4 Draw [3 marks]
A radio mast transmits to a distance of 30 km. A scale drawing uses 1 cm for 10 km. Describe the locus of the points that are exactly 30 km from the mast, and say how to draw it. [3 marks]
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Model answer
The locus is a circle with the mast at the centre. On the scale drawing the radius is \(30 \div 10 = 3\) cm.
Mark scheme
- A circle — M1
- Centre at the mast — A1
- Radius 3 cm on the drawing — B1
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5 Shade [4 marks]
\(A\) and \(B\) are two points 8 cm apart. Show how to find the region of points that are less than 5 cm from \(A\) and closer to \(B\) than to \(A\). [4 marks]
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Model answer
Draw a circle of radius 5 cm around \(A\) and the perpendicular bisector of \(AB\), which is 4 cm from \(A\). The region is inside the circle and on the \(B\) side of the bisector, a small segment of the circle.
Mark scheme
- Circle of radius 5 cm centred on \(A\) — M1
- Perpendicular bisector with arcs — M1
- The \(B\) side chosen — A1
- The segment shaded — A1
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6 Work out [3 marks]
\(P\) and \(Q\) are 12 cm apart. Is there a point that is closer to \(Q\) than to \(P\) and less than 5 cm from \(P\)? Explain your answer. [3 marks]
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Model answer
No. The perpendicular bisector is 6 cm from \(P\), and points closer to \(Q\) are on the far side of it. A point less than 5 cm from \(P\) is inside a circle that does not reach the bisector.
Mark scheme
- No — B1
- The bisector is 6 cm from \(P\) — M1
- The circle of radius 5 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) west, (b) north-east. [2 marks]
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Model answer
(a) \(270^\circ\). (b) North-east is halfway between north and east, \(045^\circ\).
Mark scheme
- (a) \(270^\circ\) — B1
- (b) \(045^\circ\) — B1
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2 Calculate [4 marks]
The diagram shows two points \(A\) and \(B\). On a map, \(AB\) is 4 cm and the scale is 1 : 50 000. The bearing of \(B\) from \(A\) is \(240^\circ\). (a) Calculate the bearing of \(A\) from \(B\). [2 marks] (b) Calculate the real distance \(AB\) in kilometres. [2 marks]
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Model answer
(a) The bearing is more than \(180^\circ\), so subtract: \(240 - 180 = 60\), written \(060^\circ\). (b) \(4 \times 50\,000 = 200\,000\) cm \(= 2\) km.
Mark scheme
- (a) \(240 - 180\) — M1
- (a) \(060^\circ\) — A1
- (b) \(4 \times 50\,000\) — M1
- (b) 2 km — A1
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3 Calculate [2 marks]
The bearing of \(B\) from \(A\) is \(250^\circ\). Calculate the bearing of \(A\) from \(B\). [2 marks]
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Model answer
\(250 - 180 = 70\), so the bearing is \(070^\circ\).
Mark scheme
- \(250 - 180\) — M1
- \(070^\circ\) — A1
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4 Calculate [3 marks]
A plan is drawn with a scale of 1 : 200. [3 marks] (a) A wall is 6 cm long on the plan. Calculate its real length in metres. [2 marks] (b) A room is 8 m long. Calculate its length on the plan. [1 mark]
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Model answer
(a) \(6 \times 200 = 1200\) cm \(= 12\) m. (b) \(8\) m \(= 800\) cm, and \(800 \div 200 = 4\) cm.
Mark scheme
- (a) \(6 \times 200\) — M1
- (a) 12 m — A1
- (b) 4 cm — B1
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5 Write down [3 marks]
A triangular prism has a base of 6 cm and a height of 4 cm in its triangular face, and the prism is 10 cm long. It lies on a rectangular face. Write down the dimensions of (a) its plan, (b) its elevation from the end, (c) its elevation from the side. [3 marks]
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Model answer
(a) The plan is a rectangle 6 cm by 10 cm. (b) The elevation from the end is the triangle, with base 6 cm and height 4 cm. (c) The elevation from the side is a rectangle 10 cm by 4 cm.
Mark scheme
- (a) A rectangle 6 cm by 10 cm — B1
- (b) A triangle with base 6 cm and height 4 cm — B1
- (c) A rectangle 10 cm by 4 cm — B1
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6 Calculate [4 marks]
\(B\) is on a bearing of \(090^\circ\) from \(A\), and \(AB = 8\) km. \(C\) is on a bearing of \(180^\circ\) from \(B\), and \(BC = 6\) km. (a) Calculate the length \(AC\). [2 marks] (b) Write down the bearing of \(A\) from \(B\). [1 mark] (c) Write down the angle \(ABC\). [1 mark]
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Model answer
(a) Angle \(ABC = 90^\circ\), so \(AC^2 = 8^2 + 6^2 = 100\) and \(AC = 10\) km. (b) \(090 + 180 = 270^\circ\). (c) The bearing of \(A\) from \(B\) is \(270^\circ\) and the bearing of \(C\) from \(B\) is \(180^\circ\), so angle \(ABC = 270 - 180 = 90^\circ\).
Mark scheme
- (a) \(8^2 + 6^2\) — M1
- (a) 10 km — A1
- (b) \(270^\circ\) — B1
- (c) \(90^\circ\) — B1
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.