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Date	May 2022	Marks available	2	Reference code	22M.1.SL.TZ1.4
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 1
Command term	Find	Question number	4	Adapted from	N/A

Question

Three towns, $A$ , $B$ and $C$ are represented as coordinates on a map, where the $x$ and $y$ axes represent the distances east and north of an origin, respectively, measured in kilometres.

Town $A$ is located at $(- 6, - 1)$ and town $B$ is located at $(8, 6)$ . A road runs along the perpendicular bisector of $[AB]$ . This information is shown in the following diagram.

Find the equation of the line that the road follows.

[5]

Town $C$ is due north of town $A$ and the road passes through town $C$ .

Find the $y$ -coordinate of town $C$ .

[2]

Markscheme

midpoint $(1, 2.5)$ A1

$m_{A B} = \frac{6 - (- 1)}{8 - (- 6)} = \frac{1}{2}$ (M1)A1

Note: Accept equivalent gradient statements including using midpoint.

$m_{⊥} = - 2$ M1

Note: Award M1 for finding the negative reciprocal of their gradient.

$y - 2.5 = - 2 (x - 1)$ OR $y = - 2 x + \frac{9}{2}$ OR $4 x + 2 y - 9 = 0$ A1

[5 marks]

substituting $x = - 6$ into their equation from part (a) (M1)

$y = - 2 (- 6) + \frac{9}{2}$

$y = 16.5$ A1

Note: Award M1A0 for $(- 6, 16.5)$ as their final answer.

[2 marks]

Examiners report

A large proportion of candidates seemed to be well drilled into finding the gradient of a line and the subsequent gradient of the normal. But without finding the coordinates of the midpoint of AB, no more marks were gained.

Many candidates worked out the value of $y$ correctly (or “correct” following the value they found in part (a)) but then incorrectly gave their answer as a coordinate pair.

Syllabus sections

Topic 2—Functions » SL 2.1—Equations of straight lines, parallel and perpendicular

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22M.2.AHL.TZ2.4c:
(i) the gradient of this line in terms of $c$ ;

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22M.2.AHL.TZ2.4d:
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18N.2.SL.TZ0.T_4b.iii:
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22M.1.SL.TZ2.11c:
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21N.1.SL.TZ0.4a:
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22M.1.SL.TZ1.4a:
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21N.1.SL.TZ0.4b:
Write down the gradient of the line through $B$ and $D$ .
18N.2.SL.TZ0.T_4b.i:
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18M.1.SL.TZ2.T_5b:
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19N.2.SL.TZ0.T_2a:
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19N.2.SL.TZ0.T_2d:
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18N.2.SL.TZ0.T_4b.ii:
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17M.1.SL.TZ2.T_4c:
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19N.2.SL.TZ0.T_2e:
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17N.1.SL.TZ0.T_2c.ii:
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19N.2.SL.TZ0.T_2c:
Find the equation of $L_{2}$ . Give your answer in the form $a x + b y + d = 0$ , where $a, b, d \in ℤ$ .
18M.1.SL.TZ2.T_5a.ii:
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21M.2.AHL.TZ1.4a:
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18M.1.SL.TZ2.T_5a.i:
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19N.2.SL.TZ0.T_2f:
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21N.1.SL.TZ0.4d:
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Markscheme

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