22M.1.SL.TZ1.4

pestleMathematicsAISLPaper 122M· sl-2-1-equations-of-straight-lines-parallel-and-perpendicularsource ↗

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 / solution

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.