21M.1.SL.TZ2.6

pestleMathematicsAISLPaper 121M· sl-3-5-intersection-of-lines-equations-of-perpendicular-bisectorssource ↗

Two schools are represented by points $A (2, 20)$ and $B (14, 24)$ on the graph below. A road, represented by the line $R$ with equation $- x + y = 4$ , passes near the schools. An architect is asked to determine the location of a new bus stop on the road such that it is the same distance from the two schools.

Find the equation of the perpendicular bisector of $[AB]$ . Give your equation in the form $y = m x + c$ .

[5]

Determine the coordinates of the point on $R$ where the bus stop should be located.

[2]

Markscheme / solution

gradient $AB = \frac{4}{12} (\frac{1}{3})$ (A1)

midpoint $AB : (8, 22)$ (A1)

gradient of bisector $= - \frac{1}{gradient AB} = - 3$ (M1)

perpendicular bisector: $22 = - 3 \times 8 + b$ OR $(y - 22) = - 3 (x - 8)$ (M1)

perpendicular bisector: $y = - 3 x + 46$ A1

[5 marks]

attempt to solve simultaneous equations (M1)

$x + 4 = - 3 x + 46$

$(10.5, 14.5)$ A1

[2 marks]

Examiners’ report

[N/A]