A marshal's whistle at \(M\) sits on a field plan scaled 1 cm to 8 m. Relay points \(N\) and \(O\) are 9 cm apart, and a water station is proposed at \(Q\);...

This question combines a circle, a perpendicular bisector and a distance check to decide whether the proposed water station \(Q\) lies in a shaded region.

(a) On a scale of \(1\) cm to \(8\) m, a real distance becomes a plan length by dividing by \(8\): \(40 \div 8 = 5\) cm. [1 mark]

(b) The locus of points within the whistle's range of \(M\) is a circle of radius \(5\) cm centred on \(M\), drawn accurately with the compasses fixed at that radius. [2 marks]

(c) The perpendicular bisector of \(NO\) is the set of points equidistant from \(N\) and \(O\): with the compasses opened to a radius greater than half of \(NO\) (more than \(4.5\) cm, since \(NO = 9\) cm), draw arcs above and below \(NO\) from \(N\), then repeat with the same radius from \(O\); a line through the two crossing points is the perpendicular bisector. [2 marks]

(d) A point within the whistle's range of \(M\) and closer to \(N\) than to \(O\) must lie inside the circle centred on \(M\) and on the \(N\) side of the perpendicular bisector, so only that overlap is shaded. [1 mark]

(e) Reading \(Q\)'s offset from \(M\) as \(9\) across and \(3\) up, Pythagoras' theorem gives \(MQ = \sqrt{9^2+3^2} \approx 9.5\) cm, which is more than the \(5\) cm range, so \(Q\) lies outside the whistle's range and cannot be in the shaded region, whatever side of the bisector it falls on. [1 mark]

Exam tip: if a point fails just one of the two defining conditions for a shaded region, that is enough to rule it out; there is no need to check the second condition once the first has already failed.