KCSE 2025 Maths P2 Q7 — Loci & Construction of an Angle Locus
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The Question
“The figure shows an isosceles triangle MLN with base MN, equal slanting sides LM and LN, a height of 5 cm dropping from the apex L perpendicularly to the base, and the apex angle MLN equal to 50 degrees. Using a ruler and a pair of compasses only, (a) construct the locus of a point P such that angle MPN is 50 degrees, and (b) locate a point Q such that the area of triangle MQN is half that of triangle MLN and angle MQN is also 50 degrees.”
Understand what the locus must be
You need every point P that sees the fixed base MN under an angle of exactly 50 degrees. The rule that unlocks this is that angles subtended by the same chord in the same segment of a circle are equal. So all points that view MN at 50 degrees lie on one circular arc. Because the apex L already sees MN at 50 degrees, that same arc must pass through M, L and N.
This is why a constant-angle locus is always a circular arc, never a straight line.
Construct the perpendicular bisector of MN
The centre of the circle through M, L and N lies on the perpendicular bisector of the chord MN. With the compasses set to more than half of MN, draw equal arcs above and below the base from M and from N, then join the two crossing points with a ruler. This gives the line of symmetry on which the circle's centre sits.
Draw the arc through M, L and N
Find the centre where the perpendicular bisector of MN meets the perpendicular bisector of another chord such as ML, so the point is equidistant from all three vertices. Set the compasses to that radius and draw the arc passing through M, L and N. This arc is the locus of P: every point on it sees the base MN at exactly 50 degrees.
The arc is drawn on the same side of MN as L. There is a matching arc on the other side.
Halve the area to fix the height of Q
Triangle MQN and triangle MLN share the same base MN, so their areas depend only on their heights. To make the area of MQN half that of MLN, the perpendicular height of Q above MN must be half of 5 cm, that is 2.5 cm. Draw a line parallel to MN at a perpendicular distance of 2.5 cm above the base.
Locate Q on the arc
Mark Q where the parallel line 2.5 cm above MN cuts the locus arc. Because Q lies on the arc it automatically keeps the 50-degree angle at MN, and because it is 2.5 cm above the base its triangle has exactly half the area. There are two such crossing points, one on each side of the line of symmetry, so either is acceptable.
Final Result
The locus of P is the circular arc drawn through the points M, L and N (every point on it subtends 50 degrees on MN). Point Q is where a line parallel to MN and 2.5 cm above it meets that arc, giving a triangle MQN with half the area of MLN while still keeping angle MQN equal to 50 degrees.
Why this method works
The construction works because of a fixed property of circles: a chord subtends the same angle at every point of a given arc, so the set of points that see MN under 50 degrees is precisely a circular arc, and the apex L pins down which arc by giving one known point on it. The area condition works because two triangles on a common base are in the ratio of their heights, so halving the area is the same as halving the perpendicular height, which is why a parallel line 2.5 cm from the base locates Q.
At Q the angle MQN equals 50 degrees because Q lies on the same arc as L, and the height is 2.5 cm, so the area is one half times MN times 2.5, which is exactly half of one half times MN times 5, the area of MLN.