JAMB 1995 · UME · Q31

Construction & loci

$P$ is on the locus of points equidistant from two given points $X$ and $Y$ . $UV$ is a straight line through $Y$ parallel to the locus. If $\angle PYU = 40^\circ$ , find $\angle XPY$ .

Worked solution (try it first)

The locus of points equidistant from $X$ and $Y$ is the perpendicular bisector of $XY$ .
$UV$ is parallel to it, so $UV$ is perpendicular to $XY$ .
So the angle between $YU$ and $YX$ is $90^\circ$ , and $\angle PYX = 90^\circ - 40^\circ$
$= 50^\circ$ .
$P$ is on the locus, so $PX = PY$ and triangle $PXY$ is isosceles: $\angle PXY = \angle PYX = 50^\circ$ .
Angles in a triangle add up to $180^\circ$ : $\angle XPY = 180^\circ - 100^\circ$
$= 80^\circ$ , option B.

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