JAMB 1997 · UME · Q32

Construction & loci

A point $P$ moves so that it is equidistant from points $L$ and $M$ . If $LM$ is 16 cm, find the distance of $P$ from $LM$ when $P$ is 10 cm from $L$ .

Worked solution (try it first)

$P$ is equidistant from $L$ and $M$ , so it lies on the perpendicular bisector of $LM$ .
The foot of the perpendicular from $P$ is the mid-point, 8 cm from $L$ .
In the right-angled triangle, $PL = 10$ cm is the hypotenuse and 8 cm is one side.
By Pythagoras, the distance from $P$ to $LM$ is $\sqrt{10^2 - 8^2} = \sqrt{36} = 6$ cm, option D.

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