JAMB 1987 · UME · Q44

The figure is an example of the construction of a

Worked solution (try it first)

The arc is centred at $P$ , a point off the line, and it cuts the line at $Q$ and $R$ , so $PQ = PR$ .
Equal arcs from $Q$ and $R$ meet at $X$ , so $X$ is also equidistant from $Q$ and $R$ .
The line $PX$ is then perpendicular to $QR$ .
So it is the perpendicular from a given point to a given line, option B.