Problem — Locus of the midpoints of segments from a point to a line

Text

A line $s$ and a point $B$ not lying on $s$ are given. As the point $X$ varies over $s$, consider the midpoint $M$ of the segment $BX$. Determine the locus described by the points $M$ and prove your claim.

(original figure: line $s$, point $B$ and several segments $BX$ with their midpoints)

Solution

Let $K$ be the foot of the perpendicular dropped from $B$ to $s$, and let $d = BK$ be the distance of $B$ from $s$. Let $M_0$ be the midpoint of $BK$; it lies at distance $\tfrac{d}{2}$ from $s$. We claim that the required locus is the line $s'$ through $M_0$ parallel to $s$, i.e. the parallel to $s$ at distance $\tfrac{d}{2}$, “halfway” between $B$ and $s$.

Every point $M$ of the locus lies on $s'$. Let $X$ be any point of $s$ and $M$ the midpoint of $BX$. In the triangle $BKX$, $M_0$ is the midpoint of $BK$ and $M$ is the midpoint of $BX$: by the midsegment theorem $M_0M \parallel KX$. But $KX$ lies on $s$, hence $M_0M \parallel s$: the point $M$ lies on the line through $M_0$ parallel to $s$, that is on $s'$.

Every point of $s'$ belongs to the locus. Let $M$ be any point of $s'$ and let $X$ be the point where the line $BM$ meets $s$ (the two lines are not parallel, so the intersection exists). In the triangle $BKX$ the line $s'$ passes through the midpoint $M_0$ of $BK$ and is parallel to the side $KX$ (contained in $s$); by the converse of the midsegment theorem it meets the third side $BX$ at its midpoint. Since $s'$ meets $BX$ exactly at $M$, it follows that $M$ is the midpoint of $BX$, with $X \in s$: hence $M$ belongs to the locus.

The required locus is therefore the line parallel to $s$, halfway between $B$ and $s$ (at distance $\tfrac{d}{2}$ from each). $\;\blacksquare$