geometric interpretation of inverse of the product of the length of two line segments given a line segment  with length a. pick  a random point within that line segment, would generate two line segments with length b and a-b where 0<b<a. what kind of geometric operation would generate the value of a/(b*(a-b)) ? or is it even possible?
 A: (Converting comment to answer, by request.)
As the numerator and denominator of $c:=\dfrac{a}{b(a−b)}$ are not homogeneous expressions of the same degree, constructing a segment of length $c$ from segments of length $a$ and $b$ requires an auxiliary segment of length $1$. (Otherwise, you could scale the resulting figure by, say, $2$, and the same construction would yield a segment of length $c':=2c$ from segments of length $a':=2a$ and $b':=2b$; but $c'\neq \dfrac{a'}{b'(a'−b')}$.)

Now, with segments of length $p$, $q$, $1$, it's not difficult to construct a segment of length $pq$, then one of length $1/(pq)$, and then one of length $(p+q)/pq$:


*

*First, arrange $\overline{OP}$, $\overline{OQ}$, $\overline{OR}$ (of lengths $p$, $q$, $1$, respectively) with the first two on opposite sides of $O$ and the last perpendicular to them. Construct $\bigcirc PQR$ (via well-known methods), and extend $\overline{OR}$ to meet this circle at $S$. Then, by the Power of a Point Theorem, $|OP||OQ|=|OR||OS|$; that is, $|OS|=pq$.



*

*Next, construct $\overline{OT}$ (of length $1$) perpendicular to $\overline{OS}$. Construct the perpendicular to $\overline{ST}$ at $T$, and let it meet the extension of $\overline{OS}$ at $U$. Since $\triangle SOT\sim \triangle TOU$, we have $|OU|/|OT| = |OT|/|OS|$; that is, $|OU| = 1/(pq)$.


*Finally, construct a perpendicular to $\overline{PQ}$ at $P$, and a parallel to $\overline{TU}$ at $Q$; let these lines meet at $V$. Since $\triangle UOT \sim \triangle VPQ$, we have $|PV|/|PQ|=|UO|/|OT|$, so that $|PV|=(p+q)/(pq)$.
This completes the construction. $\square$
