Divide a directed segment in a chosen ratio

Try

Let A=(−2,1) and B=(8,6). Locate P between them so that AP:PB=2:3. The whole segment contains five ratio parts, so P lies two fifths of the way from A to B, not two thirds.

The displacement B−A is (10,5). Two fifths of it is (4,2), and adding A gives P=(2,3). Check the remaining displacement B−P=(6,3): it is three halves of P−A, so the required ratio is satisfied. The point is constructed from a directed displacement even though the requested lengths are positive.

Predict

Predict what happens if the endpoint names are reversed while the ratio is still stated as AP:PB=2:3. A is the starting endpoint in that statement. Switching the names changes which part is shorter and generally gives a different point.

For positive numbers r and s, the fraction of the entire segment from A is t=r/(r+s). Thus P=A+t(B−A). Since 0 < t < 1, the point lies strictly between A and B. Its two displacements are t(B−A) and (1−t)(B−A), giving length ratio t/(1−t)=r/s.

Experiment

Construct a 2:3 division without measuring the original segment. Draw an auxiliary ray from A that is not on AB. Step off five equal compass lengths along it. Call the second mark X and the fifth Y. Join Y to B, then construct through X a parallel to YB; let its intersection with AB be P.

State the centers and repeated compass radius used to place the five marks. A marked ruler estimate can suggest the answer, but it is not the same construction argument. The auxiliary ray’s angle and step size may vary as long as the ray is noncollinear with AB and the steps are equal.

Understand

The triangles AXP and AYB share angle A and have matching corresponding angles because XP is parallel to YB. AA therefore gives AP/AB=AX/AY=2/5. Subtracting from the whole yields PB/AB=3/5, so AP:PB=2:3. This proves why the construction works; the diagram alone does not establish the ratio.

For any positive integer ratio r:s, use r+s equal steps and the r-th mark. A rational ratio can first be written as a ratio of positive integers. If arbitrary positive lengths representing r and s are supplied, lay them consecutively on the auxiliary ray instead; the same parallel-cut proof applies. We do not claim that an arbitrary unsupplied real number can be marked exactly by a compass operation.

In coordinates, expand P=A+(r/(r+s))·(B−A) to obtain

P=(sA+rB)/(r+s).

The coefficient of A is s, not r, because P is closer to A when r is small. The formula acts on both coordinates. For the worked data it gives x=(3·(−2)+2·8)/5=2 and y=(3·1+2·6)/5=3.

This also proves uniqueness: every point on segment AB has a unique fractional displacement t when A≠B, and t/(1−t)=r/s forces t=r/(r+s). Equal coordinates A=B would give a zero-length segment and no positive part-length ratio, so distinct endpoints are a necessary hypothesis.

Apply

Construct a fresh internal ratio, then independently check it with coordinates or similar-triangle ratios. Record the endpoint order, the auxiliary marks and the particular parallel line used. A complete answer distinguishes the numerical ratio r/s from the whole-segment fraction r/(r+s).

Repair the claim that the 2:3 point is A+(2/3)(B−A). That point has AP/AB=2/3 and PB/AB=1/3, hence AP:PB=2:1. Substituting the intended part-to-part ratio directly into a whole-displacement formula changes the problem.

Continue

Midpoints are the special case r=s, but the same construction handles every positive internal ratio with the stated input lengths. This is a concrete use of the parallel-cut theorem rather than a new unrelated formula.

Optional depth: a parameter t outside [0,1] locates a point beyond an endpoint on the supporting line. Such external division needs an explicit directed-sign convention. The positive internal-ratio construction here should not be applied to external points without changing and explaining that convention.