Division of a segment | Analytic Geometry | Mathematics

Theorem 1.2 (Division of a segment). Let \(P_1(x_1, y_1)\) and \(P_2(x_2, y_2)\) be distinct points.

(a): The point \(P\) on \(P_1P_2\) with \(P_1P : PP_2 = r : s\) (where \(r, s > 0\)) is
\[ P = \left ( \frac {s x_1 + r x_2}{r + s}, \frac {s y_1 + r y_2}{r + s} \right ). \]

In particular the midpoint is \(\bigl ( \frac {x_1 + x_2}{2}, \frac {y_1 + y_2}{2} \bigr )\).
(b): If \(P(x, y)\) and \(Q(x', y')\) also lie on the line \(P_1P_2\), then
\[ \frac {PQ}{P_1P_2} = \frac {\abs {x' - x}}{\abs {x_2 - x_1}} \quad \text {(line not vertical)}, \qquad \frac {PQ}{P_1P_2} = \frac {\abs {y' - y}}{\abs {y_2 - y_1}} \quad \text {(line not horizontal)}. \]

Proof. Every point of the line is \(P_1 + t\,(x_2 - x_1, y_2 - y_1)\) for exactly one real \(t\), and by Theorem 1.1 the points with parameters \(t\) and \(t'\) are \(\abs {t' - t} \cdot P_1P_2\) apart. Their \(x\)-coordinates differ by \(\abs {t' - t} \abs {x_2 - x_1}\) and their \(y\)-coordinates by \(\abs {t' - t} \abs {y_2 - y_1}\), which gives (b). For (a), \(P\) lies between \(P_1\) and \(P_2\) at \(t = r / (r + s)\), and \(x_1 + \frac {r}{r + s}(x_2 - x_1) = \frac {s x_1 + r x_2}{r + s}\), and likewise for \(y\).