301. The Section Formula · Dividing segments with precision
P always divides AB in the ratio m₁ : m₂, and its coordinates strictly follow the section formula.
What this lesson covers
Try to break it
Drag P along AB. The perpendiculars from A, P, and B to the axes carve two similar right triangles. The ratio AP : PB equals the ratio of their horizontal legs (and vertical legs), so P sits at ((m·x₂ + n·x₁) / (m+n), (m·y₂ + n·y₁) / (m+n)) where AP : PB = m : n. Try to find a position where this fails; impossible.
How you build it
Divide segment AB in a given ratio.
- Mark point A — one end of the segment to be divided.
- Mark point B — the other end of the segment.
- Draw segment AB.
- Mark point P anywhere on segment AB. The position of P determines the ratio AP : PB = m₁ : m₂, and the section formula gives P's coordinates in terms of A, B, m₁ and m₂.
The proof, step by step
Prove that P divides AB in the ratio m₁ : m₂ as given by the section formula.
- Draw perpendiculars AA', BB', and PP' to a horizontal line through the base.
- AA' ∥ PP' ∥ BB', so ∠P'PA = ∠A'AP and ∠P'PB = ∠B'BP (corresponding angles).
- By AA similarity, ΔAPP' ~ ΔABB'. Hence, AP/AB = PP'/BB'.
- Similarly, using horizontal projections, AP/PB = (x - x₁)/(x₂ - x). Equating ratios gives x = (m₁x₂ + m₂x₁)/(m₁ + m₂).
- The same logic applies to the y-coordinates, yielding y = (m₁y₂ + m₂y₁)/(m₁ + m₂). Q.E.D.
Worked example
Point P divides the line segment joining A(2, 3) and B(8, 7) in the ratio 2:3. Find the coordinates of P.
Using the section formula: x = (2×8 + 3×2)/(2+3) = 22/5 = 4.4, y = (2×7 + 3×3)/(2+3) = 23/5 = 4.6. Thus, P is (4.4, 4.6).
- (4.4, 4.6) — correct
- (5, 5)
- (3.6, 4.2)
- (4, 5)