The coordinates of the end points of the line segment AB are A(–2, –2) and B(2, –4). P is the point on AB such that $BP = \frac{4}{7} AB$. Find the coordinates of point P.
Generated by claude-sonnet-4-6 · 2026-06-15 10:31 · grounding rag
Model Answer
Given: A(–2, –2), B(2, –4), and BP = (4/7)AB.
So AP = AB – BP = (3/7)AB, meaning AP : PB = 3 : 4.
P divides AB in ratio 3 : 4 (from A).
Using the section formula:
$$x = \frac{3(2) + 4(-2)}{3+4} = \frac{6-8}{7} = \frac{-2}{7}$$
$$y = \frac{3(-4) + 4(-2)}{3+4} = \frac{-12-8}{7} = \frac{-20}{7}$$
∴ Coordinates of P = $\left(\dfrac{-2}{7},\ \dfrac{-20}{7}\right)$
Source: Chapter 7, Section 7.3 (Section Formula)
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Explanation
- Key step: BP = (4/7)AB means AP = (3/7)AB, so AP : PB = 3 : 4. Students often misread the ratio — BP is given, not AP.
- Then directly apply the section formula with A(x₁,y₁) = (–2,–2), B(x₂,y₂) = (2,–4), m₁:m₂ = 3:4.
- This is similar to Exercise 7.2, Q.8 (where AP = 3/7 AB is given instead of BP = 4/7 AB — note the difference).