Derive the Section Formula for internal division of a line segment joining two points $A(x_1, y_1)$ and $B(x_2, y_2)$ in the ratio $m : n$. Using this formula, solve:
Find the coordinates of the points of trisection of the line segment joining $A(2, 1)$ and $B(5, -8)$.
Step-by-Step Solution
Key Concept: Part 1: Proof of section formula using similar right triangles $APL \sim PBM \Rightarrow \dfrac{x - x_1}{x_2 - x} = \dfrac{m}{n} \Rightarrow x = \dfrac{m x_2 + n x_1}{m + n}$ and $y = \dfrac{m y_2 + n y_1}{m + n}$. Part 2: $P(1:2) = \left(\dfrac{5+4}{3}, \dfrac{-8+2}{3}\right) = (3, -2)$. $Q(2:1) = \left(\dfrac{10+2}{3}, \dfrac{-16+1}{3}\right) = (4, -5)$.
Part 1: Derivation of Section Formula via similar triangles $\Delta APL \sim \Delta PBM \Rightarrow \dfrac{x - x_1}{x_2 - x} = \dfrac{m}{n} \Rightarrow x = \dfrac{m x_2 + n x_1}{m + n}$ and $y = \dfrac{m y_2 + n y_1}{m + n}$. [2.5 Marks]
Part 2: Trisection of $A(2, 1)$ and $B(5, -8)$:
Point 1 ($1 : 2$ ratio): $P = \left(\dfrac{1(5) + 2(2)}{3}, \dfrac{1(-8) + 2(1)}{3}\right) = (3, -2)$. [1.25 Marks]
Point 2 ($2 : 1$ ratio): $Q = \left(\dfrac{2(5) + 1(2)}{3}, \dfrac{2(-8) + 1(1)}{3}\right) = (4, -5)$. [1.25 Marks]
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🎯 Official CBSE Marking Scheme:
Complete derivation of Section Formula: 2.5 Marks
Calculating first point of trisection $P(3, -2)$: 1.25 Marks
Calculating second point of trisection $Q(4, -5)$: 1.25 Marks
Correct Answer: