Find the coordinates of the points of trisection of the line segment joining the points $A(2, -2)$ and $B(-7, 4)$.
Step-by-Step Solution
Key Concept: Point 1 ($1 : 2$ ratio): $P\left(\dfrac{1(-7)+2(2)}{3}, \dfrac{1(4)+2(-2)}{3}\right) = \left(-1, 0\right)$. Point 2 ($2 : 1$ ratio): $Q\left(\dfrac{2(-7)+1(2)}{3}, \dfrac{2(4)+1(-2)}{3}\right) = \left(-4, 2\right)$.
First point $P$ ($1 : 2$ ratio): $P = \left(\dfrac{-7+4}{3}, \dfrac{4-4}{3}\right) = (-1, 0)$. [1.5 Marks]
Second point $Q$ ($2 : 1$ ratio): $Q = \left(\dfrac{-14+2}{3}, \dfrac{8-2}{3}\right) = (-4, 2)$. The points are $(-1, 0)$ and $(-4, 2)$. [1.5 Marks]
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🎯 Official CBSE Marking Scheme:
Finding first point $P(-1, 0)$: 1.5 Marks
Finding second point $Q(-4, 2)$: 1.5 Marks
Correct Answer: