Find the coordinates of the points of trisection of the line segment joining $(4, -1)$ and $(-2, -3)$.
Step-by-Step Solution
Key Concept: Point 1 divides in ratio $1 : 2$: $P\left(\dfrac{1(-2)+2(4)}{3}, \dfrac{1(-3)+2(-1)}{3}\right) = \left(2, -\dfrac{5}{3}\right)$. Point 2 divides in ratio $2 : 1$: $Q\left(\dfrac{2(-2)+1(4)}{3}, \dfrac{2(-3)+1(-1)}{3}\right) = \left(0, -\dfrac{7}{3}\right)$.
Point 1 ($1 : 2$ ratio): $P = \left(\dfrac{-2+8}{3}, \dfrac{-3-2}{3}\right) = \left(2, -\dfrac{5}{3}\right)$. [1.0 Mark]
Point 2 ($2 : 1$ ratio): $Q = \left(\dfrac{-4+4}{3}, \dfrac{-6-1}{3}\right) = \left(0, -\dfrac{7}{3}\right)$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Finding first point of trisection $P(2, -5/3)$: 1.0 Mark
Finding second point of trisection $Q(0, -7/3)$: 1.0 Mark
Correct Answer: