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