Problem Statement
The line segment joining $P(3, 3)$ and $Q(6, -6)$ is trisected by points $A$ and $B$, where $A$ is nearer to $P$. If $A$ lies on the line $2x + y + k = 0$, find the value of $k$.
Verified Solution & Marking Scheme
Find coordinates of point of trisection A
Since $A$ is nearer to $P$, $A$ divides $PQ$ in the ratio $1:2$.
By section formula:
$A(x, y) = \left( \frac{1(6) + 2(3)}{1 + 2}, \, \frac{1(-6) + 2(3)}{1 + 2} \right)$
$x = \frac{6 + 6}{3} = \frac{12}{3} = 4$
$y = \frac{-6 + 6}{3} = \frac{0}{3} = 0$
Thus, $A = (4, 0)$.
Substitute A(4, 0) into the given line equation
Since $A(4, 0)$ lies on the line $2x + y + k = 0$:
$2(4) + 0 + k = 0$
$8 + k = 0 \implies k = -8$