Question

The line segment joining the points $P(3,3)$ and $Q(6,-6)$ is trisected at the points $A$ and $B$ such that $A$ is nearer to $P$. If $A$ also lies on the line given by $2x+y+k=0$, find the value of $k$.

Answer 1

Let the coordinates of point A be $\left(x_{1,}{y}_1\right)$

Since A divides PQ in the ratio $1:2$

So, $\left(x_{1,}{y}_1\right)=\left(\frac{1\times 6+2\times 3}{1+2},\frac{1\times -6+2\times 3}{1+2}\right)$

$\text{=}\left(\frac{6+6}{3},\frac{-6+6}{3}\right)$

$=\left(4,0\right)$

Since A lies on $2x+y+k=0$

So, $2x_1+y_1+k=0$

$\Rightarrow 2\times 4+0+k=0$

$\Rightarrow 8+k=0$

$\Rightarrow k=-8$