Question

The line segment joining the points $P(3,3),Q(6,-6)$ 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

Calculate the value of $k$:

If a line segment $PQ$ has coordinates $P(x_1,y_1),Q(x_2,y_2)$ and a point $R$ divides the line segment in the ratio of $m:n$.

Then, the co-ordinate of point $R$ is computed by using the section formula.

$\mathbf{R}\mathbf{=}\left(\frac{{\mathrm{mx}}_2+{\mathrm{nx}}_1}{m+n},\frac{{\mathrm{my}}_2+{\mathrm{ny}}_1}{m+n}\right)$

In the given question, the point $A$divides the given line segment into $1:2$ and the diagram is shown below,

So, the co-ordinate points of $A$is,

$$\begin{gathered} \therefore A(a,b)=\left(\frac{(1\times 6)+(2\times 3)}{1+2},\frac{(1\times -6)+(2\times 3)}{1+2}\right) \\ \Rightarrow A(a,b)=\left(\frac{12}{3},0\right) \\ \Rightarrow A(a,b)=\left(4,0\right) \end{gathered}$$

Since the point $A$ also lies on the line given by $2x+y+k=0$.

So, substitute the point $A$in the equation of the line.

$$\begin{gathered} \therefore 2\left(4\right)+0+k=0 \\ \Rightarrow 8+k=0 \\ \Rightarrow k=-8 \end{gathered}$$

Hence, the value of $k$ is $-8$.