The line segment joining the points P(3, 3) and Q(6, −6) is trisected
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.

Solution:

We have two points P (3, 3) and Q (6,−6). There are two points A and B which trisect the line segment joining P and Q.

Let the co-ordinate of $\mathrm{A}$ be $\mathrm{A}\left(x_{1}, y_{1}\right)$

Now according to the section formula if any point $P$ divides a line segment joining $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ in the ratio $m$ : $n$ internally than,

$\mathrm{P}(x, y)=\left(\frac{m x_{1}+m x_{2}}{m+n}, \frac{m y_{1}+m y_{2}}{m+n}\right)$

The point A is the point of trisection of the line segment PQ. So, A divides PQ in the ratio 1: 2

Now we will use section formula to find the co-ordinates of unknown point A as,

$\mathrm{A}(x, y)=\left(\frac{2(3)+1(6)}{1+2}, \frac{2(3)+1(-6)}{1+2}\right)$

$=(4,0)$

Therefore, co-ordinates of point A is(4, 0)

It is given that point A lies on the line whose equation is

$2 x+y+k=0$

So point A will satisfy this equation.

$2(4)+0+k=0$

So,

$k=-8$