The line segment joining the points (3, −4) and (1, 2) is

Solution:

We have two points $A(3,-4)$ and $B(1,2)$. There are two points $P(p,-2)$ and $Q\left(\frac{5}{3}, q\right)$ which trisect the line segment joining $A$ and $B$.

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 P is the point of trisection of the line segment AB. So, P divides AB in the ratio 1: 2

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

$\mathrm{P}(p,-2)=\left(\frac{2(3)+1(1)}{1+2}, \frac{2(-4)+1(2)}{1+2}\right)$

$=\left(\frac{7}{3},-2\right)$

Equate the individual terms on both the sides. We get,

$p=\frac{7}{3}$

Similarly, the point Q is the point of trisection of the line segment AB. So, Q divides AB in the ratio 2: 1

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

$Q\left(\frac{5}{3}, q\right)=\left(\frac{2(1)+1(3)}{1+2}, \frac{2(2)+1(-4)}{1+2}\right)$

$=\left(\frac{5}{3}, 0\right)$

Equate the individual terms on both the sides. We get,

$q=0$