The line segment joining the points A(3, −4) and B(1, 2) is trisected at the points P(p, −2) and

Solution:

Let P and Q be the points of trisection of AB.
Then, P divides AB in the ratio 1:2.
So, the coordinates of P are

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

$\Rightarrow x=\frac{\{1 \times 1+2 \times(3)\}}{1+2}, y=\frac{\{1 \times 2+2 \times(-4)\}}{1+2}$

$\Rightarrow x=\frac{1+6}{3}, y=\frac{2-8}{3}$

$\Rightarrow x=\frac{7}{3}, y=-\frac{6}{3}$

$\Rightarrow x=\frac{7}{3}, y=-2$

Hence, the coordinates of $P$ are $\left(\frac{7}{3},-2\right)$.

But (p, −2) are the coordinates of P.

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

Also, Q divides the line AB in the ratio 2:1.
So, the coordinates of Q are

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

$\Rightarrow x=\frac{(2 \times 1+1 \times 3)}{2+1}, y=\frac{\{2 \times 2+1 \times(-4)\}}{2+1}$

$\Rightarrow x=\frac{2+3}{3}, y=\frac{4-4}{3}$

$\Rightarrow x=\frac{5}{3}, y=0$

Hence, coordinates of $Q$ are $\left(\frac{5}{3}, 0\right)$.

But the given coordinates of $Q$ are $\left(\frac{5}{3}, q\right)$.

So, q = 0

Thus, $p=\frac{7}{3}$ and $q=0$