# The line segment joining the points $(3, -4)$ and $(1, 2)$ is trisected at the points $P$ and $Q$. If the coordinates of $P$ and $Q$ are $(p, -2)$ and $(\frac{5}{3}, q)$ respectively. Find the values of $p$ and $q$.

Given:

The line segment joining the points $(3, -4)$ and $(1, 2)$ is trisected at the points $P$ and $Q$. The coordinates of $P$ and $Q$ are $(p, -2)$ and $(\frac{5}{3}, q)$ respectively.

To do:

We have to find the values of $p$ and $q$.

Solution:

Let $\mathrm{AB}$ be the line segment whose ends points are $\mathrm{A}(3,-4)$ and $\mathrm{B}(1,2)$
The coordinates of $\mathrm{P}$ and $\mathrm{Q}$ which trisect $\mathrm{AB}$ are $\mathrm{P}(p,-2)$ and $\mathrm{Q}\left(\frac{5}{3}, q\right)$.

Therefore,
$P$ divides $A B$ in the ratio $1: 2$.

Using section formula, we get,

$(x,y)=(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n})$
The coordinates of $\mathrm{P}$ are $(\frac{1 \times 1+2 \times 3}{1+2}, \frac{1 \times 2+2 \times (-4)}{1+2}$

$(p,-2)=(\frac{1+6}{3}, \frac{2-8}{3})$

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

This implies, $p=\frac{7}{3}$
Similarly,

$Q$ divides $\mathrm{AB}$ in the ratio $2: 1$.
The coordinates of $\mathrm{P}$ are $(\frac{2 \times 1+1 \times 3}{1+2}, \frac{2 \times 2+1 \times (-4)}{1+2}$

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

$=(\frac{5}{3},\frac{0}{3}$

This implies,

$q=0$
The values of $p$ and $q$ are $\frac{7}{3}$ and $0$ respectively.

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