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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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