Question

The line segment joining the points A(3, −4) and B(1, 2) is trisected at the points P(p, −2) and $Q\left(\frac{5}{3},q\right)$. Find the values of p and q.

Answer 1

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
$$\begin{gathered} x=\frac{\left(m{x}_2+n{x}_1\right)}{\left(m+n\right)},y=\frac{\left(m{y}_2+n{y}_1\right)}{\left(m+n\right)} \\ \Rightarrow x=\frac{\left\{1\times 1+2\times \left(3\right)\right\}}{1+2},y=\frac{\left\{1\times 2+2\times \left(-4\right)\right\}}{1+2} \\ \Rightarrow x=\frac{1+6}{3},y=\frac{2-8}{3} \\ \Rightarrow x=\frac{7}{3},y=-\frac{6}{3} \\ \Rightarrow x=\frac{7}{3},y=-2 \end{gathered}$$
Hence, the coordinates of P are ($\frac{7}{3}$, −2).
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
$$\begin{gathered} x=\frac{\left(m{x}_2+n{x}_1\right)}{\left(m+n\right)},y=\frac{\left(m{y}_2+n{y}_1\right)}{\left(m+n\right)} \\ \Rightarrow x=\frac{\left(2\times 1+1\times 3\right)}{2+1},y=\frac{\left\{2\times 2+1\times \left(-4\right)\right\}}{2+1} \\ \Rightarrow x=\frac{2+3}{3},y=\frac{4-4}{3} \\ \Rightarrow x=\frac{5}{3},y=0 \\ \mathrm{Hence},\text{coordinates of}Q\text{are}\left(\frac{5}{3},0\right). \end{gathered}$$
But the given coordinates of $Q\text{are}\left(\frac{5}{3},q\right).$
So, q = 0
Thus, $p=\frac{7}{3}$ and $q=0$53