Question

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 $\left(\begin{array}{c}\frac{5}{3},q\end{array}\right)$ respectively, find the values of $p$ and $q$.

• A. $p=\frac{7}{3}$
• B. $p=0$
• C. $q=0$
• D. $q=\frac{7}{3}$

Answer 1

Answer: (A), (C)

$p=\frac{7}{3}$
$q=0$

Points $P$ and $Q$ trisect the line segment joining the points $A(3,-4)$ and $B(1,2)$.
This means, $P$ divides $AB$ in the ratio $1:2$ and $Q$ divides it in the ratio $2:1$

Using the section formula, if a point $(x,y)$ divides the line joining the points $(x_1,y_1)$ and $(x_2,y_2)$ internally in the ratio $m:n$, then $(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

Substituting $(x_1,y_1)=(3,-4)$ and $(x_2,y_2)=(1,2)$ and $m=1,n=2$ in the section formula,

we get the point $P=(\frac{1\left(1\right)+2\left(3\right)}{1+2},\frac{1\left(2\right)+2(-4)}{1+2})=(\frac{7}{3},-2)$

Given $P$ as $(p,-2)$

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

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

Substituting $(x_1,y_1)=(3,-4)$ and $(x_2,y_2)=(1,2)$ and $m=2,n=1$ in the section formula,

we get the point $Q=(\frac{2\left(1\right)+1\left(3\right)}{2+1},\frac{2\left(2\right)+1(-4)}{2+1})=(\frac{5}{3},0)$

Given $Q$ as $(\frac{5}{3},0)$

$\Rightarrow (\frac{5}{3},0)=(\frac{5}{3},q)$

$\Rightarrow q=0$