Question

If A and B are $(-2,-2)$ and $(2,-4)$ respectively, find the coordinates of P such that $AP=\frac{3}{7}AB$ and P lies on the line segment AB.

• A. $\left(\begin{array}{c}-\frac{2}{7},-\frac{20}{7}\end{array}\right)$
  
• B. $\left(\begin{array}{c}-\frac{2}{7},-\frac{2}{7}\end{array}\right)$
• C. $\left(\begin{array}{c}-\frac{2}{7},-\frac{10}{7}\end{array}\right)$
• D. $\left(\begin{array}{c}-\frac{12}{7},-\frac{20}{7}\end{array}\right)$

Answer 1

The correct option is A

$\left(\begin{array}{c}-\frac{2}{7},-\frac{20}{7}\end{array}\right)$

Since $AP=\frac{3}{7}AB=>\frac{AP}{AB}=\frac{3}{7}$
Since P lies on AB, $=>AP+PB=AB$
So, $\frac{AP}{AB}=\frac{AP}{AP+PB}=\frac{3}{4+3}$
Hence, $\frac{AP}{PB}=\frac{3}{4}$
This means, P divides AB in the ratio $3:4$

Using the section formula, if a point $(x,y)$ divides the line joining the
points $(x_1,y_1)$ and $(x_2,y_2)$ 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 $(x_1,y_1)=(-2,-2)$ and $(x_2,y_2)=(2,-4)$ and $m=3,n=4$ in the section formula, we get
the point $(\frac{3\left(2\right)+4(-2)}{3+4},\frac{3(-4)+4(-2)}{3+4})=(-\frac{2}{7},-\frac{20}{7})$