Question

Find the coordinates of a point $P$ on the line segment joining $A(1,2)$ and $B(6,7)$ such that $AP=\frac{2}{5}AB$

Answer 1

$A=(1,2)B(6,7)$

$\frac{AP}{AB}=\frac{2}{5}\to \left(1\right)\Rightarrow \frac{AB-PB}{AB}=\frac{2}{5}\Rightarrow 5AB-5PB=2AB\Rightarrow 3AB=5PB\Rightarrow \frac{AB}{PB}=\frac{5}{3}\to \left(2\right)$

FRom $\left(1\right)$ and $\left(2\right)$

$\frac{AP}{AB}=\frac{2}{3}$

$\therefore$ P divides $AB$ in ratio $2:3$

$\therefore x=\frac{m{x}_2+n{x}_1}{m+n}$ and $y=\frac{m{y}_2+n{y}_1}{m+n}$

$x=\frac{2\times 6+3\times 1}{2+3}=\frac{12+3}{5}=\frac{15}{5}=3y=\frac{2\times 7+3\times 2}{2+3}=\frac{20}{4}=4$

$\therefore P$ is $(3,4)$