Question

Find the co-ordinate 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

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

$(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

Since,

$\begin{array}{c}AP=\frac{2}{5}AB\\ \Rightarrow 5AP=2\left(AP+PB\right)\\ \Rightarrow 5AP=2AP+2PB\\ \Rightarrow 3AP=2PB\\ \Rightarrow \frac{AP}{PB}=\frac{2}{3}\\ \Rightarrow AP:PB=2:3\end{array}$

Therefore,

$\begin{array}{c}\left(x,y\right)=\left(\frac{2\times 6+3\times 1}{2+3},\frac{2\times 7+3\times 2}{2+3}\right)\\ =\left(\frac{15}{5},\frac{20}{5}\right)\\ =\left(3,4\right)\end{array}$

So, coordinates of point $P$ is $(3,4)$