Question

The points $A(x_1,y_1),B(x_2,y_2)$ and $C(x_3,y_3)$ are the vertices of $ABC$. (ii) Find the coordinates of the point $P$ on $AD$ such that $AP:PD=2:1$.

Answer 1

Solution.

Use section formula

$P=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$ where $m,n$ are ratios .

As, $AP:PD=2:1$

Coordinates of $D$ are $\left(\frac{x_2+x_1}{2},\frac{y_2+y_1}{2}\right)$.

Now, Coordinates of $P$ are:

$$\begin{gathered} P=\left(\frac{2\left({\displaystyle \frac{x_2+x_1}{2}}\right)+x_1}{2+1},\frac{2\left({\displaystyle \frac{y_2+y_1}{2}}\right)+y_1}{2+1}\right) \\ =\left(\frac{x_2+x_1+x_1}{2+1},\frac{y_2+y_1+y_1}{2+1}\right) \\ =\left(\frac{x_2+2x_1}{3},\frac{y_2+2y_1}{3}\right) \end{gathered}$$

Hence, the coordinates of $P$ are $\left(\frac{x_2+2x_1}{3},\frac{y_2+2y_1}{3}\right)$.