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The points $A (x_1, y_1), B (x_2, y_2)$ and $C (x_3, y_3)$ are the vertices of $\triangle ABC$.
Find the coordinates of the point $P$ on $AD$ such that $AP : PD = 2 : 1$.
Given:
The points $A (x_1, y_1), B (x_2, y_2)$ and $C (x_3, y_3)$ are the vertices of $\triangle ABC$.
$AP : PD = 2 : 1$.
To do:
We have to find the coordinates of point $P$ on $AD$.
Solution:
$D$ is the mid-point of BC.
This implies,
Using mid-point formula, we get,
Coordinates of $D=(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2})$.
$AP : PD = 2 : 1$.
Using section formula, we get,
\( (x,y)=(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}) \)
The coordinates of \( \mathrm{P} \) are \( (\frac{2 \times \frac{x_2+x_3}{2}+1 \times x_1}{1+2}, \frac{2 \times \frac{y_2+y_3}{2}+1 \times y_1}{1+2}) \)
\( =(\frac{x_2+x_3+x_1}{3}, \frac{y_2+y_3+y_1}{3}) \)
\( =(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}) \)
The coordinates of \( \mathrm{P} \) are \( (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}) \).
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