$A (4, 2), B (6, 5)$ and $C (1, 4)$ are the vertices of $\triangle ABC$.Find the coordinates of point P on AD such that $AP : PD = 2 : 1$.

Given:

$A (4, 2), B (6, 5)$ and $C (1, 4)$ 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 $D=(\frac{6+1}{2}, \frac{5+4}{2})$

$=(\frac{7}{2},\frac{9}{2})$ 

$AP : PD = 2 : 1$.

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 \( (\frac{2 \times \frac{7}{2}+1 \times 4}{1+2}, \frac{2 \times \frac{9}{2}+1 \times 2}{1+2}) \)

\( =(\frac{7+4}{3}, \frac{9+2}{3}) \)

\( =(\frac{11}{3}, \frac{11}{3}) \)

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Updated on: 10-Oct-2022

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