The points $ A\left(x_{1}, y_{1}\right), \mathrm{B}\left(x_{2}, y_{2}\right) $ and $ \mathrm{C}\left(x_{3}, y_{3}\right) $ are the vertices of $ \Delta \mathrm{ABC} $
Find the coordinates of the point $ P $ on $ A D $ such that $ A P: P D=2: 1 $

Given:

The points \( A\left(x_{1}, y_{1}\right), \mathrm{B}\left(x_{2}, y_{2}\right) \) and \( \mathrm{C}\left(x_{3}, y_{3}\right) \) are the vertices of \( \Delta \mathrm{ABC} \)
The median from \( \mathrm{A} \) meets \( \mathrm{BC} \) at \( \mathrm{D} \).

To do:

We have to find the coordinates of the point \( P \) on \( A D \) such that \( A P: P D=2: 1 \).

Solution:

We know that,

The median bisects the line segment into two equal parts

$D$ is the mid-point of $B C$.

This implies,

Coordinate of mid-point of $B C=(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2})$

$D=(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2})$.

Let the coordinates of the point $P$ be $(x, y)$.

The point $P(x, y)$, divides the line joining $A(x_{1}, y_{1})$ and $D(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2})$ in the ratio $2: 1$

Using internal 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 $P$

 are $P(x, y)=[\frac{2(\frac{x_{2}+x_{3}}{2})+1(x_{1})}{2+1}, \frac{2(\frac{y_{2}+y_{3}}{2})+1 (y_{1})}{2+1}]$

$=(\frac{x_{2}+x_{3}+x_{1}}{3}, \frac{y_{2}+y_{3}+y_{1}}{2})$

Therefore,

The required coordinates of point $P$ are $(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3})$. 

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Updated on: 10-Oct-2022

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