# 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 $A P: P D=2: 1$

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

Updated on 10-Oct-2022 13:28:51