If (−2, 3), (4, −3) and (4, 5) are the mid-points of the sides of a triangle,

Solution:

Let $\triangle \mathrm{ABC}$ be ant triangle such that $\mathrm{P}(-2,3) ; \mathrm{Q}(4,-3)$ and $\mathrm{R}(4,5)$ are the mid-points of the sides $\mathrm{AB}, \mathrm{BC}, \mathrm{CA}$ respectively.

We have to find the co-ordinates of the centroid of the triangle.

Let the vertices of the triangle be $\mathrm{A}\left(x_{1}, y_{1}\right) ; \mathrm{B}\left(x_{2}, y_{2}\right) ; \mathrm{C}\left(x_{3}, y_{3}\right)$

In general to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ we use section formula as,

$\mathrm{P}(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

So, co-ordinates of P,

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

Equate the x component on both the sides to get,

$x_{1}+x_{2}=-4$......(1)

Similarly,

$y_{1}+y_{2}=6 \ldots \ldots(2)$

Similarly, co-ordinates of Q,

$(4,-3)=\left(\frac{x_{3}+x_{2}}{2}, \frac{y_{3}+y_{2}}{2}\right)$

Equate the x component on both the sides to get,

$x_{3}+x_{2}=8$ .....(3)

Similarly,

$y_{3}+y_{2}=-6 \ldots \ldots(4)$

Similarly, co-ordinates of R,

$(4,5)=\left(\frac{x_{3}+x_{1}}{2}, \frac{y_{3}+y_{1}}{2}\right)$

Equate the x component on both the sides to get,

$x_{3}+x_{1}=8 \ldots \ldots(5)$

Similarly,

$y_{3}+y_{1}=10$....(6)

Add equation (1) (3) and (5) to get,

$2\left(x_{1}+x_{2}+x_{3}\right)=12$

$x_{1}+x_{2}+x_{3}=6$

Similarly, add equation (2) (4) and (6) to get,

$2\left(y_{1}+y_{2}+y_{3}\right)=10$

$y_{1}+y_{2}+y_{3}=5$

We know that the co-ordinates of the centroid $\mathrm{G}$ of a triangle whose vertices are $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right)$ is-

$G\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$

So, centroid $G$ of a triangle $\triangle A B C$ is,

$G\left(2, \frac{5}{3}\right)$