If a vertex of a triangle be (1, 1) and the

Solution:

Let a  in which P and Q are the mid-points of sides AB and AC respectively. The coordinates are: A (1, 1); P (−2, 3) and Q (5, 2).

We have to find the co-ordinates of $\mathrm{B}\left(x_{1}, y_{1}\right)$ and $\mathrm{C}\left(x_{2}, y_{2}\right)$.

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,

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

Therefore mid-point P of side AB can be written as,

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

Now equate the individual terms to get,

$x_{1}=-5$

$y_{1}=5$

So, co-ordinates of B is (−5, 5)

Similarly, mid-point Q of side AC can be written as,

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

Now equate the individual terms to get,

$x_{2}=9$

 

$y_{2}=3$

So, co-ordinates of C is (9, 3)