If the coordinates of the mid-points of the sides of a triangle be $(3, -2), (-3, 1)$ and $(4, -3)$, then find the coordinates of its vertices.

 Given:

The coordinates of the mid-points of the sides of a triangle are $(3, -2), (-3, 1)$ and $(4, -3)$.

To do:

We have to find the coordinates of its vertices.

Solution:

Let $A (x_1, y_1), B (x_2, y_2)$ and $C (x_3, y_3)$ be the vertices of the $\triangle ABC$ and $D(3,-2), E(-3,1)$ and $F(4,-3)$ are the mid-points of $BC, CA$ and $AB$. 

$D$ is the mid-point of $BC$.

This implies,

\( \frac{x_{2}+x_{3}}{2}=3 \)

\( \Rightarrow x_{2}+x_{3}=6 \).....(i)

\( \frac{y_{2}+y_{3}}{2}=-2 \)

\( \Rightarrow y_{2}+y_{3}=-4 \)......(a)

Similarly,

\( E \) is the mid-point of \( A C \).

\( \frac{x_{3}+x_{1}}{2}=-3 \)

\( \Rightarrow x_{3}+x_{1}=-6 \).......(ii)

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

\( \Rightarrow y_{3}+y_{1}=2 \).......(b)

\( \mathrm{F} \) is the mid-point of \( \mathrm{AB} \).

\( \frac{x_{2}+x_{1}}{2}=4 \)

\( \Rightarrow x_{2}+x_{1}=8 \)........(iii)

\( \frac{y_{2}+y_{1}}{2}=-3 \)

\( \Rightarrow y_{2}+y_{1}=-6 \).......(c)

Adding (i), (ii) and (iii), we get,

\( 2\left(x_{1}+x_{2}+x_{3}\right)=8 \)

\( \Rightarrow x_{1}+x_{2}+x_{3}=4 \)......(iv)

Subtracting (i), (ii) and (iii) from (iv), we get,

\( x_{1}=-2, x_{2}=10, x_{3}=-4 \)

Similarly, 

Adding (a), (b) and (c), we get,

\( 2\left(y_{1}+y_{2}+y_{3}\right)=-8 \)

\( \Rightarrow y_{1}+y_{2}+y_{3}=-4 \).......(d)

Subtracting (a),(b) \) and (c) from (d), we get,

\( y_{1}=0 \) \( y_{2}=-6 \) \( y_{3}=2 \)

Therefore, vertices of \( \Delta \mathrm{ABC} \) are \( \mathrm{A}(-2,0), \mathrm{B}(10,-6), \mathrm{C}(-4,2) \)

The vertices of the triangle are $(-2,0), (10,-6)$ and $(-4,2)$.