Three consecutive vertices of a parallelogram are $(-2, -1), (1, 0)$ and $(4, 3)$. Find the fourth vertex.

Given:

Three consecutive vertices of a parallelogram are $(-2, -1), (1, 0)$ and $(4, 3)$. 

To do:

We have to find the fourth vertex.

Solution:

Let the coordinates of the three vertices are $A (-2, -1), B (1, 0)$ and $C (4, 3)$.
Let the fourth vertex be $D(x,y)$ and the diagonals $AC$ and $BD$ bisect each other at $O$.

This implies,

\( \mathrm{O} \) is the mid-point of \( \mathrm{AC} \).

The coordinates of \( \mathrm{O} \) are \( (\frac{-2+4}{2}, \frac{-1+3}{2}) \)

\( =(\frac{2}{2}, \frac{2}{2}) \)

\( =(1,1) \)

\( \mathrm{O} \) is the mid-point of \( \mathrm{BD} \).

The coordinates of \( \mathrm{O} \) are \( (\frac{1+x}{2}, \frac{0+y}{2}) \)

Therefore,

\( (1,1)=(\frac{1+x}{2}, \frac{y}{2}) \)

On comparing, we get,

\( \frac{1+x}{2}=1 \)

\( 1+x=1(2) \)

\( x=2-1=1 \)

Similarly,

\( \frac{y}{2}=1 \)

\( y=1(2) \)

\( y=2 \)

Therefore, the coordinates of the fourth vertex are $(1,2)$.