If three consecutive vertices of a parallelogram are $(1, -2), (3, 6)$ and $(5, 10)$, find its fourth vertex.

Given:

Three consecutive vertices of a parallelogram are $(1, -2), (3, 6)$ and $(5, 10)$. 

To do:

We have to find the fourth vertex.

Solution:

Let the coordinates of the three vertices are $A (1, -2), B (3, 6)$ and $C (5, 10)$.
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 \( (\frac{1+5}{2}, \frac{-2+10}{2}) \)

\( =(\frac{6}{2}, \frac{8}{2}) \)

\( =(3,4) \)

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

The coordinates of \( \mathrm{O} \) are \( (\frac{3+x}{2}, \frac{6+y}{2}) \)

Therefore,

\( (3,4)=(\frac{3+x}{2}, \frac{6+y}{2}) \)

On comparing, we get,

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

\( 3+x=3(2) \)

\( x=6-3=3 \)

Similarly,

\( \frac{6+y}{2}=4 \)

\( 6+y=4(2) \)

\( y=8-6=2 \)

Therefore, the coordinates of the fourth vertex are $(3,2)$.