$ABCD$ is a parallelogram with vertices $A (x_1, y_1), B (x_2, y_2)$ and $C (x_3, y_3)$. Find the coordinates of the fourth vertex $D$ in terms of $x_1, x_2, x_3, y_1, y_2$ and $y_3$.

Given:

$ABCD$ is a parallelogram with vertices $A (x_1, y_1), B (x_2, y_2)$ and $C (x_3, y_3)$.

To do:

We have to find the fourth vertex $D$ in terms of $x_1, x_2, x_3, y_1, y_2$ and $y_3$.

Solution:

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 \( (\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2}) \)

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

The coordinates of \( \mathrm{O} \) are \( (\frac{x_2+x}{2}, \frac{y_2+y}{2}) \)

Therefore,

\( (\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2})=(\frac{x_2+x}{2}, \frac{y_2+y}{2}) \)

On comparing, we get,

\( \frac{x_1+x_3}{2}=\frac{x_2+x}{2} \)

\( \Rightarrow x_1+x_3=x_2+x \)

\( \Rightarrow x=x_1+x_3-x_2 \)

Similarly,

\( \frac{y_1+y_3}{2}=\frac{y_2+y}{2} \)

\( \Rightarrow y_1+y_3=y_2+y \)

\( \Rightarrow y=y_1+y_3-y_2 \)

Therefore, the coordinates of the fourth vertex are $(x_1+x_3-x_2, y_1+y_3-y_2)$.