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$ABCD$ 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 $(x_1+x_3-x_2, y_1+y_3-y_2)$.
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