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If two vertices of a parallelogram are $(3, 2), (-1, 0)$ and the diagonals cut at $(2, -5)$, find the other vertices of the parallelogram.
Given:
Two vertices of a parallelogram are $(3, 2), (-1, 0)$ and the diagonals cut at $(2, -5)$
To do:
We have to find the other vertices of the parallelogram.
Solution:
Let the coordinates of the two vertices are $A (3, 2)$ and $B (-1, 0)$.
Let the third and fourth vertices be $C(x_1,y_1)$ and $D(x_2,y_2)$ and the diagonals $AC$ and $BD$ bisect each other at $O(2,-5)$.
This implies,
\( \mathrm{O} \) is the mid-point of \( \mathrm{AC} \).
The coordinates of \( \mathrm{O} \) are \( (\frac{3+x_1}{2}, \frac{2+y_1}{2}) \)
\( (2,-5)=(\frac{3+x_1}{2}, \frac{2+y_1}{2}) \)
\( \Rightarrow \frac{3+x_1}{2}=2 \) and \( \frac{2+y_1}{2}=-5 \)
\( \Rightarrow 3+x_1=2(2) \) and \( 2+y_1=2(-5) \)
\( \Rightarrow x_1=4-3=1 \) and \( y_1=-10-2=-12 \)
\( \mathrm{O} \) is the mid-point of \( \mathrm{BD} \).
The coordinates of \( \mathrm{O} \) are \( (\frac{-1+x_2}{2}, \frac{0+y_2}{2}) \)
\( (2,-5)=(\frac{-1+x_2}{2}, \frac{y_2}{2}) \)
\( \Rightarrow \frac{-1+x_2}{2}=2 \) and \( \frac{y_2}{2}=-5 \)
\( \Rightarrow -1+x_2=2(2) \) and \( y_2=2(-5) \)
\( \Rightarrow x_2=4+1=5 \) and \( y_2=-10 \)
Therefore, the coordinates of the other two vertices are $(1,-12)$ and $(5,-10)$.
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