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The points $(3, -4)$ and $(-6, 2)$ are the extremities of a diagonal of a parallelogram. If the third vertex is $(-1, -3)$. Find the coordinates of the fourth vertex.
Given:
The points $(3, -4)$ and $(-6, 2)$ are the extremities of a diagonal of a parallelogram. The third vertex is $(-1, -3)$.
To do:
We have to find the coordinates of the fourth vertex.
Solution:
Let the extremities of the diagonal $AC$ of a parallelogram $ABCD$ are $A (3, -4)$ and $C (-6, 2)$. The third vertex $B$ be $(-1,-3)$.
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 \( (\frac{3-6}{2}, \frac{-4+2}{2}) \)
\( =(\frac{-3}{2}, \frac{-2}{2}) \)
\( =(\frac{-3}{2}, -1) \)
\( \mathrm{O} \) is the mid-point of \( \mathrm{BD} \).
The coordinates of \( \mathrm{O} \) are \( (\frac{-1+x}{2}, \frac{-3+y}{2}) \)
Therefore,
\( (\frac{-3}{2}, -1)=(\frac{-1+x}{2}, \frac{-3+y}{2}) \)
On comparing, we get,
\( \frac{-1+x}{2}=\frac{-3}{2} \)
\( -1+x=-3 \)
\( x=-3+1=-2 \)
Similarly,
\( \frac{-3+y}{2}=-1 \)
\( -3+y=-1(2) \)
\( y=-2+3 \)
\( y=1 \)
Therefore, the coordinates of the fourth vertex are $(-2,1)$.
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