Three consecutive vertices of a parallelogram are $(-2, -1), (1, 0)$ and $(4, 3)$. Find the fourth vertex.
Given:
Three consecutive vertices of a parallelogram are $(-2, -1), (1, 0)$ and $(4, 3)$.
To do:
We have to find the fourth vertex.
Solution:
Let the coordinates of the three vertices are $A (-2, -1), B (1, 0)$ and $C (4, 3)$.
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 \( (\frac{-2+4}{2}, \frac{-1+3}{2}) \)
\( =(\frac{2}{2}, \frac{2}{2}) \)
\( =(1,1) \)
\( \mathrm{O} \) is the mid-point of \( \mathrm{BD} \).
The coordinates of \( \mathrm{O} \) are \( (\frac{1+x}{2}, \frac{0+y}{2}) \)
Therefore,
\( (1,1)=(\frac{1+x}{2}, \frac{y}{2}) \)
On comparing, we get,
\( \frac{1+x}{2}=1 \)
\( 1+x=1(2) \)
\( x=2-1=1 \)
Similarly,
\( \frac{y}{2}=1 \)
\( y=1(2) \)
\( y=2 \)
Therefore, the coordinates of the fourth vertex are $(1,2)$.
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