The fourth vertex D of a parallelogram ABCD w | vedclass

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(B) Let the fourth vertex of the parallelogram be $D(x_4, y_4)$.In a parallelogram,the diagonals bisect each other,which means the midpoint of diagona

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$(0,-1)$

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(B) Let the fourth vertex of the parallelogram be $D(x_4, y_4)$.In a parallelogram,the diagonals bisect each other,which means the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$.The midpoint formula for a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.Midpoint of $AC = \left(\frac{-2+8}{2}, \frac{3+3}{2}\right) = \left(\frac{6}{2}, \frac{6}{2}\right) = (3,3)$.Midpoint of $BD = \left(\frac{6+x_4}{2}, \frac{7+y_4}{2}\right)$.Since the midpoints are equal:$\frac{6+x_4}{2} = 3 \Rightarrow 6+x_4 = 6 \Rightarrow x_4 = 0$.$\frac{7+y_4}{2} = 3 \Rightarrow 7+y_4 = 6 \Rightarrow y_4 = -1$.Thus,the fourth vertex $D$ is $(0,-1)$.

The fourth vertex $D$ of a parallelogram $ABCD$ whose three vertices are $A(-2,3)$,$B(6,7)$,and $C(8,3)$ is

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