If three vertices of a parallelogram ABCD are | vedclass

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(D) In a parallelogram $ABCD$,the diagonals $AC$ and $BD$ bisect each other at the same midpoint.Let the coordinates of vertex $D$ be $(x, y)$.The mid

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$(1,0)$

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(D) In a parallelogram $ABCD$,the diagonals $AC$ and $BD$ bisect each other at the same midpoint.Let the coordinates of vertex $D$ be $(x, y)$.The midpoint of diagonal $AC$ is given by $(\frac{x_A + x_C}{2}, \frac{y_A + y_C}{2}) = (\frac{3+2}{2}, \frac{2+3}{2}) = (2.5, 2.5)$.The midpoint of diagonal $BD$ is given by $(\frac{x_B + x_D}{2}, \frac{y_B + y_D}{2}) = (\frac{4+x}{2}, \frac{5+y}{2})$.Since the midpoints are the same,we equate the coordinates:$\frac{4+x}{2} = 2.5 \implies 4+x = 5 \implies x = 1$.$\frac{5+y}{2} = 2.5 \implies 5+y = 5 \implies y = 0$.Therefore,the coordinates of vertex $D$ are $(1, 0)$.

If three vertices of a parallelogram $ABCD$ are $A(3,2)$,$B(4,5)$,and $C(2,3)$,then find the coordinates of the fourth vertex $D$.

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