The extremities of a diagonal of a parallelog | vedclass

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(B) Let $A = (3, -4)$ and $C = (-6, 5)$ be the endpoints of one diagonal of the parallelogram $ABCD$. Let $B = (-2, 1)$ be the third vertex and $D = (

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$(-1, 0)$

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(B) Let $A = (3, -4)$ and $C = (-6, 5)$ be the endpoints of one diagonal of the parallelogram $ABCD$. Let $B = (-2, 1)$ be the third vertex and $D = (x, y)$ be the fourth vertex.The diagonals of a parallelogram bisect each other,meaning the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$.Midpoint of $AC = \left( \frac{3 + (-6)}{2}, \frac{-4 + 5}{2} \right) = \left( -\frac{3}{2}, \frac{1}{2} \right)$.Midpoint of $BD = \left( \frac{-2 + x}{2}, \frac{1 + y}{2} \right)$.Equating the midpoints:$\frac{-2 + x}{2} = -\frac{3}{2} \implies -2 + x = -3 \implies x = -1$.$\frac{1 + y}{2} = \frac{1}{2} \implies 1 + y = 1 \implies y = 0$.Therefore,the fourth vertex $D$ is $(-1, 0)$.

The extremities of a diagonal of a parallelogram are the points $(3, -4)$ and $(-6, 5)$. If the third vertex is $(-2, 1)$,then the fourth vertex is:

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