If the points A (6, 1),B (8, 2),C (9, 4),and | vedclass

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(D) We know that the diagonals of a parallelogram bisect each other.Therefore,the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $B

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$7$

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(D) We know that the diagonals of a parallelogram bisect each other.Therefore,the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$.The coordinates of the midpoint of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ are given by $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.Midpoint of $AC = \left(\frac{6+9}{2}, \frac{1+4}{2}\right) = \left(\frac{15}{2}, \frac{5}{2}\right)$.Midpoint of $BD = \left(\frac{8+p}{2}, \frac{2+3}{2}\right) = \left(\frac{8+p}{2}, \frac{5}{2}\right)$.Equating the midpoints,we get $\frac{15}{2} = \frac{8+p}{2}$.Multiplying both sides by $2$,we get $15 = 8 + p$.Thus,$p = 15 - 8 = 7$.

If the points $A (6, 1)$,$B (8, 2)$,$C (9, 4)$,and $D (p, 3)$ are the vertices of a parallelogram,taken in order,find the value of $p$.

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