The diagonals AC and BD of a rhombus ABCD int | vedclass

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(D) Since $O(3,4)$ is the midpoint of $AC$, $C = (2(3)-1, 2(4)-2) = (5,6)$.Since $O(3,4)$ is the midpoint of $BD$, $\alpha+\gamma=6$ and $\beta+\delta

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$6$

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(D) Since $O(3,4)$ is the midpoint of $AC$, $C = (2(3)-1, 2(4)-2) = (5,6)$.Since $O(3,4)$ is the midpoint of $BD$, $\alpha+\gamma=6$ and $\beta+\delta=8$.$BD = 2\sqrt{2}$, so $OB = OD = \sqrt{2}$.Vector $\vec{OB} = (\alpha-3, \beta-4)$. Since $|\vec{OB}|^2 = 2$, $(\alpha-3)^2 + (\beta-4)^2 = 2$.Since $AC \perp BD$, the slope of $AC$ is $m_{AC} = \frac{6-2}{5-1} = 1$. Thus, the slope of $BD$ is $m_{BD} = -1$.So, $\frac{\beta-4}{\alpha-3} = -1 \implies \beta-4 = -(\alpha-3) \implies \beta = 7-\alpha$.Substituting into the distance equation: $(\alpha-3)^2 + (7-\alpha-4)^2 = 2 \implies 2(\alpha-3)^2 = 2 \implies (\alpha-3)^2 = 1$.So $\alpha-3 = 1$ or $\alpha-3 = -1$, giving $\alpha=4$ or $\alpha=2$.If $\alpha=4$, then $\beta=3$, $\gamma=2$, $\delta=5$. Condition $\alpha If $\alpha=2$, then $\beta=5$, $\gamma=4$, $\delta=3$. Condition $2 Thus, $\beta+\gamma-\delta = 5+4-3 = 6$.

The diagonals $AC$ and $BD$ of a rhombus $ABCD$ intersect at the point $O(3,4)$. If $BD=2 \sqrt{2}$, $A=(1,2)$, $B=(\alpha, \beta)$, $D=(\gamma, \delta)$ and $\alpha < \delta < \gamma < \beta$, then the value of $\beta+\gamma-\delta$ is:

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