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Let $\mathrm{A}(-2,-1), \mathrm{B}(1,0), \mathrm{C}(\alpha, \beta)$ and $\mathrm{D}(\gamma, \delta)$ be the vertices of a parallelogram $A B C D$. If the point $C$ lies on $2 x-y=5$ and the point $D$ lies on $3 x-2 y=6$, then the value of $|\alpha+\beta+\gamma+\delta|$ is equal to_____.
$30$
$31$
$32$
$33$
Solution
$\mathrm{P} \equiv\left(\frac{\alpha-2}{2}, \frac{\beta-1}{2}\right) \equiv\left(\frac{\gamma+1}{2}, \frac{\delta}{2}\right) $
$ \frac{\alpha-2}{2}=\frac{\gamma+1}{2} \text { and } \frac{\beta-1}{2}=\frac{\delta}{2} $
$\Rightarrow \alpha-\gamma=3 \ldots . .(1), \beta-\delta=1 \ldots ….(2)$
Also, $(\gamma, \delta)$ lies on $3 x-2 y=6$
$3 \gamma-2 \delta=6$
and $(\alpha, \beta)$ lies on $2 x-y=5$
$\Rightarrow 2 \alpha-\beta=5 \text {……(4) }$
Solving $(1), (2), (3), (4)$
$\alpha=-3, \beta=-11, \gamma=-6, \delta=-12$
$|\alpha+\beta+\gamma+\delta|=32$
