Let A − 2 , − 1 , B 1 , 0 , C α , β and D γ , δ be the vertices of a parallelogram A B C D . If the point C…

Given,

$A\left(-2,-1\right), B\left(1,0\right), C\left(\alpha ,\beta \right)$ and $D\left(\gamma ,\delta \right)$ be the vertices of a parallelogram $ABCD$

And the point $C$ lies on $2x-y=5$ and the point $D$ lies on $3x-2y=6$,

Now, plotting the diagram we get,

Now, using the midpoint formula we get,

$P\equiv \left(\frac{\alpha -2}{2},\frac{\beta -1}{2}\right)\equiv \left(\frac{\gamma +1}{2},\frac{\delta }{2}\right)$

Now, on comparing we get,

$\frac{\alpha -2}{2}=\frac{\gamma +1}{2}$ and $\frac{\beta -1}{2}=\frac{\delta }{2}$

$\Rightarrow \alpha -\gamma =3 ... \left(1\right), \beta -\delta =1 ... \left(2\right)$

Also, given $\left(\gamma ,\delta \right)$ lies on $3x-2y=6$

$\Rightarrow 3\gamma -2\delta =6 ...\left(3\right)$

And $\left(\alpha ,\beta \right)$ lies on $2x-y=5$

$\Rightarrow 2\alpha -\beta =5 ...\left(4\right)$

On solving equations $\left(1\right), \left(2\right), \left(3\right), \left(4\right)$ we get,

$\alpha =-3, \beta =-11, \gamma =-6, \delta =-12$

Hence, $\left|\alpha +\beta +\gamma +\delta \right|=\left|-3-11-6-12\right|=32$