JEE Main 2024 — Mathematics Coordinate Geometry

Given,

$A(-2,-1),B(1,0),C(\alpha ,\beta )$A(−2,−1),B(1,0),C(α,β) and $D(\gamma ,\delta )$D(γ,δ) be the vertices of a parallelogram $ABCD$ABCD

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

Now, plotting the diagram we get,

Now, using the midpoint formula we get,

$$P\equiv (\frac{\alpha -2}{2},\frac{\beta -1}{2})\equiv (\frac{\gamma +1}{2},\frac{\delta }{2})$$P≡(2α−2​,2β−1​)≡(2γ+1​,2δ​)

Now, on comparing we get,

$\frac{\alpha -2}{2}=\frac{\gamma +1}{2}$2α−2​=2γ+1​ and $\frac{\beta -1}{2}=\frac{\delta }{2}$2β−1​=2δ​

$$\Rightarrow \alpha -\gamma =3...(1),\beta -\delta =1...(2)$$⇒α−γ=3...(1),β−δ=1...(2)

Also, given $(\gamma ,\delta )$(γ,δ) lies on $3x-2y=6$3x−2y=6

$$\Rightarrow 3\gamma -2\delta =6...(3)$$⇒3γ−2δ=6...(3)

And $(\alpha ,\beta )$(α,β) lies on $2x-y=5$2x−y=5

$$\Rightarrow 2\alpha -\beta =5...(4)$$⇒2α−β=5...(4)

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

$$\alpha =-3,\beta =-11,\gamma =-6,\delta =-12$$α=−3,β=−11,γ=−6,δ=−12

Hence, $|\alpha +\beta +\gamma +\delta |=|-3-11-6-12|=32$∣α+β+γ+δ∣=∣−3−11−6−12∣=32