Verified Solution • 2025–26 Session moderate QJEEM24Ma1d50: JEE Main 2024 Mathematics - Class 11 JEEM24Ma1d50 Solution: Let and be the vertices of a parallelogram . If the point lies… Let A(−2,−1),B(1,0),C(α,β)A(-2,-1), B(1,0), C(\alpha, \beta)A(−2,−1),B(1,0),C(α,β) and D(γ,δ)D(\gamma, \delta)D(γ,δ) be the vertices of a parallelogram ABCDA B C DABCD . If the point CCC lies on 2x−y=52 x-y=52x−y=5 and the point DDD lies on 3x−2y=63 x-2 y=63x−2y=6 , then the value of ∣α+β+γ+δ∣|\alpha+\beta+\gamma+\delta|∣α+β+γ+δ∣ is equal to ___________. Step-by-Step Verified Solution amp;P≡(α−22,β−12)≡(γ+12,δ2)amp;α−22=γ+12 and β−12=δ2amp;⇒α−γ=3….(1),β−δ=1……(2)\begin{aligned}& \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 \ldots (2)\end{aligned}amp;P≡(2α−2,2β−1)≡(2γ+1,2δ)amp;2α−2=2γ+1 and 2β−1=2δamp;⇒α−γ=3….(1),β−δ=1……(2) Also, (γ,δ)(\gamma, \delta)(γ,δ) lies on 3x−2y=63 x-2 y=63x−2y=6 3γ−2δ=63 \gamma-2 \delta=63γ−2δ=6 ..... (3) and (α,β)(\alpha, \beta)(α,β) lies on 2x−y=52 x-y=52x−y=5 ⇒2α−β=5. \Rightarrow 2 \alpha-\beta=5 \text {. }⇒2α−β=5. Solving (1), (2), (3), (4) amp;α=−3,β=−11,γ=−6,δ=−12amp;∣α+β+γ+δ∣=32\begin{aligned}& \alpha=-3, \beta=-11, \gamma=-6, \delta=-12 \\& |\alpha+\beta+\gamma+\delta|=32\end{aligned}amp;α=−3,β=−11,γ=−6,δ=−12amp;∣α+β+γ+δ∣=32 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring
