In a triangle ABC, A≡(α,β),B≡(2,3) and C≡(1,3) and point A lies on line y=2x+3 whereα∈I , area of ΔABC , Δ , is such that [Δ]=5 . Possible coordinates of A are (where [.] represents greatest integer function)

A≡(α.2α+3),BC=1 Unit. Equation of BC is y−3=0 Distance of A from BC is p⇒|2α+3−3| Area of ΔABC=Δ=|α|;5≤Δ<6⇒5≤|α|<6

A≡(α.2α+3),BC=1 Unit. Equation of BC is y−3=0 Distance of A from BC is p⇒|2α+3−3| Area of ΔABC=Δ=|α|;5≤Δ<6⇒5≤|α|<6

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In a triangle ABC, $A \equiv (α, β), B \equiv (2, 3)$ and $C \equiv (1, 3)$ and point A lies on line $y = 2 x + 3$ where $α \in I$ , area of $Δ A B C$ , $Δ$ , is such that $[Δ] = 5$ . Possible coordinates of A are (where [.] represents greatest integer function)

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$(2, 3)$

$(5, 13)$

$(- 5, - 7)$

$(- 3, - 5)$

answer is B.

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$A \equiv (α .2 α + 3), B C = 1$ Unit. Equation of BC is $y - 3 = 0$

Distance of A from BC is $p \Rightarrow | 2 α + 3 - 3 |$

Area of $Δ A B C = Δ = | α |; 5 \leq Δ < 6 \Rightarrow 5 \leq | α | < 6$

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