In a △ABC, A≡(α,β),B≡(2,3) and C≡(1,3) and point A lies on line y=2x+3 where α∈I. Area of △ABC=Δ, is such that [Δ]=5. Possible coordinates of A are (where [.] represents greatest integer function)
(−5,−7)
(5,13)
(−3,−5)
(2,3)
A
(2,3)
B
(−3,−5)
C
(−5,−7)
D
(5,13)
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Solution
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Let A be (a,2a+3).
Length of BC is 1 unit.
Equation of BC: y−3=3−32−1×(x−2)
⇒y=3 is the perpendicular distance of A from
BC=∣∣∣2a+3−31∣∣∣=|2a|
Using the formula of perpendicular distance of any point (x0,y0) from line ax+by+c=0 is |ax0+by0+c|√a2+b2
Area of triangle ABC =12× base × height =12×|2a|×1=|a|
Given: [Δ] = 5
Thus 5≤|a|<6
Therefore, co-ordinates are (5,13).
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