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Question

In a ABC, A(α,β),B(2,3) and C(1,3) and point A lies on line y=2x+3 where αI. Area of ABC=Δ, is such that [Δ]=5. Possible coordinates of A are (where [.] represents greatest integer function)
  1. (5,7)
  2. (5,13)
  3. (3,5)
  4. (2,3)

A
(2,3)
B
(3,5)
C
(5,7)
D
(5,13)
Solution
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Let A be (a,2a+3).
Length of BC is 1 unit.
Equation of BC: y3=3321×(x2)
y=3 is the perpendicular distance of A from
BC =2a+331=|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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