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Q.

Let X=(x,y)×:x28+y220<1 and y2<5x. Three distinct point P, Q and R are randomly chosen from X. Then the probability that P, Q and R form a triangle whose area is a positive integer, is

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a

73220

b

71220

c

83220

d

79220

answer is B.

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Detailed Solution

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x28+y220<1    y2<5xx28+5x20=1    x28+x4=1 x2+2x=8(x+1)2=9    x=2x=1,y2<5y<5    (1,0)(1,1)(1,2)(1,1)(1,2)(2,0)(2,1)(2,2)(2,3)(2,1)(2,2)(2,3)
Number of ways to select 3 points =12C3
=12×11×106=2×11×10 =220
Area of  PQR is an integer 
Triangle has two vertices on x = 1 or x = 2 
1,y11,y22,y3 (or) 1,y32,y12,y2
Area of Δℓe=12(1)y1y2
y1y2  is an even integer 
Number of ways =  3C2+2C27C1 for x = 1
= 4C2+3C25C1 for x=2
Number of triangles = 28 + 45 = 73 
Probability = 73220

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Let X=(x,y)∈ℤ×ℤ:x28+y220<1 and y2<5x. Three distinct point P, Q and R are randomly chosen from X. Then the probability that P, Q and R form a triangle whose area is a positive integer, is