Q. 275.0( 1 Vote )

# Using integration Let the vertices be A(−2,1), B(0,4), and C(2,3).

The equation of AB is; 3x − 2y + 8 = 0

The equation of BC is; x + 2y − 8 = 0

The equation of AC is; x − 2y + 4 = 0

Required area =  = 0 + 0 − 3 + 8 + 1 + 8 − 0 − 0 − (1 + 4 − 1 + 4)

= 4 sq.unit

OR By solving the equations x2 + y2 = 16 and x = √3 y

3y2 + y2 = 16

y = 2 x = √3 y = x√3

So the point of intersection is (2√3,2)

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