Q. 275.0( 1 Vote )

Using integration

Answer :


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)


Required Area






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