Area of a triangle whose vertices are (a cos θ, b sinθ), (–a sin θ, b cos θ) and (–a cos θ, –b sin θ) is
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1.The area of the quadrilateral ABCD formed by the points a(4,3),b(6,4),c(5,6),d(3,7)
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Given :
A triangle ABC with coordinates of vertex A (a cos θ, b sinθ), B(–a sin θ, b cos θ) and C(–a cos θ, –b sin θ).
To Find : Area of given triangle
Solution :
•In coordinate geometry area of triangle coordinates of whose vertex are ( X1 , Y1 ) , ( X2 , Y2 ) ( X3 , Y3 ) is given by
•Area of triangle = 1/2[X1(Y2-Y3) + X2(Y3-Y2) + X3(Y1-Y2) ]
•So area of triangle coordinates of whose vertex are A (a cos θ, b sinθ), B(–a sin θ, b cos θ) and C(–a cos θ, –b sin θ) will be
•Area of triangle = 1/2[ acosθ(bcosθ + bsinθ) + (–asinθ)( –bsinθ–bsinθ) + (-acosθ)(bsinθ-bcosθ) ]
•Area of triangle = 1/2[ abcos²θ + abcosθsinθ +2absin²θ - abcosθsinθ + abcos²θ]
•Area of triangle = 1/2[ 2abcos²θ+2absin²θ ]
•Area of triangle = 1/2(2ab)(cos²θ+sin²θ )
•as , cos²θ+sin²θ = 1
•Area of triangle = ab square units
• So, area of triangle whose coordinates are A (a cos θ, b sinθ), B(–a sin θ, b cos θ) and C(–a cos θ, –b sin θ) is ab square units