Math, asked by triggerAVS2776, 1 year ago

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

Answers

Answered by MaheswariS
1

\textbf{Formula used:}

\text{Area of the triangle having vertices}

\text{$(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$ is}

\boxed{\bf\triangle=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]}

\textbf{Given points are}

(x_1,y_1)=(a\,cos\theta,b\,sin\theta)

(x_2,y_2)=(-a\,sin\theta,b\,cos\theta)

(x_3,y_3)=(-a\,cos\theta,-b\,sin\theta)

\textbf{Area of the triangle}

=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]

=\frac{1}{2}[a\,cos\theta(b\,cos\theta+b\,sin\theta)-a\,sin\theta(-b\,sin\theta-b\,sin\theta)-a\,cos\theta(b\,sin\theta-b\,cos\theta)]

=\frac{1}{2}[ab\,cos^2\theta+ab\,cos\theta\,sin\theta+2ab\,sin^2\theta-ab\,cos\theta\,sin\theta+ab\,cos^2\theta]

=\frac{1}{2}[ab\,cos^2\theta+2ab\,sin^2\theta+ab\,cos^2\theta]

=\frac{1}{2}[2ab\,cos^2\theta+2ab\,sin^2\theta]

=\frac{1}{2}[2ab(cos^2\theta+sin^2\theta)]

=\frac{1}{2}[2ab(1)]

=\text{ab square units}

\therefore\textbf{Area of the triangle is ab square units}

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Answered by AnkitaSahni
1

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

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