Find the area of the triangle with vertices ( | vedclass

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(D) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1)

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$ab$

Vedclass · Answer

(D) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. Substituting the given vertices: $x_1 = a \cos 	heta, y_1 = b \sin 	heta$ $x_2 = -a \sin 	heta, y_2 = b \cos 	heta$ $x_3 = -a \cos 	heta, y_3 = -b \sin 	heta$ $\Delta = \frac{1}{2} |a \cos 	heta (b \cos 	heta - (-b \sin 	heta)) + (-a \sin 	heta) (-b \sin 	heta - b \sin 	heta) + (-a \cos 	heta) (b \sin 	heta - b \cos 	heta)|$ $\Delta = \frac{1}{2} |a \cos 	heta (b \cos 	heta + b \sin 	heta) - a \sin 	heta (-2b \sin 	heta) - a \cos 	heta (b \sin 	heta - b \cos 	heta)|$ $\Delta = \frac{1}{2} |ab \cos^2 	heta + ab \sin 	heta \cos 	heta + 2ab \sin^2 	heta - ab \sin 	heta \cos 	heta + ab \cos^2 	heta|$ $\Delta = \frac{1}{2} |ab \cos^2 	heta + 2ab \sin^2 	heta + ab \cos^2 	heta|$ $\Delta = \frac{1}{2} |2ab \cos^2 	heta + 2ab \sin^2 	heta| = \frac{1}{2} |2ab(\cos^2 	heta + \sin^2 	heta)|$ Since $\cos^2 	heta + \sin^2 	heta = 1$,we get $\Delta = \frac{1}{2} |2ab| = |ab|$.

Find the area of the triangle with vertices $(a \cos \theta, b \sin \theta)$,$(-a \sin \theta, b \cos \theta)$,and $(-a \cos \theta, -b \sin \theta)$.

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