The area of a triangle whose vertices are (a | vedclass

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

Vedclass · Accepted Answer

$ab$

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(D) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.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$Area $= \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)|$$= \frac{1}{2} |a \cos 	heta (b \cos 	heta + b \sin 	heta) - a \sin 	heta (-2b \sin 	heta) - a \cos 	heta (b \sin 	heta - b \cos 	heta)|$$= \frac{ab}{2} |\cos^2 	heta + \cos 	heta \sin 	heta + 2 \sin^2 	heta - \sin 	heta \cos 	heta + \cos^2 	heta|$$= \frac{ab}{2} |2 \cos^2 	heta + 2 \sin^2 	heta|$$= \frac{ab}{2} 	imes 2(\cos^2 	heta + \sin^2 	heta) = ab(1) = ab$.

The area of a triangle whose vertices are $(a \cos \theta, b \sin \theta)$,$(-a \sin \theta, b \cos \theta)$,and $(-a \cos \theta, -b \sin \theta)$ is:

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