Coordinate Geometry Question 141
Question: 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
Options:
A) $ a\cos \theta \sin \theta $
B) $ ab\sin \theta \cos \theta $
C) $ \frac{1}{2}ab $
D) $ ab $
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Answer:
Correct Answer: D
Solution:
Area $ =\frac{1}{2} \begin{vmatrix} a\cos \theta & b\sin \theta & 1 \\ -a\sin \theta & b\cos \theta & 1 \\ -a\cos \theta & -b\sin \theta & 1 \\ \end{vmatrix} $
$ =\frac{1}{2}(a\times b)\begin{vmatrix} \cos \theta & \sin \theta & 1 \\ -\sin \theta & \cos \theta & 1 \\ -\cos \theta & -\sin \theta & 1 \\ \end{vmatrix} $
$ =\frac{ab}{2}[\cos \theta (\cos \theta +\sin \theta )-\sin \theta (-\sin \theta +\cos \theta ) $
$ +1({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )] $
$ =\frac{ab}{2}(1+1)=ab $ .
BETA
