Complex Numbers And Quadratic Equations question 351
Question: The area of the triangle whose vertices are represented by the complex numbers 0, z, $ z{e^{i\alpha }}, $ $ (0<\alpha <\pi ) $ equals [AMU 2002]
Options:
A) $ \frac{1}{2}|z{{|}^{2}}\cos \alpha $
B) $ \frac{1}{2}|z{{|}^{2}}\sin \alpha $
C) $ \frac{1}{2}|z{{|}^{2}}\sin \alpha \cos \alpha $
D) $ \frac{1}{2}|z{{|}^{2}} $
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Answer:
Correct Answer: B
Solution:
Vertices are $ 0=0+i0, $ $ z=x+iy $ and $ z{e^{i\alpha }}=(x+iy)(\cos \alpha +i\sin \alpha ) $ $ =(x\cos \alpha -y\sin \alpha )+i(y\cos \alpha +x\sin \alpha ) $
$ \therefore $ Area $ =\frac{1}{2}| \begin{matrix} 0 & 0 & 1 \\ x & y & 1 \\ (x\cos \alpha -y\sin \alpha ) & (y\cos \alpha +x\sin \alpha ) & 1 \\ \end{matrix} | $ $ =\frac{1}{2}[xy\cos \alpha +x^{2}\sin \alpha -xy\cos \alpha +y^{2}\sin \alpha ] $ $ =\frac{1}{2}\sin \alpha (x^{2}+y^{2}) $ $ =\frac{1}{2}|z{{|}^{2}}\sin \alpha $ $ [\because |z|=\sqrt{x^{2}+y^{2}}] $ .
BETA
