Complex Numbers And Quadratic Equations question 343
Question: If $ z=x+iy, $ then area of the triangle whose vertices are points $ z,iz $ and $ z+iz $ is [MP PET 1997; IIT 1986; AMU 2000; UPSEAT 2002]
Options:
A) $ 2|z{{|}^{2}} $
B) $ \frac{1}{2}|z{{|}^{2}} $
C) $ |z{{|}^{2}} $
D) $ \frac{3}{2}|z{{|}^{2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ z=x+iy $ ; $ z+iz=(x-y)+i(x+y) $ and $ iz=-y+ix $ If A denotes the area of the triangle formed by $ z,z+iz $ and $ iz $ , then $ A=\frac{1}{2} \begin{vmatrix} x & y & 1 \\ x-y & x+y & 1 \\ -y & x & 1 \\ \end{vmatrix} $ Applying transformation $ R_2\to R_2-R_1-R_3 $ , we get $ A=\frac{1}{2} \begin{vmatrix} x & y & 1 \\ 0 & 0 & -1 \\ -y & x & 0 \\ \end{vmatrix} =\frac{1}{2}(x^{2}+y^{2})=\frac{1}{2}|z{{|}^{2}} $
BETA
