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VITEEE 2014 Paper

Section: Mathematics

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Question : 112 of 120

Marks: +1, -0

If the area of the triangle on the complex plane formed by the points z, z + iz and iz is 200, then the value of 3 | z | must be equal to

20
40
60
80

Solution:

Let z = x + iy, then

z + iz = x + iy + i (x+ iy) = (x-y) + i (x + y) and iz = i (x + iy) = -y + ix,

Then, the area of the triangle formed by these lines is

Δ =

\frac{1}{2} \begin{vmatrix} x & y & 1 \\ (x-y) & (x+y) & 1 \\ -y & x & 1 \end{vmatrix}

Applying

R_2

→

R_2

-

(R_1+R_3)

Δ =

\frac{1}{2} \begin{vmatrix} x & y & 1 \\ 0 & 0 & -1 \\ -y & x & 1 \end{vmatrix}

=

\frac{1}{2} (x^2+y^2)

⇒

\frac{1}{2} \left|z\right|^2

= 200 (given)

\left|z\right|^2

= 400 ⇒ |z| = 20

∴ 3 |z| = 3 × 20 = 60

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