Let z_1 and z_2 be two distinct complex numbers and let z=(1-t) z_1+t z_2 for some real number t with 0 < t < 1 . If (w) denotes the principal argument of a non-zero complex number w , then

Correct answer: A. |z-z_1|+|z-z_2|=|z_1-z_2| Given, z=((1-t) z_1+t z_2)/((1-t)+t) Clearly, z divides z_1 and z_2 in the ratio of t:(1-t), 0 < t < 1 aligned & A P+B P=A B & ie, |z-z_1|+|z-z_2|=|z_1-z_2| & Option (a) is true. & and aligned (z-z_1) & = (z_2-z) & = (z_2-z_1) aligned aligned (b) is false and (d) is true. Also, (z-z_1)= (z_2-z_1) aligned & ((z-z_1)/(z_2-z_1))=0 & (z-z_1)/(z_2-z_1) is purely real. & (z-z_1)/(z_2-z_1)= z - z _1 z _2- z _1 & or | array cc z-z_1 & z - z _1 z_2-z_1 & z _2- z _1 array |=0 & aligned Option (c) is correct. Hence, (a, c, d) is the correct option.

JEE Advanced 2010Paper 1

Exam: JEE Advanced
Year: 2010
Paper: Paper 1
Subject: Mathematics
Chapter: Complex Number

JEE Advanced 2010 · Paper 1 · Mathematics · Complex Number

Let z_1 and z_2 be two distinct complex numbers and let z=(1-t)...

Let

z_1

and

z_2

be two distinct complex numbers and let

z=(1-t) z_1+t z_2

for some real number

t

with

0 < t < 1

. If

\arg (w)

denotes the principal argument of a non-zero complex number

w

, then

Options

A $\left|z-z_1\right|+\left|z-z_2\right|=\left|z_1-z_2\right|$
B $\arg \left(z-z_1\right)=\arg \left(z-z_2\right)$
C $\left|$ $\begin{array}{cc}z-z_1 & \bar{z}-\bar{z}_1 \\ z_2-z_1 & \bar{z}_2-\bar{z}_1\end{array}$ $\right|=0$
D $\arg \left(z-z_1\right)=\arg \left(z_2-z_1\right)$

Correct answer

Step-by-step solution

Given,

z=\frac{(1-t) z_1+t z_2}{(1-t)+t}

Clearly,

z

divides

z_1 and z_2

in the ratio of

t:(1-t), 0 < t < 1

\begin{aligned} & \Rightarrow \quad A P+B P=A B \\ & \text{ ie, }\left|z-z_1\right|+\left|z-z_2\right|=\left|z_1-z_2\right| \\ & \Rightarrow \text{ Option (a) is true. } \\ & \text{\text{ and } } \begin{aligned} \arg \left(z-z_1\right) & =\arg \left(z_2-z\right) \\ & =\arg \left(z_2-z_1\right) \end{aligned}

\end{aligned}

\Rightarrow

(b) is false and (d) is true. Also,

\arg \left(z-z_1\right)=

\arg \left(z_2-z_1\right)

\begin{aligned} & \Rightarrow \quad \arg \left(\frac{z-z_1}{z_2-z_1}\right)=0 \\ & \therefore \frac{z-z_1}{z_2-z_1} \text{ is purely real. } \\ & \Rightarrow \quad \frac{z-z_1}{z_2-z_1}=\frac{\bar{z}-\bar{z}_1}{\bar{z}_2-\bar{z}_1} \\ & \text{\text{ or } }\left|\begin{array}{cc} z-z_1 & \bar{z}-\bar{z}_1 \\ z_2-z_1 & \bar{z}_2-\bar{z}_1 \end{array}\right|=0 \\ & \end{aligned}

\therefore

Option (c) is correct. Hence, (a, c, d) is the correct option.

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