You visited us 1 times! Enjoying our articles? Unlock Full Access!

Rotation Concept

If | z 2+ iz ...

Question

If $| z_{2} + i z_{1} | = | z_{1} | + | z_{2} |$ and $| z_{1} | = 3$ and $| z_{2} | = 4$ , then area of $△ A B C$ , if affixes of $A, B$ and $C$ are $z_{1}, z_{2}$ and $[\frac{z_{2} - i z_{1}}{1 - i}]$ respectively, is,

\frac{5}{2}

sq. units

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0

sq. unit

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{25}{2}

sq. units

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{25}{4}

sq. units

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{25}{4}$ sq. units
$| z_{2} + i z_{1} | = | z_{1} | + | z_{2} | \Rightarrow | z_{2} + i z_{1} | = | z_{2} | + | i z_{1} |$
$\Rightarrow i z_{1}, 0 + i 0$ and $z_{2}$ are collinear
$\Rightarrow arg (i z_{1}) = arg (z_{2})$
$\Rightarrow arg (z_{2}) - arg (z_{1}) = \frac{π}{2}$

Let,
$z_{3} = \frac{z_{2} - i z_{1}}{1 - i}$
$\Rightarrow (1 - i) z_{3} = z_{2} - i z_{1}$
$\Rightarrow z_{2} - z_{3} = i (z_{1} - z_{3})$
$∴ ∠ A C B = \frac{π}{2}$ and
$| z_{1} - z_{3} | = | z_{2} - z_{3} |$
$\Rightarrow A C = B C$
$∵ A B^{2} = A C^{2} + B C^{2}$
$\Rightarrow A C = \frac{5}{\sqrt{2}} (∵ A B = 5)$
Therefore area of $△ A B C$
$= (\frac{1}{2}) A C \times B C = \frac{25}{4}$ sq.unit

Suggest Corrections

Similar questions

Q. If

| z_{2} + i z_{1} | = | z_{1} | + | z_{2} |

and

| z_{1} | = 3

and

| z_{2} | = 4

, then area of

△ A B C

, if affixes of

A, B

and

C

are

z_{1}, z_{2}

and

[\frac{z_{2} - i z_{1}}{1 - i}]

respectively, is,

Q. If

| z_{2} + i z_{1} | = | z_{1} | + | z_{2} |

and

| z_{1} | = 3

and

| z_{2} | = 4

, then the area of

Δ A B C

, if affixes of A, B, and C are

z_{1}, z_{2},

and

[(z_{2} - i z_{1}) / (1 - i)]

respectively, is

Q. If

| z_{1} | = | z_{2} | = | z_{3} | = 1

and

z_{1} + z_{2} + z_{3} = 0

, then area of the triangle whose vertices are

z_{1}, z_{2}, z_{3}

Q. If

z_{1}, z_{2}

z_{3}

are the affixes of three points

A, B

C

respectively and satisfy the condition

| z_{1} - z_{2} | = | z_{1} | + | z_{2} |

and

| (2 - i) z_{1} + i z_{3} | = | z_{1} | + | (1 - i) z_{1} + i z_{3} |

then prove that

△ A B C

in a right angled.

Let $Z_{1}, Z_{2}, Z_{3}$ be three points $A, B$ and $P$ respectively in the argand plane. Let $P$ moves in the plane such that $| Z - Z_{1} | + | Z - Z_{2} | = K .$ Let $Z_{1} = - Z_{2} = i$ Maximum area of the triangle ABP is (if K = 4)

Join BYJU'S Learning Program

Explore more

Rotation Concept

Standard XII Mathematics

Join BYJU'S Learning Program

If |z2+iz1|=|z1|+|z2| and |z1|=3 and |z2|=4, then area of △ABC, if affixes of A,B and C are z1,z2 and [z2−iz11−i] respectively, is,

If $| z_{2} + i z_{1} | = | z_{1} | + | z_{2} |$ and $| z_{1} | = 3$ and $| z_{2} | = 4$ , then area of $△ A B C$ , if affixes of $A, B$ and $C$ are $z_{1}, z_{2}$ and $[\frac{z_{2} - i z_{1}}{1 - i}]$ respectively, is,