If Az1- Bz2- Cz3 are vertices of - ABC in which - ABC-4 and AB-BC-2- then find z2 in terms of z1 and z3

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

Area Method to Find Condition for Co-Linearity

If Az1, Bz2...

Question

If $A (z_{1}), B (z_{2}), C (z_{3})$ are vertices of $Δ A B C$ in which $∠ A B C = \frac{π}{4}$ and $\frac{A B}{B C} = \sqrt{2}$ , then find $z_{2}$ in terms of $z_{1}$ and $z_{3}$

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

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

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

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

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

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

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

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

Open in App

Solution

The correct option is C $z_{2} = z_{3} + i (z_{1} - z_{3})$
Rotating the side CB about the point B by $45^{o}$ anticlockwise direction, alongwith increasing its modulus $\sqrt{2}$ times will yield AB,
Thus, $z_{1} = (z_{3} - z_{2}) e^{i \frac{π}{4}} \times \sqrt{2} + z_{2}$
$\Rightarrow z_{1} = (z_{3} - z_{2}) \times (1 + i) + z_{2}$
$\Rightarrow z_{1} = z_{3} + i (z_{3} - z_{2})$
$\Rightarrow (z_{1} - z_{3}) = i (z_{3} - z_{2})$
$\Rightarrow z_{2} = z_{3} + i (z_{1} - z_{3})$

Suggest Corrections

Similar questions

If $A (z_{1}), B (z_{2}), C (z_{3})$ be the vertices of triangle ABC in which $| A B C - ---- -$ = $\frac{π}{4}$ and $\frac{A B}{B C}$ = $\sqrt{2}$ then $z_{2}$ is equal to

A (z_{1}), B (z_{2}), C (z_{3})

are the vertices of the triangle

A B C

(in anticlockwise order). If

∠ A B C = \frac{π}{4}

and

A B = \sqrt{2} (B C)

, then

A (z_{1}), B (z_{2}), C (z_{3})

are the vertices of the triangle ABC (in anticlockwise order). If

∠ A B C = π / 4

and

A B = \sqrt{2} (B C)

Then prove that

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

Q. Statement 1: If

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

and

(z_{1}), B (z_{2}), C (z_{3})

are the vertices of

Δ A B C

, then one of the values of

a r g (\frac{(z_{2} + z_{3} - 2 z_{1})}{(z_{3} - z_{2})})

\frac{π}{2}

.
Statement 2: In equilateral triangle, orthocenter coincides with centroid.

Q. If

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

represent the vertices of an equilateral

Δ A B C

and

\sqrt{\frac{| z_{2} - z_{1} |^{2} + | z_{3} - z_{1} |^{2}}{| z_{2} + z_{3} - 2 z_{1} |^{2} + | z_{3} - z_{2} |^{2}}} = s i n θ

(where

0 < θ < π

) then

θ

Join BYJU'S Learning Program

If A(z1),B(z2),C(z3) are vertices of ΔABC in which ∠ABC=π4 and ABBC=√2, then find z2 in terms of z1 and z3

The correct option is C z2=z3+i(z1−z3)Rotating the side CB about the point B by 45o anticlockwise direction, alongwith increasing its modulus √2 times will yield AB,Thus, z1=(z3−z2)eiπ4×√2+z2⇒z1=(z3−z2)×(1+i)+z2⇒z1=z3+i(z3−z2)⇒(z1−z3)=i(z3−z2)⇒z2=z3+i(z1−z3)

If $A (z_{1}), B (z_{2}), C (z_{3})$ are vertices of $Δ A B C$ in which $∠ A B C = \frac{π}{4}$ and $\frac{A B}{B C} = \sqrt{2}$ , then find $z_{2}$ in terms of $z_{1}$ and $z_{3}$