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Equation of Circle in Complex Form

If z1, z2, ...

Question

If $z_{1}, z_{2}, z_{3}$ are three complex numbers such that $4 z_{1} - 7 z_{2} + 3 z_{3} = 0$ , then $z_{1}, z_{2}, z_{3}$ are

A

Vertices of a scalene traingle

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B

Vertices of a right triangle

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C

Points on a circle

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D

Collinear points

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Solution

The correct option is D Collinear points
$4 z_{1} - 7 z_{2} + 3 z_{3} = 0$
$7 z_{2} = 4 z_{1} + 3 z_{3}$
$z_{2} = \frac{4 z_{1} + 3 z_{3}}{7}$
Hence
$z_{2}, z_{3}, z_{1}$ are collinear points, where $z_{2}$ divides the line joining $z_{1}$ and $z_{3}$

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Similar questions

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

are three distinct complex numbers and

p, q, r

are three positive real numbers such that

\frac{p}{| z_{2} - z_{3} |} = \frac{q}{| z_{3} - z_{1} |} = \frac{r}{| z_{1} - z_{2} |}

\frac{p^{2}}{z_{2} - z_{3}} + \frac{q^{2}}{z_{3} - z_{1}} + \frac{r^{2}}{z_{1} - z_{2}} = 0

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

are three complex numbers whose moduli are

a, b, c

respectively and are such that

∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣ = 0

z_{1} \neq z_{2}

, then prove that

arg {(\frac{z_{3} - z_{1}}{z_{2} - z_{1}})}^{2} = arg (\frac{z_{3}}{z_{2}})

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

are complex numbers such that

| z_{1} | = | z_{2} | = | z_{3} | = ∣ ∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} ∣ ∣ ∣ = 1

| z_{1} + z_{2} + z_{3} |

z_{1}

z_{2}

z_{3}

are three distinct complex numbers such that

\frac{1}{∣ z_{1} - z_{2} ∣} = \frac{3}{∣ z_{2} - z_{3} ∣} = \frac{5}{∣ z_{1} - z_{3} ∣}

, then the value of

\frac{1}{z_{1} - z_{2}} + \frac{9}{z_{2} - z_{3}} + \frac{25}{z_{3} - z_{1}}

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

be the complex numbers such that

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

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

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

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