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Byju's Answer

Standard XII

Mathematics

Combination

A point ' z '...

Question

A point ‘z’ is equidistant from three distinct points $z_{1}, z_{2}, z_{3}$ in argand plane,if $z, z_{1} a n d z_{2}$ are collinear, then $a r g (\frac{z_{3} - z_{1}}{z_{3} - z_{2}})$ will be (z_1,z_2,z_3 in anti-clockwise sense)

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Solution

The correct option is A
$| z - z_{1} | = | z - z_{2} | = | z - z_{3} |$
$∴$ z must be the midpoint of $z_{1}$ and $z_{2}$ $| z - z_{1} | = | z - z_{2} |$ and $z, z_{1}, z_{2}$ are collinear
$\Rightarrow$ z is circumcentre of $Δ$ formed by $z_{1}, z_{2} a n d z_{3}$
$∴ a r g (\frac{z_{3} - z_{1}}{z_{3} - z_{2}}) = \pm \frac{π}{2} (n e g l e c t i n g - v e)$

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A point ‘z’ is equidistant from three distinct points z1,z2,z3 in argand plane,if z,z1andz2 are collinear, then arg(z3−z1z3−z2) will be (z_1,z_2,z_3 in anti-clockwise sense)

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A point ‘z’ is equidistant from three distinct points $z_{1}, z_{2}, z_{3}$ in argand plane,if $z, z_{1} a n d z_{2}$ are collinear, then $a r g (\frac{z_{3} - z_{1}}{z_{3} - z_{2}})$ will be (z_1,z_2,z_3 in anti-clockwise sense)