Question
The complex number $${z_1},{z_2}\,{\text{and }}{z_3}$$ satisfying $$\frac{{{z_1} - {z_3}}}{{{z_2} - {z_3}}} = \frac{{1 - i\sqrt 3 }}{2}$$ are the vertices of a triangle which is
A.
of area zero
B.
right - angled isosceles
C.
equilateral
D.
obtuse - angled isosceles
Answer :
equilateral
Solution :
$$\eqalign{
& \frac{{{z_1} - {z_3}}}{{{z_2} - {z_3}}} = \frac{{1 - i\sqrt 3 }}{2} \cr
& \Rightarrow \,\,\arg \left( {\frac{{{z_1} - {z_3}}}{{{z_{_2}} - {z_3}}}} \right) = \arg \left( {\frac{{1 - i\sqrt 3 }}{2}} \right) \cr
& \Rightarrow \,\,\arg \left( {\cos \left( { - \frac{\pi }{3}} \right) + i\sin \left( { - \frac{\pi }{3}} \right)} \right) \cr
& \Rightarrow \,\,{\text{angle}}\,{\text{between}}\,{\text{ }}{z_1} - {z_3}\,{\text{and}}\,{z_2} - {z_3}{\text{ is }}{60^ \circ }. \cr
& {\text{and }}\left| {\frac{{{z_1} - {z_3}}}{{{z_2} - {z_3}}}} \right| = \left| {\frac{{1 - i\sqrt 3 }}{2}} \right| \cr
& \Rightarrow \,\,\left| {\frac{{{z_1} - {z_3}}}{{{z_2} - {z_3}}}} \right| = 1 \cr} $$
$$ \Rightarrow \,\,\left| {{z_1} - {z_3}} \right| = \left| {{z_2} - {z_3}} \right|$$ NOTE THIS STEP
⇒ The $$\Delta $$ with vertices $${z_1},{z_2}{\text{ and }}{z_3}$$ is isosceles with
vertical $$\angle {60^ \circ }.$$ Hence rest of the two angles should
also be 60° each.
⇒ Req. $$\Delta $$ is an equilateral $$\Delta .$$