If the complex numbers z1, z2 and z3 satisfying z1-z3/z2-z3=1-iroot3/2 then triangle is (a) an equilateral triangle (b) a right angles triangle (c) an acute angled triangle (d) an obtuse angled isosceles triangle. Please give me step by step solution.

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \frac{{\mathrm{z}}_1-{\mathrm{z}}_3}{{\mathrm{z}}_2-{\mathrm{z}}_3}=\frac{1-\mathrm{i}\sqrt{3}}{2} \\ \left|\frac{1-\mathrm{i}\sqrt{3}}{2}\right|=\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(-\frac{\sqrt{3}}{2}\right)}^2}=\sqrt{\frac{1}{4}+\frac{3}{4}}=1 \\ \arg \left(\frac{1-\mathrm{i}\sqrt{3}}{2}\right)={\tan }^{-\text{'}}\left(\frac{-{\displaystyle \frac{\sqrt{3}}{2}}}{{\displaystyle \frac{1}{2}}}\right)=-{\tan }^{-1}\left(\sqrt{3}\right)=-\frac{\mathrm{\pi }}{3} \\ \frac{{\mathrm{z}}_1-{\mathrm{z}}_3}{{\mathrm{z}}_2-{\mathrm{z}}_3}={\mathrm{e}}^{\mathrm{i}\left(-\frac{\mathrm{\pi }}{3}\right)} \\ \Rightarrow {\mathrm{z}}_1-{\mathrm{z}}_3=\left({\mathrm{z}}_2-{\mathrm{z}}_3\right){\mathrm{e}}^{\mathrm{i}\left(-\frac{\mathrm{\pi }}{3}\right)} ;\left(\mathrm{i}\right) \\ \mathrm{By} \mathrm{standard} \mathrm{equation} \mathrm{of} \mathrm{rotation} \mathrm{we} \mathrm{have}, \\ \frac{\mathrm{q}-\mathrm{p}}{\left|\mathrm{q}-\mathrm{p}\right|}=\frac{\mathrm{r}-\mathrm{p}}{\left|\mathrm{r}-\mathrm{p}\right|}{\mathrm{e}}^{\mathrm{i\theta }} , \mathrm{where} \mathrm{p},\mathrm{q},\mathrm{r} \mathrm{are} \mathrm{complex} \mathrm{numbers}. \\ \mathrm{Hence} \mathrm{comparing} \mathrm{it} \mathrm{with} \left(\mathrm{i}\right), \mathrm{we} \mathrm{get}, \\ \left|{\mathrm{z}}_1-{\mathrm{z}}_3\right|=\left|{\mathrm{z}}_2-{\mathrm{z}}_3\right| \\ \Rightarrow \mathrm{Side} \left({\mathrm{z}}_1-{\mathrm{z}}_3\right) \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{is} \mathrm{obtained} \mathrm{by} \mathrm{rotating} \mathrm{side} \left({\mathrm{z}}_2-{\mathrm{z}}_3\right) \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{in} \\ \mathrm{clockwise} \mathrm{direction} \mathrm{by} \mathrm{an} \mathrm{angle} \mathrm{of} \frac{\mathrm{\pi }}{3}, \mathrm{and} \mathrm{also} \mathrm{length} \mathrm{of} \mathrm{sides} \left({\mathrm{z}}_1-{\mathrm{z}}_3\right) \mathrm{and} \left({\mathrm{z}}_2-{\mathrm{z}}_3\right) \mathrm{are} \\ \mathrm{equal} \left[\mathrm{As} \left|{\mathrm{z}}_1-{\mathrm{z}}_3\right|=\left|{\mathrm{z}}_2-{\mathrm{z}}_3\right|\right]. \\ \mathbf{\Rightarrow }{\mathbf{z}}_{\mathbf{1}}\mathbf{,}{\mathbf{z}}_{\mathbf{2}}\mathbf{,}{\mathbf{z}}_{\mathbf{3}}\mathbf{ }\mathbf{represents}\mathbf{ }\mathbf{vertices}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{an}\mathbf{ }\mathbf{equilateral}\mathbf{ }\mathbf{triangle}\mathbf{.} \\ \mathbf{Hence}\mathbf{ }\mathbf{option}\left(\mathbf{a}\right)\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{correct}\mathbf{.} \end{gathered}$$

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