Question

If a,b,c and u,v,w are complex numbers representing the vertices of two triangles such that c=(1-r)a+rb and (1-r)u+rv , where r is a complex number, then the two triangles

• A. Have the same area
• B. Are similar
• C. Are congruent
• D. None of these
  
  

Answer 1

The correct option is B Are similar

Let the complex number a,b,c and u,v,wrepresent the vertices A,B,C and D,E,Fof the two triangle ABCand DEFrespectively
Put $b-a=r_1{e}^{i{\theta }_l}$
$c-a=r_2{e}^{i{\theta }_2}$
$v-u={\rho }_1{e}^{i{\varphi }_l}$,$w-u={\rho }_2{e}^{i{\varphi }_2}andr=\lambda e^{ia}$
Substituting these values in the given relations
$c-a=r(b-a)andw-u=(v-u)r,$ we have
$r_2{e}^{i{\theta }_2}=\lambda e^{i\alpha }{r}_1{e}^{i{\theta }_1}=\lambda r_1{e}^{i(\alpha +thet{a}_l)}$ .....(i)
and ${\rho }_2{e}^{i{\Phi }_2}={\rho }_1{e}^{i{\Phi }_1}\lambda e^{i\alpha }=\left(\lambda {\rho }_1\right)e^{i({\Phi }_1+\alpha )}$ .......(ii)
Equating moduli and arguments of the complex numbers on both sides (i),
we get $r_2=\lambda r_1,{\theta }_2=\alpha +{\theta }_1$
i.e., $AC=\lambda AB$ and $\angle CAB={\theta }_2-{\theta }_1=\alpha$
Similarly from (ii), we shall get $DF=\lambda DE$ and $\angle FDE={\Phi }_2-{\Phi }_1=\alpha$
Thus we get $\frac{AC}{DF}=\frac{AB}{DE}$ and $\angle CAB=\angle FDE$
Hence the triangle ABC and DEF are similar.