MCQ
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 $w = (1 - r)u + rv$, where $r$ is a complex number, then the two triangles
  • A
    Have the same area
  • Are similar
  • C
    Are congruent
  • D
    None of these

Answer

Correct option: B.
Are similar
b
(b)Let the complex number $a,b,c$and $u,v,w$ represent the vertices $A,B,C$and $D,E,F$ of the two triangle $ABC$ and $DEF$ respectively.
Put $b - a = {r_1}{e^{i{\theta _1}}}$
$c - a = {r_2}{e^{i{\theta _2}}}$
$v - u = {\rho _1}{e^{i{\phi _1}}},w - u = {\rho _2}{e^{i{\phi _2}}}$and $r = \lambda {e^{i\alpha }}$
Substituting these values in the given relations
$c - a = r(b - a)$and $w - u = (v - u)r,$ we have
${r_2}{e^{i{\theta _2}}} = \lambda {e^{i\alpha }}{r_1}{e^{i{\theta _1}}} = \lambda {r_1}{e^{i(\alpha + {\theta _1})}}$ .......$(i)$
and ${\rho _2}{e^{i{\phi _2}}} = {\rho _1}{e^{i{\phi _1}}}\lambda {e^{i\alpha }} = (\lambda {\rho _1}){e^{i({\phi _1} + \alpha )}}$ .......$(ii)$
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.

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