Question

If $a,b,candu,v,w$ are complex numbers representing the vertices of two triangles such that $c=(1-r)a+rbandw=(1-r)u+rv$, 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

Step1. Finding sides of the triangle

Vertices of $△ABC$ are represented by $a,b\&c$ then, sides of the $△ABC$

$\begin{array}{rcl}AB& =& b-a\\ BC& =& c-b\\ & =& (1-r)a+rb-b\\ & =& (a-b)(1-r)\\ CA& =& a-c\\ & =& a-(1+r)a-rb\\ & =& r(a-b)\end{array}$

Vertices of $△UVW$ are represented by sides of the $△UVW$

$\begin{array}{rcl}UV& =& u-w\\ VW& =& w-v\\ & =& (1-r)u+rv-v\\ & =& u-v-r(u-v)\\ & =& (u-v)(1-r)\\ WU& =& u-w\\ & =& u-(1-r)u-rv\\ & =& r(u-v)\end{array}$

Step2. Find the Relation between the triangle.

Now,

$\begin{array}{rcl}\frac{AB}{UV}& =& \frac{(b-a)}{(v-u)}\\ & =& \frac{(a-b)}{(u-v)}\dots \left(1\right)\\ \frac{BC}{VW}& =& \frac{(a-b)(1-r)}{(u-v)(1-r)}=\frac{(a-b)}{(u-v)}\dots \left(2\right)\\ \frac{CA}{WU}& =& \frac{r(a-b)}{r(u-v)}=\frac{(a-b)}{(u-v)}\dots \left(3\right)\\ & & Comparing\left(1\right),\left(2\right)and\left(3\right)\\ \frac{AB}{UV}& =& \frac{BC}{VW}=\frac{CA}{WU}\end{array}$

So, $△ABC$ similar to $△UVW$

Hence, the correct option is $\mathbf{\left(}\mathit{B}\mathbf{\right)}\mathbf{.}$