Question

If $A,B,C$ are represented by $3+4i,5-2i,-1+16i$, then $A,B,C$ are
A. Collinear
B. Vertices of equilateral triangle
C. Vertices of isosceles triangle
D. Vertices of right-angled triangle

Answer 1

Hint: In this question, we are to find what the given points represent. To do this, the points are assumed as vertices of a triangle. Then, by calculating their side lengths and area, we are able to find their type.

Formula used: The complex number $(x,y)$ is represented by $x+iy$.
If $z=x+iy\in C$, then $x$ is called the real part and $y$ is called the imaginary part of $z$. These are represented by $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$ respectively.
$z=x+iy$ be a complex number such that $\left| z \right|=r$ and $\theta $ be the amplitude of $z$. So, $\cos \theta =\frac{x}{r},\sin \theta =\frac{b}{r}$
And we can write the magnitude as
 $\begin{align}
  & \left| z \right|=\left| x+iy \right| \\
 & \Rightarrow r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
\end{align}$
Thus, we can write
$z=x+iy=r\cos \theta +ir\sin \theta =r(\cos \theta +i\sin \theta )$
This is said to be the mod amplitude form or the polar form of $z$.
Where $\cos \theta +i\sin \theta $ is denoted by $cis\theta $ and the Euler’s formula is $\cos \theta +i\sin \theta ={{e}^{i\theta }}$

Complete step by step solution: It is given that, there are three points such as
$\begin{align}
  & A(3+4i) \\
 & B(5-2i) \\
 & C(-1+16i) \\
\end{align}$
If these are the vertices of a triangle, then their sides are $\left| AB \right|,\left| BC \right|,\left| CA \right|$
So,
$\begin{align}
  & \left| AB \right|=\left| 5-2i-3-4i \right| \\
 & \text{ }=\left| 2-6i \right| \\
 & \text{ }=\sqrt{{{2}^{2}}+{{6}^{2}}} \\
 & \text{ }=2\sqrt{10} \\
\end{align}$
$\begin{align}
  & \left| BC \right|=\left| -1+16i-5+2i \right| \\
 & \text{ }=\left| -6+18i \right| \\
 & \text{ }=\sqrt{{{6}^{2}}+{{18}^{2}}} \\
 & \text{ }=6\sqrt{10} \\
\end{align}$
$\begin{align}
  & \left| CA \right|=\left| 3+4i+1-16i \right| \\
 & \text{ }=\left| 4-12i \right| \\
 & \text{ }=\sqrt{{{4}^{2}}+{{12}^{2}}} \\
 & \text{ }=4\sqrt{10} \\
\end{align}$
Since the sides are $\left| AB \right|\ne \left| BC \right|\ne \left| CA \right|$, then the triangle is not an equilateral nor an isosceles.
Then,
$\begin{align}
  & {{\left| BC \right|}^{2}}={{\left| AB \right|}^{2}}+{{\left| CA \right|}^{2}} \\
 & \text{ }={{\left( 2\sqrt{10} \right)}^{2}}+{{\left( 4\sqrt{10} \right)}^{2}} \\
 & \text{ }=200 \\
 & \Rightarrow \left| BC \right|=10\sqrt{2}\ne 6\sqrt{10} \\
\end{align}$
Thus, the triangle is not a right-angled triangle.
So, the area of the triangle is
$\begin{align}
  & A=\frac{1}{2}\left| \begin{matrix}
   {{x}_{1}} & {{y}_{1}} & 1 \\
   {{x}_{2}} & {{y}_{2}} & 1 \\
   {{x}_{3}} & {{y}_{3}} & 1 \\
\end{matrix} \right| \\
 & \text{ }=\frac{1}{2}\left| \begin{matrix}
   3 & 4 & 1 \\
   5 & -2 & 1 \\
   -1 & 16 & 1 \\
\end{matrix} \right| \\
 & \text{ }=\frac{1}{2}\left[ 3(-2-16)-4(5+1)+1(80-2) \right] \\
 & \text{ }=\frac{1}{2}[-54-24-78] \\
 & \text{ }=0 \\
\end{align}$
Since the area of the vertices is zero, the given points are collinear.

Thus, Option (A) is correct.

Note: Here we need to assume the points to be vertices of a triangle, then we can use them and apply the properties of the triangle in order to know the type of the triangle. Since they are unequal and all the properties are failed, we can directly choose the remaining option, that is they are collinear. If the area of the three points is zero, then they lie on the same line. So, they are collinear.