# Prove that the triangle formed by the points representing the complex numbers $(10+8i),(-2+4i)$ and $(-11+31i)$ on the Argand plane is right angled.

Toolbox:
• If $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ are represented by the points $A(x_1,y_1),B(x_2,y_2)$ on the Argand plane,their $AB=\mid z_1-z_2\mid$ that on $OA=\mid z_1\mid,OB=\mid z_2\mid$
Step 1:
Let $z_1=10+8i,z_2=-2+4i,z_3=-11+31i$ be represented by the points $A,B,C$ is the Argand plane.
$AB=\mid z_1-z_2\mid=\mid (10+2)+(8-4)i\mid$
$\qquad\quad\qquad\;\;=\mid 12+4i\mid$
$\qquad\quad\qquad\;\;=\sqrt{144+16}$
$\qquad\quad\qquad\;\;=\sqrt{160}$
$\qquad\quad\qquad\;\;=4\sqrt{10}$
Step 2:
$BC=\mid z_2-z_3\mid=\mid (-2+11)+(4-31)i\mid$
$\qquad\quad\qquad\;\;=\mid 9-27i\mid$
$\qquad\quad\qquad\;\;=\sqrt{9^2+27^2}$
$\qquad\quad\qquad\;\;=\sqrt{810}$
$\qquad\quad\qquad\;\;=9\sqrt{10}$
Step 3:
$CA=\mid z_3-z_1\mid=\mid (-11-10)+(31-8)i\mid$
$\qquad\quad\qquad\;\;=\mid -21+23i\mid$
$\qquad\quad\qquad\;\;=\sqrt{21^2+23^2}$
$\qquad\quad\qquad\;\;=\sqrt{970}$
Step 4:
Now $AB^2+BC^2=160+810=970$
$AC^2=970$
$AB^2+BC^2=AC^2\Rightarrow ABC$ is a right triangle,right angled at B.