Condition for Points in Complex Plane to form Isosceles Triangle
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Theorem
Let $A = z_1 = x_1 + i y_1$, $B = z_2 = x_2 + i y_2$ and $C = z_3 = x_3 + i y_3$ represent on the complex plane the vertices of a triangle.
Then $\triangle ABC$ is isosceles, where $A$ is the apex, if and only if:
- ${x_2}^2 + {y_2}^2 - 2 \paren {x_1 x_2 + y_1 y_2} = {x_3}^2 + {y_3}^2 - 2 \paren {x_1 x_3 + y_1 y_3}$
Proof
By definition of isosceles triangle:
- $\triangle ABC$ is isosceles, where $A$ is the apex, if and only if $AB = AC$.
Hence:
\(\ds \cmod {z_1 - z_2}\) | \(=\) | \(\ds \cmod {z_1 - z_3}\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \cmod {x_1 + i y_1 - x_2 + i y_2}^2\) | \(=\) | \(\ds \cmod {x_1 + i y_1 - x_3 + i y_3}^2\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2\) | \(=\) | \(\ds \paren {x_1 - x_3}^2 + \paren {y_1 - y_3}^2\) | Definition of Complex Modulus | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds {x_1}^2 - 2 x_1 x_2 + {x_2}^2 + {y_1}^2 - 2 y_1 y_2 + {y_2}^2\) | \(=\) | \(\ds {x_1}^2 - 2 x_1 x_3 + {x_3}^2 + {y_1}^2 - 2 y_1 y_3 + {y_3}^2\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds {x_2}^2 + {y_2}^2 - 2 \paren {x_1 x_2 + y_1 y_2}\) | \(=\) | \(\ds {x_3}^2 + {y_3}^2 - 2 \paren {x_1 x_3 + y_1 y_3}\) |
$\blacksquare$
Examples
Vertices at $1 + 2 i$, $4 - 2 i$, $1 - 6 i$
Let $A = z_1 = 1 + 2 i$, $B = z_2 = 4 - 2 i$ and $C = z_3 = 1 - 6 i$ represent on the complex plane the vertices of a triangle.