Question:
If the vertices \(A\), \(B\) and \(C\) of an isosceles triangle \(ABC\) are respectively \(z_1\), \(z_2\) and \(z_3\), and if \(\angle C=90^\circ\), then
If the vertices \(A\), \(B\) and \(C\) of an isosceles triangle \(ABC\) are respectively \(z_1\), \(z_2\) and \(z_3\), and if \(\angle C=90^\circ\), then
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In complex geometry, if two sides are equal and perpendicular, then their corresponding complex vectors differ by multiplication with \(i\) or \(-i\).
Updated On: Jun 26, 2026
- \((z_1-z_2)=(z_1-z_3)(z_3-z_2)\)
- \((z_1-z_2)^2=(z_1-z_3)(z_3-z_2)\)
- \((z_1-z_2)^2=2(z_1-z_3)(z_3-z_2)\)
- \(z_1^2+z_2^2+z_3^2=z_1z_2z_3+2\)
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The Correct Option is C
Solution and Explanation
Step 1: Use the given geometrical condition.
Since \(ABC\) is an isosceles triangle and \[ \angle C=90^\circ, \] the equal sides are \[ CA=CB \] Therefore, \[ |z_1-z_3|=|z_2-z_3| \] Also, since the angle at \(C\) is \(90^\circ\), the two complex vectors \[ z_1-z_3 \] and \[ z_2-z_3 \] are perpendicular.
Step 2: Express perpendicular equal vectors in complex form.
If two vectors are perpendicular and have equal magnitude, then one is obtained from the other by multiplication by \(i\) or \(-i\).
So, we can write \[ z_1-z_3=i(z_2-z_3) \] Squaring both sides, \[ (z_1-z_3)^2=i^2(z_2-z_3)^2 \] \[ (z_1-z_3)^2=-(z_2-z_3)^2 \]
Step 3: Use the relation for the hypotenuse.
Now, \[ z_1-z_2=(z_1-z_3)+(z_3-z_2) \] Since \[ z_3-z_2=-(z_2-z_3), \] and the two perpendicular equal vectors form a right isosceles triangle, the square of the hypotenuse vector gives \[ (z_1-z_2)^2=2(z_1-z_3)(z_3-z_2) \]
Step 4: Final conclusion.
Hence, \[ \boxed{(z_1-z_2)^2=2(z_1-z_3)(z_3-z_2)} \]
Since \(ABC\) is an isosceles triangle and \[ \angle C=90^\circ, \] the equal sides are \[ CA=CB \] Therefore, \[ |z_1-z_3|=|z_2-z_3| \] Also, since the angle at \(C\) is \(90^\circ\), the two complex vectors \[ z_1-z_3 \] and \[ z_2-z_3 \] are perpendicular.
Step 2: Express perpendicular equal vectors in complex form.
If two vectors are perpendicular and have equal magnitude, then one is obtained from the other by multiplication by \(i\) or \(-i\).
So, we can write \[ z_1-z_3=i(z_2-z_3) \] Squaring both sides, \[ (z_1-z_3)^2=i^2(z_2-z_3)^2 \] \[ (z_1-z_3)^2=-(z_2-z_3)^2 \]
Step 3: Use the relation for the hypotenuse.
Now, \[ z_1-z_2=(z_1-z_3)+(z_3-z_2) \] Since \[ z_3-z_2=-(z_2-z_3), \] and the two perpendicular equal vectors form a right isosceles triangle, the square of the hypotenuse vector gives \[ (z_1-z_2)^2=2(z_1-z_3)(z_3-z_2) \]
Step 4: Final conclusion.
Hence, \[ \boxed{(z_1-z_2)^2=2(z_1-z_3)(z_3-z_2)} \]
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