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The area of the triangle with vertices z, iz, z+iz is 50, then $\left| z \right|=$
  & A.0 \\
 & B.5 \\
 & C.10 \\
 & D.15 \\

Last updated date: 13th Jun 2024
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Hint: In this question, we need to find value of $\left| z \right|$ if the area of the triangle has vertices z, iz, z+iz is 50. For this, we will first suppose vertices as A, B and C. Then, we will find length of sides of triangle using property that, $AB=\left| \text{vertexB}-\text{vertexA} \right|$. We will suppose z as x+iy and evaluate AB, AC and BC. Using them, we will derive formula of area of triangle and put given value to find value of $\left| z \right|$. We will use the property of isosceles triangles that the line drawn from the common point of equal sides towards the midpoint of the third side will be perpendicular to the third side. Area of a triangle is given by $\dfrac{1}{2}\times \text{base}\times \text{height}$. Midpoint of any line AB is given by $\dfrac{A+B}{2}$. Magnitude x+iy is given by $\sqrt{{{x}^{2}}+{{y}^{2}}}$.

Complete step by step answer:
Here, we are given vertices of a triangle as z, iz, z+iz. Let us suppose A as z, B as iz and C as z+iz.
Let us suppose z = x+iy.
Let us find the magnitude of sides of triangle ABC. Side AB will be given by,
  & \Rightarrow \left| \text{vertexB}-\text{vertexA} \right| \\
 & \Rightarrow \left| iz-z \right| \\
 & \Rightarrow \left| i\left( x+iy \right)-\left( x+iy \right) \right| \\
 & \Rightarrow \left| ix+{{i}^{2}}y-x-iy \right| \\
 & \Rightarrow \left| ix-y-x-iy \right| \\
 & \Rightarrow \left| -x-y+i\left( x-y \right) \right| \\
So, magnitude is given by $\sqrt{{{\left( -x-y \right)}^{2}}+{{\left( x-y \right)}^{2}}}\Rightarrow AB=\sqrt{{{\left( x+y \right)}^{2}}+{{\left( x-y \right)}^{2}}}$.
Now, side BC will be,
  & \Rightarrow \left| z+iz-iz \right| \\
 & \Rightarrow \left| z \right| \\
 & \Rightarrow \left| x+iy \right| \\
 & BC=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
Side AC will be,
  & \Rightarrow \left| z+iz-z \right| \\
 & \Rightarrow \left| iz \right| \\
 & \Rightarrow \left| i\left( x+iy \right) \right| \\
 & \Rightarrow \left| ix+{{i}^{2}}y \right| \\
 & \Rightarrow \left| ix-y \right| \\
 & \Rightarrow \sqrt{{{\left( x \right)}^{2}}+{{\left( -y \right)}^{2}}} \\
 & AC=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
As we can see, AC = BC, therefore triangle ABC is an isosceles triangle.
We know that, in an isosceles triangle, a line drawn from a common point of equal side towards the midpoint of the third side will be perpendicular to the third side. So, let us suppose midpoint of AB as P. Therefore, CP is perpendicular to AB.
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Now, let us first point P. Since, P is the midpoint of A and B so P will be equal to $\dfrac{A+B}{2}$. Hence, $P=\dfrac{z+iz}{2}$.
So, line PC becomes equal to,
  & \Rightarrow \left| z+iz-\dfrac{z+iz}{2} \right| \\
 & \Rightarrow \left| \dfrac{2z+2iz-z-iz}{2} \right| \\
 & \Rightarrow \left| \dfrac{z+iz}{2} \right| \\
Now, we know that, $AB=\left| iz-z \right|$ or we can write it as $AB=\left| z-iz \right|$.
Since, the area of any triangle is $\dfrac{1}{2}\times \text{base}\times \text{perpendicular}$.
Here, base is AB and perpendicular is PC so,
  & \text{Area}=\dfrac{1}{2}\times \left| z-iz \right|\left| \dfrac{z+iz}{2} \right| \\
 & \Rightarrow \dfrac{1}{4}\left| z-iz \right|\left| z+iz \right| \\
Using $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$,
  & \Rightarrow \dfrac{1}{4}\left| {{z}^{2}}-{{\left( iz \right)}^{2}} \right| \\
 & \Rightarrow \dfrac{1}{4}\left| {{z}^{2}}-{{i}^{2}}{{z}^{2}} \right| \\
 & \Rightarrow \dfrac{1}{4}\left| {{z}^{2}}+{{z}^{2}} \right| \\
 & \Rightarrow \dfrac{1}{2}\left| {{z}^{2}} \right| \\
 & \Rightarrow \dfrac{1}{2}{{\left| z \right|}^{2}} \\
Hence, area of triangle is $\dfrac{{{\left| z \right|}^{2}}}{2}$.
Given area of triangle is 50, so, $\dfrac{{{\left| z \right|}^{2}}}{2}=50$.
Cross multiplying, we get: ${{\left| z \right|}^{2}}=100$.
Taking square root both sides, we get: $\left| z \right|=\pm 10$.
But, we are taking magnitude only, so $\left| z \right|=10$.

So, the correct answer is “Option C”.

Note: Students should note that, we have taken ${{i}^{2}}=1$ as $i=\sqrt{-1}\text{ and }{{\left( \sqrt{-1} \right)}^{2}}=-1$. While calculating the magnitude of complex numbers, we take squares of only x and y and do not include 'i'. Students can also remember the formula of area of triangle with these three vectors as $\dfrac{1}{2}{{\left| z \right|}^{2}}$ and apply it directly to solve for $\left| z \right|$.