The area of the triangle with vertices \((0, 0)\), \((\cos\theta, \sin\theta)\) and \((-\sin\theta, \cos\theta)\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{2}\) square unit
The area of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is the absolute value of
\(\displaystyle \Delta = \frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix}\)
Here:
\(\displaystyle \Delta = \frac{1}{2}\begin{vmatrix}0 & 0 & 1\\ \cos\theta & \sin\theta & 1\\ -\sin\theta & \cos\theta & 1\end{vmatrix}\)
Expanding along the first row, only the third entry is non-zero:
\(\displaystyle \Delta = \frac{1}{2}\left[1 \cdot (\cos\theta \cdot \cos\theta - \sin\theta \cdot (-\sin\theta))\right]\)
\(\displaystyle \Delta = \frac{1}{2}(\cos^{2}\theta + \sin^{2}\theta)\)
\(\displaystyle \Delta = \frac{1}{2}\)
The area is \(\frac{1}{2}\) square unit for every value of \(\theta\).