JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
Let the area of the triangle with vertices \(A (1, \alpha)\), \(B (\alpha, 0)\) and \(C (0, \alpha)\) be \(4\, sq.\) units. If the point \((\alpha,-\alpha),(-\alpha, \alpha)\) and \(\left(\alpha^{2}, \beta\right)\) are collinear, then \(\beta\) is equal to
- A \(64\)
- B \(-8\)
- C \(-64\)
- D \(512\)
Answer & Solution
Correct Answer
(C) \(-64\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{2}\left|\begin{array}{lll}\alpha & 0 & 1 \\ 1 & \alpha & 1 \\ 0 & \alpha & 1\end{array}\right|=\pm 4\) \(\alpha=\pm 8\) Now given points \((8,-8),(-8,8),(64, \beta)\) \(OR (-8,8),(8,-8),(64, \beta)\) are collinear \(\Rightarrow\) Slope \(=-1\). \(\beta=-64\)
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