ExamBro
ExamBro
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

  1. A \(64\)
  2. B \(-8\)
  3. C \(-64\)
  4. D \(512\)
Verified Solution

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 \(=-1\). \(\beta=-64\)
From JEE Mains
Explore more questions on app