Let the area of the triangle with vertices A (1, α), B (α, 0) and C(0, α) be 4 sq. units. If the points (α, - α), (- α, α) and ( α ^ 2 , β) are collinear, then β is equal to

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Correct answer: C. - 64

Given, the area of the triangle with vertices A

\left(1, \alpha\right), B \left(\alpha, 0\right)

and

C\left(0, \alpha\right)

be

4

sq. units

Now applying formula for area we get,

$\dfrac{1}{2}\left|$ \begin{cases}\alpha & 0 & 1 \\ 1 & \alpha & 1 \\ 0 & \alpha & 1\end{cases} $\right| = \pm 4$

$⇒ \alpha = \pm 8$

Now given points will be \left(8, - 8\right), \left(- 8, 8\right), \left(64, \beta\right) $OR \left(- 8, 8\right), \left(8, - 8\right), \left(64, \beta\right)Given above points are collinear, so their slope will be equal, so \text{slope=}\dfrac{8 + 8}{- 8 - 8} =$ - 1 $$ ⇒ $\dfrac{\beta - 8}{64 + 8} =$ - 1 $$ ⇒ \beta = - 64 $$

Let the area of the triangle with vertices A (1, α), B (α, 0) and...

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