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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Given, the area of the triangle with vertices $A (1, α), B (α, 0)$ and $C (0, α)$ be $4$ sq. units

Now applying formula for area we get,

$\frac{1}{2} |\begin{array}{l} α & 0 & 1 \\ 1 & α & 1 \\ 0 & α & 1 \end{array}| = \pm 4$

$\Rightarrow α = \pm 8$

Now given points will be $(8, - 8), (- 8, 8), (64, β)$

OR $(- 8, 8), (8, - 8), (64, β)$

Given above points are collinear, so their slope will be equal, so $slope= \frac{8 + 8}{- 8 - 8} = - 1$

$\Rightarrow \frac{β - 8}{64 + 8} = - 1$

$\Rightarrow β = - 64$

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