Question

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

• A. 64
• B. -8
• C. -64
• D. 512

Answer 1

Answer: (C)

-64

Answer 2

Vertices of $ΔABC$ are $∵ A(1, α), B(α, 0) \;and C(0, α)$

The Area of $ΔABC = 4$

$∴ |\frac{1}{2}| \begin {vmatrix}1 &α& 1\\\ α &0 &1 \\\0 &α &1 \end{vmatrix}=4$

$⇒ | 1 (1- α)-α(α)+α^2|=8$

$⇒ α = +- 8$

Then, $(α, –α), (–α, α) \;and\; (α^2, β)$

$∴ \begin{vmatrix}8& -8& 1\\\ -8& 8& 1\\\ 64 &β &1 \end{vmatrix} = 0 = \begin{vmatrix} -8& 8 &1 \\\8 &-8 &1 \\\64 &α &1 \end{vmatrix}$\

$⇒ 8(8- β)+8(-8-64)+1(-8β-8 ×64) = 0$
$⇒ 8- β - 72 - β -64 = 0$
$⇒ β = -64$

Hence, the correct option is (C): $-64$