Let the area of the triangle with vertices A (1, α), B (α, 0) and C (0, α) be 4 sq. units. If the point (α,-α),(-α, α) and (α2, β) are collinear, then β is equal to

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 point (α,-α),(-α, α) and (α2, β) are collinear, then β is equal to (A) 64 (B) -8 (C) -64 (D) 512

Tardigrade Admin · Accepted Answer

(1/2)|α 0 1 1 α 1 0 α 1|=± 4 α=± 8 Now given points (8,-8),(-8,8),(64, β) OR (8,8),(8,8),(64, β) are collinear ⇒ Slope =-1. β=-64

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

64

-8

-64

512

$\frac{1}{2} ∣ ∣ α 10 0 α α 111 ∣ ∣ = \pm 4$
$α = \pm 8$
Now given points $(8, - 8), (- 8, 8), (64, β)$
OR $(8, 8), (8, 8), (64, β)$
are collinear $\Rightarrow$ Slope $= - 1$ .
$β = - 64$

Q. Let the area of the triangle with vertices A(1,α), B(α,0) and C(0,α) be 4 sq. units. If the point (α,−α),(−α,α) and (α2,β) are collinear, then β is equal to

64

-8

-64

512

21​∣∣​α10​0αα​111​∣∣​=±4 α=±8 Now given points (8,−8),(−8,8),(64,β) OR (8,8),(8,8),(64,β) are collinear ⇒ Slope =−1. β=−64

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

$\frac{1}{2} ∣ ∣ α 10 0 α α 111 ∣ ∣ = \pm 4$
$α = \pm 8$
Now given points $(8, - 8), (- 8, 8), (64, β)$
OR $(8, 8), (8, 8), (64, β)$
are collinear $\Rightarrow$ Slope $= - 1$ .
$β = - 64$