Question

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

Answer 1

\(\because \) \(A(1, \alpha), B(\alpha , 0)\) and \(C(0, \alpha)\) are the vertices of \(\Delta\) and area of \(\Delta ABC\) = 4

Answer 2

https://www.sarthaks.com/?qa=blob&qa_blobid=1475664176304034445

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

Triangle with vertices A(1 , α) , B(α , 0) and C(0 , α) 

Shifting the Orgin to (0, α) then new vertices will be 

A’ , B’ and C’ A’ (1, 0) , B’ (α, - α) and C’(0, 0)

Area of Triange A’ B’ C’ =

4 = 1/2 |-α| therefore α = +/- 8

Now 3 points are collinear (α , - α) , (- α , α) and (α² , β)

(8 , - 8) , (- 8 , 8) and (64 , β) ... area formed by collinear points is 0. 

Shifting Origin to (8, 8) then point will be (0 , − 16) , (- 16 , 0) and (56 , β-8) Area formed by above points

0 = 1/2 | 16(-16) + (-16)( β-8) + 56(-16) | 

0 = | 16 + ( β-8 ) + 56 | = | β+ 64 | 

β = - 64