JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Straight Lines

The area of the triangle with vertices $A(1,\alpha )$, $B(\alpha ,0)$ and $C(0,\alpha )$ is $\frac{1}{2}|1(0-\alpha )+\alpha (\alpha -\alpha )+0(\alpha -0)|=\frac{1}{2}|-\alpha |=\frac{|\alpha |}{2}$. Setting this equal to $4$ gives $|\alpha |=8$, so $\alpha =\pm 8$. For the three points $(\alpha ,-\alpha )$, $(-\alpha ,\alpha )$ and $({\alpha }^2,\beta )$, the first two have slope $\frac{\alpha -(-\alpha )}{-\alpha -\alpha }=-1$ and lie on the line $y=-x$. Collinearity forces the third point onto that same line, so $\beta =-{\alpha }^2$, and since $\alpha =\pm 8$ gives ${\alpha }^2=64$ in either case, $\beta =-64$.