Question

If the point $(\alpha ,{\alpha }^2)$ lies inside the triangle formed by the lines $2x+3y-1=0,x+2y-3=0$ and $5x-6y-1=0$, then

• A. $-\frac{1}{2}<\alpha <\frac{1}{2}$
• B. $-\frac{3}{2}<\alpha <-1$
• C. $-\frac{3}{2}<\alpha <\frac{3}{2}$
• D. $-\frac{3}{2}<\alpha <-\frac{1}{2}$

Answer 1

The correct option is B $-\frac{3}{2}<\alpha <-1$

At $(0,0)$ $2x+3y-1=-1<0$
$\Rightarrow 2\alpha +3{\alpha }^2-1>0$ ...(1)
At $(0,0)$ $x+2y-3=-3<0$
$\Rightarrow \alpha +3{2\alpha }^2-3<0$ ...(2)
From (1) and (2)
$-\frac{3}{2}<\alpha <-1$