Question

Determine all values of $\alpha$ for which the point $(\alpha ,{\alpha }^2)$ lies inside the triangles formed by the lines $2x+3y-1=0,x+2y-3=0,5x-6y-1=0$

Answer 1

Given lines are
$2x+3y-1=\mathrm{0....}$ (i)
$x+2y-3=\mathrm{0.....}$ (ii)
$5x-6y-1=\mathrm{0.....}$(iii)

On solving Eqs. (i), (ii) and (iii), we get the vertices of a triangle are $A(-7,5),B\left(\frac{1}{3},\frac{1}{9}\right)$ and $C\left(\frac{5}{4},\frac{7}{8}\right)$.
Let $P\left(\alpha ,{\alpha }^2\right)$ be a point inside the $△ABC$. Since, $A$ and $P$ are on the same side of $5x-6y-1=0$, both $5(-7)-6(5)-1$ and $5\alpha -6{\alpha }^2-1$ must have the same sign, therefore
$5\alpha -6{\alpha }^2-1<0$
$\Rightarrow 6{\alpha }^2-5\alpha +1>0$
$\Rightarrow (3\alpha -1)(2\alpha -1)>0$
$\Rightarrow \alpha <\frac{1}{3}$ or $\alpha >\frac{1}{2}....$(iv)
Also, since $P\left(\alpha ,{\alpha }^2\right)$ and $C\left(\frac{5}{4},\frac{7}{8}\right)$ lie on the same side of
$2x+3y-1=0$, therefore both $2\left(\frac{5}{4}\right)+3\left(\frac{7}{8}\right)-1$ and $2\alpha +3{\alpha }^2-1$ must have the same sign.
Therefore, $2\alpha +3{\alpha }^2-1>0$
$\Rightarrow (\alpha +1)\left(\alpha -\frac{1}{3}\right)>0$
$\Rightarrow \alpha <-1\cup \alpha >1/3\dots$ (v)
and lastly $\left(\frac{1}{3},\frac{1}{9}\right)$ and $P\left(\alpha ,{\alpha }^2\right)$ lie on the same side of the line therefore, $\frac{1}{3}+2\left(\frac{1}{9}\right)-3$ and $\alpha +2{\alpha }^2-3$ must have the same sign.
Therefore, $2{\alpha }^2+\alpha -3<0$
$\Rightarrow 2\alpha (\alpha -1)+3(\alpha -1)<0$
$\Rightarrow (2\alpha +3)(\alpha -1)<0\Rightarrow -\frac{2}{3}<\alpha <1$
On solving Eqs. (i), (ii) and (iii), we get the common answer is
$-\frac{3}{2}<\alpha <-1\cup \frac{1}{2}<\alpha <1$.