Question

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

Answer 1

Given lines are
$2 x+3 y-1=0 ....$ (i)
$x+2 y-3=0 .....$ (ii)
$5 x-6 y-1=0.....$(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 $\triangle A B C$. Since, $A$ and $P$ are on the same side of $5 x-6 y-1=0$, 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
$2 x+3 y-1=0$, 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$
$\Rightarrow (\alpha+1)\left(\alpha-\frac{1}{3}\right) > 0$
$\Rightarrow \alpha < -1 \cup \alpha > 1 / 3 \ldots$ (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 $.