Determine all values of $\alpha$ for which the point $\left(\alpha, \alpha^{2}\right)$ lies inside t

Question

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

Professor PC Thomas Classes · Answer

The points of intersection of given lines are\ $P\left(\frac{1}{3}, \frac{1}{9} ight) ; Q(-7,5) ; R\left(\frac{5}{4}, \frac{7}{8} ight)$\ If ( $\alpha, \alpha^{2}$ ) lies inside the triangle formed by the given lines.\ $\left(\frac{1}{3}, \frac{1}{9} ight)$ and $\left(\alpha, \alpha^{2} ight)$ lie on the same side of\ the line $x+2 y-3=0$\ Similarly $\left(\frac{5}{4}, \frac{7}{8} ight)$ and $\left(\alpha, \alpha^{2} ight)$ lie on the same\ side of the line $2 x+3 y-1=0$\ $(-7,5) ;\left(\alpha, \alpha^{2} ight)$ lie on the same side of the line $$\begin{equation*}5 x-6 y-1=0 ag{2} \end{equation*}$$ In case of (1) $\alpha+2 \alpha^{2}-3<0$ $$\begin{equation*}\Rightarrow \quad-\frac{3}{2}<\alpha<1 ag{I} \end{equation*}$$ In case of (2) $2 \alpha+3 \alpha^{2}-1>0$\ $\Rightarrow$ Either $\alpha>\frac{1}{3}$ or $\alpha<-1$\ In case of (3)\ $5 \alpha-6 \alpha^{2}-1<0 \Rightarrow \alpha>\frac{1}{2}$ or $\alpha<\frac{1}{3}$\ From (I), (II) and (III) we have either\ $-\frac{3}{2}<\alpha<-1$ or $\frac{1}{2}<\alpha<1$\ $\alpha \in\left(-\frac{3}{2},-1 ight) \cup\left(\frac{1}{2}, 1 ight)$.\