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

Question

If the point $(\alpha,\alpha^2)$ lies inside the triangle formed by the lines $2x+3y-1=0,\hspace{1cm}x+2y-3=0,\hspace{1cm}5x-6y-1=0$.Then prove that $\alpha\in\left(\frac{-3}{2},-1\right)\cup\left({\frac{1}{2},1}\right)$.

I could not solve this problem, I dont know how to start with. Please help me. Thanks.

This is very related. See my answer in that link. – najayaz Oct 20 '15 at 06:22 — najayaz, Oct 20 '15 at 06:22
See this post for the more general case. – Hosein Rahnama Nov 21 '15 at 13:35 — Hosein Rahnama, Nov 21 '15 at 13:35

score 2 · Accepted Answer · answered Oct 20 '15 at 06:28

Hint:

Find the points of intersection of the triangle by simultaneously solving the equations. The points come out as $(-7,5), (\frac {5}{4},\frac{7}{8}), (\frac{1}{3},\frac{1}{9})$

Now use this property to ensure that the point $(\alpha,\alpha^2)$ lies on the same side of the line, as the third vertex by ensuring that on substitution of the coordinates in the equation of the line, you get the same sign for both the equations.

Hence, you'll get 3 equations and hence 3 solution sets for $\alpha$, and the required solution will be the intersection of the three solution sets.

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

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