Values of $\alpha$? Determine values of $\alpha$ for which the point $(\alpha,\alpha^2)$ lies inside the triangle formed by the lines
$ax+by+c=0$
$dx+ey+f=0$
$gx+hy+i=0$
Any idea on how to go about this one? Any help is appreciated thanks!
 A: Every line delimits two half-planes such that the LHS of its equation keeps a constant sign (it is $0$ on the edge). The triangle is formed by the intersection of three of theses half-planes, i.e. the system of inequations where $=$ is replaced by $\le$ or $\ge$ depending on the appropriate signature.
You can determine the signs by computing the pairwise line intersections (giving the triangle vertices), and plugging them in the equation of the third lines.
Then plugging $(\alpha,\alpha^2)$, you must obtain the same signs.

For one of the vertices,
$$-g\frac{\begin{vmatrix}c&b\\f&e\end{vmatrix}}{\begin{vmatrix}a&b\\d&e\end{vmatrix}}-h\frac{\begin{vmatrix}a&c\\d&f\end{vmatrix}}{\begin{vmatrix}a&b\\d&e\end{vmatrix}}+i\text{, and }g\alpha+h\alpha^2+i$$
must have the same sign. Repeat for the two others.
Note that the first expression is the determinant of the $3\times3$ matrix formed by the $9$ coefficients, over some minor. If I am right, it is the inverse of the third element of the third column of the inverse matrix.
So the condition is
$$\left(M^{-1}\right)_{13}(a\alpha+b\alpha^2+c)>0\land \left(M^{-1}\right)_{23}(d\alpha+e\alpha^2+f)>0\land \left(M^{-1}\right)_{33}(g\alpha+h\alpha^2+i)>0.$$
