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

Explanation

The equations of sides of a triangle ABC are given by
`2+3y-1=0,x+2y-3=0and 5x-6y-1=0.`
The coordinates of the point of intersection (vertices) taken two by two are `A((5)/(4),(7)/(8)),B(-7,5)and C((1)/(3),(1)/(9)).`
Since `P(alpha,alpha^(2))` lies inside the `DeltaABC,` then

(i) A and P must lie on the same side of BC
(ii) P and B must lie on the same side of CA and C & P must lie on the same side of AB, hence `((5)/(2)+(21)/(8)-1)(2alpha+3alpha^(2)-1)gt0implies3alpha^(2)+2alpha-1gt0`
`implies(alpha+1)(alpha-(1)/(3))gt0`
`implies alpha in (-oo,-1)uu((1)/(3)=oo)...(i)`

`and(-35-30-1)(5alpha-6alpha^(2)-1)gt0`
`implies5alpha-6alpha^(2)-1lt0`
`implies6alpha^(2)-5alpha+1gt0`

`implies(alpha-(1)/(2))(alpha-(1)/(3))gt0`
`impliesalphain(-oo,(1)/(3))uu((1)/(2),oo)" "...(ii)`
Also `((1)/(3)+(2)/(9)-3)(alpha+2alpha^(2)-3)gt0`
`implies2alpha^(2)+alpha-3lt0`
`implies(2alpha+3)(alpha-1)lt0`

`impliesa in (-(3)/(2),1)" "...(iii)`
From (i),(ii) and (iii) we get,
`alphain(-(3)/(2)-1)uu((1)/(2),1).`