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

Given lines are `2x+3y-1=0`…….`(i)`
`x+2y-3=0`……..`(ii)`
`5x-6y-1=0`……….`(iii)`

On solving Eqs. `(i),(ii)` and `(iii)` , we get the vertices of a triangle are `A(-7,5)`, `B((1)/(3),(1)/(9))` and `C((5)/(4),(7)/(8))`.
Let `P(alpha,alpha^(2))` be a point inside the `DeltaABC`. Since, `A` and `P` are on the same side of `5x-6y-1=0`, both `5(-7)-6(5)-1` and `5alpha-6alpha^(2)-1` must have the same sign, therefore
`5alpha-6alpha^(2)-1 lt 0`
`implies6alpha^(2)-5alpha+1 gt 0`
`implies(3alpha-1)(2alpha-1) gt 0`
`impliesalpha lt (1)/(3)` or `alpha gt (1)/(2)`..........`(iv)`
Also, since `P(alpha,alpha^(2))` and `C((5)/(4),(7)/(8))` lie on the same side of `2x+3y-1=0`, therefore both `2((5)/(4))+3((7)/(8))-1` and `2alpha+3alpha^(2)-1` must have the same sign.
Therefore, `2alpha+3alpha^(2)-1 gt 0`
`implies(alpha+1)(alpha-(1)/(3)) gt 0`
`implies alpha lt -1 uu alpha gt 1//3`.......`(v)`
and lastly `((1)/(3),(1)/(9))` and `P(alpha,alpha^(2))` lie on the same side of the line therefore, `(1)/(3)+2((1)/(9))-3` and `alpha+2alpha^(2)-3` must have the same sign.
Therefore, `2alpha^(2)+alpha-3 lt 0`
`implies2alpha(alpha-1)+3(alpha-1) lt 0`
`implies (2alpha+3)(alpha-1) lt 0 implies -(2)/(3) lt alpha lt 1`
On solving Eqs. `(i),(ii)` and `(iii)`, we get the common answer is `-(3)/(2) lt alpha lt -1 uu (1)/(2) lt alpha lt 1`.