Question

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

Answer 1

Solving the equations in pairs, we get the vertices of the triangles as $A(5/4,7/8),B(1/3,1/9)<C(-7,5)$ if the given lines be $BC,CA$ and $AB$ respectively. Putting the coordinates in the opposite sides, we get the results $+,-,-$ respectively. hence we write the equations by multiplying by - sign so that all the results are $+,+,+$
$BC:2x+3y-1=0;CAL-x-2y+3=0;AB:-5x+6y+1=0$
Now the points $A,B,C$ are on the positive sides of the triangle whose revised forms of equations are written above. If the point $(\alpha ,{\alpha }^2)$ lies inside the triangle then the results obtained by putting the coordinates in the revised form of equations will be all +ive
$\therefore 2\alpha +3{\alpha }^2-1$ is +ive
or $(\alpha +1)(3\alpha -1)=$+ive.......(1)
$-\alpha -2{\alpha }^2+3=$+ive
$(2\alpha +3)(\alpha -1)=$-ive.....(2)
$-5\alpha +6{\alpha }^2+1$
$(3\alpha -1)(2\alpha -1)=$+ive......(3)
(1)$\Rightarrow$ $\alpha <-1$ or $\alpha >1/3$
(2) $\Rightarrow$ $-3/2<\alpha <-1$
(3) $\Rightarrow$ $\alpha <1/3$ or $\alpha >1/2$
From above three relations we conclude that
$-3/2<\alpha <-1$ and $1/2<\alpha <1$
$\therefore$ $\alpha$ belongs to $(-\frac{3}{2},-1)\cup (\frac{1}{2},1)$