Question

Find the values of $\alpha$ so that the point $P({\alpha }^2,\alpha )$ lies inside or on the triangle formed by the lines $x-5y+6=0$, $x-3y+2=0$ and $x-2y-3=0$

Answer 1

Let ABC be the triangle of the equations whose sides AB, BC and CA are respectively.

$x-5y+6=0$, $x-3y+2=0$ and $x-2y-3=0$

On solving equations, we get

A(9, 3), B(4, 2) and C(13, 5)

If the point $P(\alpha ,{\alpha }^2)$ lies n side the $\Delta ABC$, then

(i) A and P must be on the same side of BC

(ii) B and P must be on the same side of AC

(iii) C and P must be on the same side of AB.

Now,

A and P are on the same side of BC if

$\left\{9\right(1)+3(-3)+2\}({\alpha }^2-3\alpha +2)>0$

$(9-9+2)({\alpha }^2-3\alpha +2)>0$

${\alpha }^2-3\alpha +2>0$

$(\alpha -1)(\alpha -2)>0$

$\alpha ϵ(-\infty ,1)v(2,\infty )\dots \dots \left(i\right)$

B and P will lie on the same side of CA, if

$\left\{13\right(1)+5(-5)+6\}({\alpha }^2-5\alpha +6)>0$

$\Rightarrow (-6)({\alpha }^2-5\alpha +6)<0$

$\Rightarrow {\alpha }^2-5\alpha +6<0$

$\Rightarrow (\alpha -2)(\alpha -3)<0$

$\Rightarrow \alpha ϵ(2,3)\dots \dots \left(ii\right)$

C and P will lie on the same side of AB, if

$\left\{4\right(1)+2(-2)-3\}({\alpha }^2-2\alpha -3)>0$

$(-3)({\alpha }^2-2\alpha -3)>0$

${\alpha }^2-2\alpha -3<0$

$(\alpha -3)(\alpha +1)<0$

$\alpha ϵ(-1,3)\dots \dots \left(iii\right)$

From (i), (ii), (iii)

$\alpha ϵ[2,3]$