Question

The range of 'alpha'for which the point (alpha, alpha + 2)and (3aplha÷2, alpha2
) lie on opposite sides of the line 2x + 3y - 6 = 0​

Answer 1

Given:

The equation of the line (L) is 2x+3y-6 = 0

To find:

The range of ‘α’

Calculation:

Substitute the point P₁ (α,α+2) in the line equation 2x+3y-6 = 0

        $L(\alpha,\alpha+2) = 2(\alpha)+3(\alpha+2)-6$

                           $=2\alpha+3 \alpha+6-6$

                           $=5 \alpha$

Substitiute the point P₂ (3α/2,α²) in the line equation 2x+3y-6 = 0            

       $L(\frac{3\alpha}{2},\alpha^{2})=2(\frac{3\alpha}{2})+3(\alpha^{2})-6$

                       $=3\alpha+3\alpha^{2}-6$

For the points to lie on the different sides they should be of opposite signs.

Therefore when we multiply both the values the resultant value should be less than zero

       $\Rightarrow(5 \alpha)(3 \alpha^{2}+3 \alpha-6)<0$

           $(5 \alpha)(3 \alpha^{2}+6 \alpha-3 \alpha-6)<0$

           $(5 \alpha)[3 \alpha(\alpha+2)-3(\alpha+2)]<0$

           $(5 \alpha)(3 \alpha-3)(\alpha+2)<0$

           $15(\alpha)(\alpha-1)(\alpha+2)<0$

           $\alpha \in(-\infty,-2)\cup(0,1)$

Final answer:

The range of α is  $\alpha \in(-\infty,-2)\cup(0,1)$