QJEEM22Ma06e3: JEE Main 2022 Mathematics - Class 11 JEEM22Ma06e3 Solution: Let 1 , 2 ( 1 < 2 ) be the values of fo the points ( , 3), (2,…

Points A( $\alpha$ , $-$ 3), B(2, 0) and C(1, $\alpha$ ) are collinear.

$\therefore$ Slope of AB = Slope of BC

$\Rightarrow {{0 + 3} \over {2 - \alpha }} = {{\alpha - 0} \over {1 - 2}}$

$\Rightarrow - 3 = \alpha (2 - \alpha )$

$\Rightarrow - 3 = 2\alpha - {\alpha ^2}$

$\Rightarrow {\alpha ^2} - 2\alpha - 3 = 0$

$\Rightarrow {\alpha ^2} - 3\alpha + \alpha - 3 = 0$

$\Rightarrow \alpha (\alpha - 3) + 1(\alpha - 3) = 0$

$\Rightarrow (\alpha + 1)(\alpha - 3) = 0$

$\Rightarrow \alpha = - 1,\,3$

Given, {\alpha _1} &lt; {\alpha _2}

$\therefore$ ${\alpha _1} = -1$ and ${\alpha _2} = 3$

$\therefore$ $\left( {{\alpha _1},\,{\alpha _2}} \right) = ( - 1,\,3)$

Now, equation of the line passing through ( $-$ 1, 3) and making angle ${\pi \over 3}$ with positive x-axis is

$(y - {y_1}) = m(x - {x_1})$

$\Rightarrow y - 3 = \left( {\tan {\pi \over 3}} \right)(x + 1)$

$\Rightarrow y - 3 = \sqrt 3 (x + 1)$

$\Rightarrow \sqrt 3 x - y + \sqrt 3 + 3 = 0$