JEE Main 2022 (Online) 30th June Morning ShiftQuestion 1
Let $\alpha_1,\alpha_2$ ($\alpha_1<\alpha_2$) be the values of $\alpha$ for the points $(\alpha,-3)$, $(2,0)$ and $(1,\alpha)$ to be collinear. Then the equation of the line, passing through $(\alpha_1,\alpha_2)$ and making an angle of $\frac{\pi}{3}$ with the positive direction of the $x$-axis, is :
Points and Straight Lines
Solution
Collinearity: $\begin{vmatrix}\alpha&-3&1\\2&0&1\\1&\alpha&1\end{vmatrix}=0$ gives $\alpha^2-2\alpha-3=0\Rightarrow\alpha=-1,3$. So $\alpha_1=-1,\ \alpha_2=3$. Line through $(-1,3)$ with slope $\tan\frac\pi3=\sqrt3$: $y-3=\sqrt3(x+1)\Rightarrow\sqrt3x-y+\sqrt3+3=0$. Answer: (B).