Key Concepts and Formulas

• Collinearity: Points $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ are collinear if the slope of $AB$ equals the slope of $BC$ (or $AC$).
• Slope Formula: The slope $m$ of a line through $(x_a, y_a)$ and $(x_b, y_b)$ is $m = \frac{y_b - y_a}{x_b - x_a}$.
• Point-Slope Form: The equation of a line through $(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$.
• Slope from Angle: The slope $m$ of a line making an angle $\theta$ with the positive x-axis is $m = \tan \theta$.

Step-by-Step Solution

• Step 1: Calculate the slope of AB. We are given $A(\alpha, -3)$ and $B(2, 0)$. We calculate the slope of line segment $AB$ using the slope formula. $m_{AB} = \frac{0 - (-3)}{2 - \alpha} = \frac{3}{2 - \alpha}$ Explanation: This step applies the slope formula to points A and B to express the slope of the line segment AB in terms of $\alpha$.

• Step 2: Calculate the slope of BC. We are given $B(2, 0)$ and $C(1, \alpha)$. We calculate the slope of line segment $BC$ using the slope formula. $m_{BC} = \frac{\alpha - 0}{1 - 2} = \frac{\alpha}{-1} = -\alpha$ Explanation: This step applies the slope formula to points B and C to express the slope of the line segment BC in terms of $\alpha$.

• Step 3: Equate the slopes and solve for $\alpha$. For $A$, $B$, and $C$ to be collinear, the slopes must be equal: $m_{AB} = m_{BC}$. $\frac{3}{2 - \alpha} = -\alpha$ $3 = -\alpha(2 - \alpha)$ $3 = -2\alpha + \alpha^2$ $\alpha^2 - 2\alpha - 3 = 0$ Explanation: This step sets the two calculated slopes equal to each other, creating an equation that can be solved to find the values of $\alpha$ that satisfy the collinearity condition.

• Step 4: Factor the quadratic equation. We factor the quadratic equation to find the roots. $\alpha^2 - 3\alpha + \alpha - 3 = 0$ $\alpha(\alpha - 3) + 1(\alpha - 3) = 0$ $(\alpha + 1)(\alpha - 3) = 0$ Explanation: This step factorizes the quadratic equation to simplify finding the roots.

• Step 5: Determine the values of $\alpha$. The roots of the equation are the possible values of $\alpha$. $\alpha + 1 = 0 \implies \alpha = -1$ $\alpha - 3 = 0 \implies \alpha = 3$ Thus, $\alpha = -1$ or $\alpha = 3$. Explanation: This step determines the values of $\alpha$ that satisfy the collinearity condition.

• Step 6: Identify $\alpha_1$ and $\alpha_2$. We are given that $\alpha_1 < \alpha_2$. Comparing the two values, $-1 < 3$. Therefore, $\alpha_1 = -1$ and $\alpha_2 = 3$. The point is $(\alpha_1, \alpha_2) = (-1, 3)$. Explanation: This step identifies the two $\alpha$ values and assigns them to $\alpha_1$ and $\alpha_2$ based on the given condition.

• Step 7: Determine the slope of the line. The line makes an angle of $\frac{\pi}{3}$ with the positive x-axis. $m = \tan\left(\frac{\pi}{3}\right) = \sqrt{3}$ Explanation: This step calculates the slope of the line using the given angle.

• Step 8: Use the point-slope form to find the equation of the line. We have the point $(-1, 3)$ and the slope $\sqrt{3}$. $y - 3 = \sqrt{3}(x - (-1))$ $y - 3 = \sqrt{3}(x + 1)$ $y - 3 = \sqrt{3}x + \sqrt{3}$ Explanation: This step substitutes the calculated slope and the coordinates of the point into the point-slope form of a linear equation.

• Step 9: Simplify and rearrange the equation. We rearrange the equation into the standard form. $\sqrt{3}x - y + \sqrt{3} + 3 = 0$ Explanation: This step simplifies the equation into the standard form.

Common Mistakes & Tips

• Sign Errors: Pay close attention to signs, especially when dealing with negative values and distributing.
• Slope Formula Application: Ensure the correct order of subtraction in the numerator and denominator of the slope formula.
• Angle Mode: Ensure your calculator is in the correct mode (degrees or radians) when calculating the tangent of an angle.

Summary

We found the values of $\alpha$ for which the points are collinear by equating the slopes of line segments formed by those points. Then, using the condition $\alpha_1 < \alpha_2$, we identified the point $(\alpha_1, \alpha_2)$. Finally, we used the point-slope form and the given angle to find the equation of the line. The equation of the line is $\sqrt{3}x - y + \sqrt{3} + 3 = 0$.

The final answer is \boxed{\sqrt{3} x - y + \sqrt{3} + 3 = 0}, which corresponds to option (B).