Key Concepts and Formulas

• Area of a Triangle: Given vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, the area of the triangle is given by: $\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| = \frac{1}{2} \left| \det \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix} \right|$
• Perpendicular Distance from a Point to a Line: The perpendicular distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by: $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

Step-by-Step Solution

We are given the points (5, 6), (3, 2), and ($\alpha$, $\beta$), and the area of the triangle formed by these points is 12. We need to find the minimum distance from the origin to a point ($\alpha$, $\beta$).

Step 1: Set up the area equation Using the determinant formula for the area of the triangle: $\text{Area} = \frac{1}{2} \left| \begin{vmatrix} 5 & 6 & 1 \\ 3 & 2 & 1 \\ \alpha & \beta & 1 \end{vmatrix} \right| = 12$ Why this step? We are given the area and the coordinates, so we substitute them into the appropriate formula to form an equation that relates $\alpha$ and $\beta$.

Step 2: Evaluate the determinant Expanding the determinant along the third column: $\begin{vmatrix} 5 & 6 & 1 \\ 3 & 2 & 1 \\ \alpha & \beta & 1 \end{vmatrix} = 1 \cdot \begin{vmatrix} 3 & 2 \\ \alpha & \beta \end{vmatrix} - 1 \cdot \begin{vmatrix} 5 & 6 \\ \alpha & \beta \end{vmatrix} + 1 \cdot \begin{vmatrix} 5 & 6 \\ 3 & 2 \end{vmatrix}$ $= (3\beta - 2\alpha) - (5\beta - 6\alpha) + (10 - 18)$ $= 3\beta - 2\alpha - 5\beta + 6\alpha - 8$ $= 4\alpha - 2\beta - 8$ Why this step? We need to simplify the determinant expression to get a linear relationship between $\alpha$ and $\beta$.

Step 3: Solve for the relationship between $\alpha$ and $\beta$ Substituting the determinant value back into the area equation: $\frac{1}{2} |4\alpha - 2\beta - 8| = 12$ Multiply by 2: $|4\alpha - 2\beta - 8| = 24$ Why this step? We isolate the absolute value term to prepare for handling it, as it will lead to two possible cases.

Step 4: Handle the absolute value We have two cases: Case 1: $4\alpha - 2\beta - 8 = 24$ $4\alpha - 2\beta = 32$ Divide by 2: $2\alpha - \beta = 16$ Rearranging into standard form: $2\alpha - \beta - 16 = 0 \quad \text{ (Equation 1) }$

Case 2: $4\alpha - 2\beta - 8 = -24$ $4\alpha - 2\beta = -16$ Divide by 2: $2\alpha - \beta = -8$ Rearranging into standard form: $2\alpha - \beta + 8 = 0 \quad \text{ (Equation 2) }$

Why this step? The set A consists of all points $(\alpha, \beta)$ that satisfy the area condition. The two equations represent two lines: Line 1: $2x - y - 16 = 0$ Line 2: $2x - y + 8 = 0$

Step 5: Find the least possible length from the origin to a point in A We need to find the shortest distance from the origin (0, 0) to either of the two lines. Using the perpendicular distance formula:

For Line 1: $2x - y - 16 = 0$ $d_1 = \frac{|2(0) - (0) - 16|}{\sqrt{2^2 + (-1)^2}} = \frac{|-16|}{\sqrt{5}} = \frac{16}{\sqrt{5}}$

For Line 2: $2x - y + 8 = 0$ $d_2 = \frac{|2(0) - (0) + 8|}{\sqrt{2^2 + (-1)^2}} = \frac{|8|}{\sqrt{5}} = \frac{8}{\sqrt{5}}$

Why this step? We are looking for the minimum distance. The shortest distance from the origin to any point on either line is the perpendicular distance from the origin to each line. We calculate both and then compare.

Step 6: Determine the least length Comparing the two distances: $\min\left( \frac{16}{\sqrt{5}}, \frac{8}{\sqrt{5}} \right) = \frac{8}{\sqrt{5}}$

Why this step? We identify the shorter of the two distances, which represents the minimum distance from the origin to the set A.

Common Mistakes & Tips

• Missing Absolute Value: Forgetting the absolute value in the area formula will result in only one line and an incorrect answer.
• Distance Formula Confusion: Ensure you use the perpendicular distance formula from a point to a line, not just the distance between two points.
• Calculation Errors: Double-check your determinant calculations and arithmetic to avoid simple mistakes.

Summary This problem involves finding the locus of a point given the area of a triangle, resulting in two lines. Then, we found the minimum distance from the origin to these lines using the perpendicular distance formula. The least possible length is $\frac{8}{\sqrt{5}}$.

The final answer is $\boxed{\frac{8}{\sqrt{5}}}$, which corresponds to option (C).