Question

Let \( A \) be the set of all points \( (\alpha, \beta) \) such that the area of triangle formed by the points \( (5, 6), (3, 2) \) and \( (\alpha, \beta) \) is 12 square units. Then the least possible length of a line segment joining the origin to a point in \( A \), is :

• A. \( \frac{8}{\sqrt{5}} \)
• B. \( \frac{4}{\sqrt{5}} \)
• C. \( \frac{16}{\sqrt{5}} \)
• D. \( \frac{12}{\sqrt{5}} \)

Hint:

The locus of a point moving such that the area of a triangle with two fixed points is constant consists of two parallel lines.
The distance of a point from these lines is always minimized at the perpendicular projection.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The set of points \( (\alpha, \beta) \) forms two parallel lines.
The "least possible length of a line segment joining the origin to a point in \( A \)" is simply the perpendicular distance from the origin to these lines.
Step 2: Key Formula or Approach:
Area of triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is \( \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \).
Distance from \( (0, 0) \) to \( ax + by + c = 0 \) is \( \frac{|c|}{\sqrt{a^2 + b^2}} \).
Step 3: Detailed Explanation:
Area \( = \frac{1}{2} |5(2 - \beta) + 3(\beta - 6) + \alpha(6 - 2)| = 12 \)
\[ |10 - 5\beta + 3\beta - 18 + 4\alpha| = 24 \]
\[ |4\alpha - 2\beta - 8| = 24 \implies |2\alpha - \beta - 4| = 12 \]
Case 1: \( 2\alpha - \beta - 4 = 12 \implies 2\alpha - \beta - 16 = 0 \)
Case 2: \( 2\alpha - \beta - 4 = -12 \implies 2\alpha - \beta + 8 = 0 \)
Distance from origin to line 1: \( d_1 = \frac{|-16|}{\sqrt{2^2 + (-1)^2}} = \frac{16}{\sqrt{5}} \).
Distance from origin to line 2: \( d_2 = \frac{|8|}{\sqrt{2^2 + (-1)^2}} = \frac{8}{\sqrt{5}} \).
The minimum distance is \( \frac{8}{\sqrt{5}} \).
Step 4: Final Answer:
The least possible length is \( \frac{8}{\sqrt{5}} \).