Question

Let \( ABC \) be an equilateral triangle. If the coordinates of \( A \) are \( (-2,2) \) and the side \( BC \) is along the line \( x + y = 6 \), then the length of the side of the triangle is

• A. \(2\sqrt{3} \)
• B. \(3\sqrt{2} \)
• C. \(4\sqrt{6} \)
• D. \(6\sqrt{6} \)
• E. \(2\sqrt{6} \)

Hint:

Distance from vertex to opposite side gives height in equilateral triangle problems.

Answer 1

Answer: (E)

Concept: Distance from vertex to opposite side in equilateral triangle: \[ \text{height} = \frac{\sqrt{3}}{2} \times \text{side} \]

Step 1: Find distance from point to line.
Line: \( x + y - 6 = 0 \) Point: \( (-2,2) \) \[ \text{Distance} = \frac{| -2 + 2 - 6 |}{\sqrt{1^2 + 1^2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2} \]

Step 2: Use height relation.
\[ \frac{\sqrt{3}}{2} \cdot \text{side} = 3\sqrt{2} \] \[ \text{side} = \frac{2 \cdot 3\sqrt{2}}{\sqrt{3}} = \frac{6\sqrt{2}}{\sqrt{3}} \] \[ = 2\sqrt{6} \]