Question

Let B and C be the two points on the line \(y + x = 0\) such that B and C are symmetric with respect to the origin. Suppose A is a point on \(y – 2x = 2\) such that \(\Delta ABC\) is an equilateral triangle. Then, the area of the \(\Delta ABC\) is

• A. 3√3
• B. 2√3
• C. 10/√3
• D. 8/√3

Answer 1

The Correct Option is D

Step 1: Analyzing the points. 
Given that \( B \) and \( C \) are symmetric with respect to the origin, the coordinates of \( B \) and \( C \) lie on the line \( y + x = 0 \), meaning that \( y = -x \). Let the coordinates of \( B \) be \( (-t, 0) \) and the coordinates of \( C \) be \( (t, 0) \), as they are symmetric about the origin.
Step 2: Equation of point \( A \). 
Point \( A \) lies on the line \( y - 2x = 2 \), so we substitute \( y = x \) (since \( A \) is on the line \( y = x \)): \[ x - 2x = 2 \quad \Rightarrow \quad -x = 2 \quad \Rightarrow \quad x = -2, \, y = -2. \] Thus, the coordinates of point \( A \) are \( (-2, -2) \).
Step 3: Calculating the height from line \( x + y = 0 \).
The height of the equilateral triangle from the line \( x + y = 0 \) is the perpendicular distance from point \( A(-2, -2) \) to the line. The formula for the distance from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is: \[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}. \] For the line \( x + y = 0 \), \( A = 1, B = 1, C = 0 \), so the distance is: \[ d = \frac{|1(-2) + 1(-2) + 0|}{\sqrt{1^2 + 1^2}} = \frac{| -2 - 2 |}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 4\sqrt{2}. \] Step 4: Finding the area of the equilateral triangle. 
The area \( A \) of an equilateral triangle is given by: \[ A = \frac{\sqrt{3}}{4} s^2, \] where \( s \) is the side length of the triangle. From the height of the triangle \( h = 4 \), we know that the relationship between the side length \( s \) and the height \( h \) of an equilateral triangle is: \[ h = \frac{s\sqrt{3}}{2}. \] So, solving for \( s \): \[ 4 = \frac{s\sqrt{3}}{2} \quad \Rightarrow \quad s = \frac{8}{\sqrt{3}}. \] Now, substituting into the formula for the area of the equilateral triangle: \[ A = \frac{\sqrt{3}}{4} \left( \frac{8}{\sqrt{3}} \right)^2 = \frac{\sqrt{3}}{4} \times \frac{64}{3} = \frac{64\sqrt{3}}{12} = \frac{8}{\sqrt{3}}. \] Thus, the area of the equilateral triangle is \( \frac{8}{\sqrt{3}} \), and the correct answer is option (4).