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

To solve this problem, we need to find the area of the equilateral triangle \( \Delta ABC \) where points B and C lie on the line \( y + x = 0 \) and are symmetric with respect to the origin. Point A lies on the line \( y - 2x = 2 \).

1. Since B and C are symmetric with respect to the origin and lie on the line \( y + x = 0 \), we can write the coordinates of B and C as \( (a, -a) \) and \( (-a, a) \), respectively, following the condition \( y = -x \).
2. Point A lies on the line \( y - 2x = 2 \). Let the coordinates of A be \( (x_1, y_1) \). Thus, it satisfies the equation \( y_1 - 2x_1 = 2 \). In the form \( y_1 = 2x_1 + 2 \).
3. For \( \Delta ABC \) to be equilateral, the distances \( AB = BC = CA \) must all be equal. Using the distance formula, we can express the distances as follows:
   1. Distance \( AB = \sqrt{(x_1 - a)^2 + (y_1 + a)^2} \)
   2. Distance \( AC = \sqrt{(x_1 + a)^2 + (y_1 - a)^2} \)
   3. Distance \( BC = \sqrt{((-a) - a)^2 + (a + a)^2} = \sqrt{(2a)^2 + (2a)^2} = \sqrt{8a^2} = 2\sqrt{2}a \)
4. Set the lengths \( AB = BC \) and \( AC = BC \). Using the condition of equilateral triangle:
   • From \( AB = BC \), \( \sqrt{(x_1 - a)^2 + (y_1 + a)^2} = 2\sqrt{2}a \)
   • From \( AC = BC \), \( \sqrt{(x_1 + a)^2 + (y_1 - a)^2} = 2\sqrt{2}a \)
5. Since point A is also on line \( y - 2x = 2 \), substitute the expression \( y_1 = 2x_1 + 2 \) into the equations for \( AB \) and \( AC \).
6. After calculations and symmetry considerations of triangle properties, solve for \( a \) and find the distance to be the length of any side.
7. Use the area formula for an equilateral triangle: \[ \text{Area of } \Delta ABC = \frac{\sqrt{3}}{4} \times \text{side}^2 \] Since we find \( a \) such that \( BC = 2\sqrt{2}a \), plug into the area formula.
8. The calculated area of \( \Delta ABC \) is \( \frac{8}{\sqrt{3}} \).

The area of the equilateral triangle \( \Delta ABC \) is therefore \(\frac{8}{\sqrt{3}}\).