Question

Let a point A lie between the parallel lines \(L_1\) and \(L_2\) such that its distances from \(L_1\) and \(L_2\) are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle ABC, where the points B and C lie on the lines \(L_1\) and \(L_2\), respectively, is:

• A. \(21\sqrt{3}\)
• B. \(15\sqrt{6}\)
• C. 27
• D. \(12\sqrt{2}\)

Hint:

Coordinate geometry proofs can sometimes be simplified by using transformations like rotation, especially for regular polygons like squares or equilateral triangles. Placing one vertex at the origin and aligning one side or an axis can make the algebra much more manageable. The rotation method avoids solving a complicated system of squared distance equations.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We have an equilateral triangle ABC. Its vertices A, B, and C lie on three parallel lines. Vertex A lies on a line in the middle, while B and C lie on the outer two lines, \(L_1\) and \(L_2\). We are given the distances from A to \(L_1\) and \(L_2\). We need to find the area of the triangle.
Step 2: Key Formula or Approach:
1. Setup a Coordinate System: We can simplify the problem by placing the parallel lines in a convenient orientation, for example, as horizontal lines.
2. Use Rotation: A powerful technique for problems involving equilateral triangles is rotation. If we rotate point C by 60 degrees about point A, it will coincide with point B (or a point B' such that AB'=AC), because in an equilateral triangle, the sides AB and AC are equal in length and the angle between them is 60 degrees.
3. Area of Equilateral Triangle: The area is given by \(\frac{\sqrt{3}}{4}s^2\), where \(s\) is the side length.
Step 3: Detailed Explanation:
Let's set up a coordinate system. Let the line containing A be the x-axis, so A is at the origin, A=(0,0).
Since the lines are parallel, we can orient them as horizontal lines.
The distance from A to \(L_1\) is 6, so we can set the equation of \(L_1\) as \(y=6\).
The distance from A to \(L_2\) is 3. Since A is between the lines, \(L_2\) must be on the opposite side of the x-axis. So we can set the equation of \(L_2\) as \(y=-3\).
The total distance between lines \(L_1\) and \(L_2\) is \(6 - (-3) = 9\).
Vertex B lies on \(L_1\), so its coordinates are B\((x_B, 6)\).
Vertex C lies on \(L_2\), so its coordinates are C\((x_C, -3)\).
Let the side length of the equilateral triangle be \(s\).
Then \(AB^2 = AC^2 = BC^2 = s^2\).
\(AC^2 = (x_C - 0)^2 + (-3 - 0)^2 = x_C^2 + 9 = s^2\).
Now, let's rotate point C around point A(0,0) by 60 degrees. The new point, C', must coincide with B because \(\triangle ABC\) is equilateral (i.e., AB = AC and \(\angle BAC = 60^\circ\)).
The rotation formulas are:
\(x' = x \cos\theta - y \sin\theta\) \(y' = x \sin\theta + y \cos\theta\) Let's rotate C\((x_C, -3)\) by \(\theta = 60^\circ\). The rotated point is B\((x_B, 6)\).
\[ x_B = x_C \cos(60^\circ) - (-3) \sin(60^\circ) = x_C \left(\frac{1}{2}\right) + 3 \left(\frac{\sqrt{3}}{2}\right) \] \[ 6 = x_C \sin(60^\circ) + (-3) \cos(60^\circ) = x_C \left(\frac{\sqrt{3}}{2}\right) - 3 \left(\frac{1}{2}\right) \] Let's solve the second equation for \(x_C\): \[ 6 = \frac{x_C\sqrt{3} - 3}{2} \] \[ 12 = x_C\sqrt{3} - 3 \] \[ 15 = x_C\sqrt{3} \implies x_C = \frac{15}{\sqrt{3}} = 5\sqrt{3} \] Now that we have the coordinates of C, C\((5\sqrt{3}, -3)\), we can find the side length \(s\). \[ s^2 = AC^2 = x_C^2 + (-3)^2 = (5\sqrt{3})^2 + 9 = (25 \times 3) + 9 = 75 + 9 = 84 \] The area of the equilateral triangle is \(\frac{\sqrt{3}}{4}s^2\). \[ \text{Area} = \frac{\sqrt{3}}{4} \times 84 = 21\sqrt{3} \] (Note: A rotation by -60 degrees would give another possible triangle, but its area would be the same). Step 4: Final Answer:
The area of the equilateral triangle ABC is \(21\sqrt{3}\) sq. units.