Straight Lines
Straight Line
star_batch_jee_advanced_2025
Grade None
Question:
The distance between the two parallel lines is 1 unit. A point 'A' is chosen to lie between the lines at a distance 'd' from one of them. Triangle $ABC$ is equilateral with $B$ on one line and $C$ on the other parallel line. The length of the side of the equilateral triangle is:
\frac{2}{3}\sqrt{d^2+d+1}
2\sqrt{\frac{d^2-d+1}{3}}
2\sqrt{d^2-d+1}
\sqrt{d^2-d+1}
Step-by-Step Solution
Key Concept: Use the constraint $\cos^2\theta + \sin^2\theta = 1$ to eliminate the trigonometric variable and solve for the unknown length.
Given $l\sin(60° - \theta) = \frac{\sqrt{3}}{2}\cos\theta - \frac{1}{2}\sin\theta = 1 - d$, we extract $\cos\theta = \frac{2-d}{\sqrt{3}}$ and $\sin\theta = d$. Squaring both equations and adding: $\cos^2\theta + \sin^2\theta = 1$ yields $\frac{(2-d)^2}{3} + d^2 = 1$. Solving gives $l = 2\sqrt{\frac{d^2 - d + 1}{3}}$.
Correct Answer: 2