Question

The joint equation of the pair of lines through the origin and making an equilateral triangle with the line $x = 3$ is

• A. $3x^2 - y^2 = 0$
• B. $3x^2 - 2xy + y^2 = 0$
• C. $x^2 - 3y^2 = 0$
• D. $x^2 + 2xy - 3y^2 = 0$

Hint:

Whenever a pair of lines from the origin forms an equilateral triangle with a vertical line $x = k$, the joint equation is always uniquely fixed as $x^2 - 3y^2 = 0$, completely independent of the value of the constant $k$!

Answer 1

The Correct Option is C

Step 1: Use the symmetry.
The line $x = 3$ is vertical. For an equilateral triangle with two lines from the origin, the x-axis is the axis of symmetry, so the lines make $30^\circ$ on each side.

Step 2: Find the slopes.
$\tan 30^\circ = \dfrac{1}{\sqrt 3}$ and $\tan(-30^\circ) = -\dfrac{1}{\sqrt 3}$. The lines are $x - \sqrt 3 y = 0$ and $x + \sqrt 3 y = 0$.

Step 3: Multiply.
\[ (x - \sqrt 3 y)(x + \sqrt 3 y) = x^2 - 3y^2 = 0 \]
\[ \boxed{x^2 - 3y^2 = 0,\ \text{option 3}} \]