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: Understanding the Question:
We need to find the combined (joint) second-degree equation of two lines passing through the origin that form a closed equilateral triangle when intersected by the vertical line $x = 3$.

Step 2: Detailed Explanation:
The line $x = 3$ is a vertical line parallel to the y-axis. For lines passing through the origin to form an equilateral triangle with a vertical line, the x-axis acts as an axis of symmetry for the triangle. \includegraphics[width=0.4\linewidth]{images/58sol.png}
Since the total internal angle of an equilateral triangle is $60^\circ$, the axis of symmetry splits this angle perfectly into two halves of $30^\circ$. Thus, the two lines from the origin make angles of $+30^\circ$ and $-30^\circ$ with the positive direction of the x-axis. Let's compute the slopes ($m_1$ and $m_2$) of these lines: $$ m_1 = \tan(30^\circ) = \frac{1}{\sqrt{3}} $$ $$ m_2 = \tan(-30^\circ) = -\frac{1}{\sqrt{3}} $$ Using the slope-intercept equation for lines passing through the origin ($y = mx$), the individual line equations are: $$ y = \frac{1}{\sqrt{3}}x \implies x - \sqrt{3}y = 0 $$ $$ y = -\frac{1}{\sqrt{3}}x \implies x + \sqrt{3}y = 0 $$ To find the joint homogeneous equation, we multiply these two linear factors together using the difference of squares identity $(a-b)(a+b) = a^2 - b^2$: $$ (x - \sqrt{3}y)(x + \sqrt{3}y) = 0 $$ $$ x^2 - (\sqrt{3}y)^2 = 0 \implies x^2 - 3y^2 = 0 $$

Step 3: Final Answer:
The joint equation of the pair of lines is $x^2 - 3y^2 = 0$, which matches option (C).