Question

The joint equation of the lines through the origin, trisecting angles in first and third quadrants is

• A. $\sqrt{3}(x^2-y^2)-4xy=0$
• B. $\sqrt{3}(x^2+y^2)-4xy=0$
• C. $\sqrt{3}(x^2+y^2)+4xy=0$
• D. $\sqrt{3}(x^2-y^2)+4xy=0$

Answer 1

Answer: (B)

(B)

Answer 2

The lines through the origin trisecting the first quadrant are given by:

$$y = \tan 30^\circ \, x = \frac{x}{\sqrt{3}} \quad \text{and} \quad y = \tan 60^\circ \, x = \sqrt{3}\,x.$$

Their equations can be written as:

$$x - \sqrt{3} \, y = 0 \quad \text{and} \quad \sqrt{3}\,x - y = 0.$$

Multiplying these, we obtain the joint equation:

$$(x - \sqrt{3} y)(\sqrt{3}x - y)=0.$$

Expanding:

$$\sqrt{3}x^2 - x y - 3xy + \sqrt{3}y^2 = \sqrt{3}(x^2+y^2)-4xy=0.$$

Thus, the correct option is (B).

Core Explanation:

The trisectors in the first quadrant give the lines $y = \frac{x}{\sqrt{3}}$ and $y = \sqrt{3}\,x$. Their joint equation is obtained by multiplying the factors:

$$(x-\sqrt{3}y)(\sqrt{3}x-y)=0 \quad \Rightarrow \quad \sqrt{3}(x^2+y^2)-4xy=0.$$