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)

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

Answer 2

The lines through the origin trisecting the angles in the first (and similarly, in the third) quadrant have slopes

$$m_1 = \tan 30^\circ = \frac{1}{\sqrt{3}},\quad m_2 = \tan 60^\circ = \sqrt{3}.$$

Thus, their equations are

$$y = \frac{1}{\sqrt{3}}x \quad \Rightarrow \quad \sqrt{3}x - y = 0,$$ $$y = \sqrt{3}x \quad \Rightarrow \quad x - \sqrt{3}y = 0.$$

Their combined (joint) equation is obtained by multiplying the two:

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

Expanding:

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

which can be written as

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