Question

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

• A. a pair of intersecting lines
• B. a pair of parallel lines with distance between them $\frac{5}{2}$
• C. a pair of parallel lines with distance between them $5 \sqrt{2}$
• D. a conic section, which is not a pair of straight lines

Answer 1

Answer: (B)

a pair of parallel lines with distance between them $\frac{5}{2}$

Answer 2

We have $a = 1, h = - \sqrt{3} , b = 3$, $g = -\frac{3}{2} , f = \frac{3 \sqrt{3}}{2} , c = - 4$ . Thus $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ Hence the equation represents a pair of straight lines. Again $\frac{a}{h} = \frac{h}{b} = \frac{g}{f} = - \frac{1}{\sqrt{3}}$ $\therefore$ the lines are parallel. The distance between them $= 2 \sqrt{\frac{g^{2} -ac}{a\left(a+b\right)}} = 2 \sqrt{\frac{\frac{9}{4} +4}{1\left(1+3\right)}} = \frac{5}{2}$