Question

If the line y = mx bisects one of the angles between the lines $x^2 +4xy -2y^2 = 0$ then 'm' can be equal to

• A. $\frac{\sqrt{5}+1}{2}$
• B. -2
• C. $\frac{\sqrt{5}-1}{2}$
• D. $\frac{1}{2}$

Answer 1

Answer: (B), (D)

B, D

Answer 2

The given equation of the pair of lines is $x^2 + 4xy - 2y^2 = 0$. Dividing by $x^2$ and setting $m = y/x$, we get $1 + 4m - 2m^2 = 0$, or $2m^2 - 4m - 1 = 0$. Let the slopes of the two lines be $m_1$ and $m_2$. By Vieta's formulas: $m_1 + m_2 = 2$ $m_1 m_2 = -1/2$

The slope $m$ of an angle bisector satisfies: $\frac{2m}{1 - m^2} = \frac{m_1 + m_2}{1 - m_1 m_2}$ Substituting the values: $\frac{2m}{1 - m^2} = \frac{2}{1 - (-\frac{1}{2})} = \frac{2}{3/2} = \frac{4}{3}$ $6m = 4(1 - m^2)$ $4m^2 + 6m - 4 = 0$ $2m^2 + 3m - 2 = 0$ Factoring the quadratic: $(2m - 1)(m + 2) = 0$. The possible values for $m$ are $m = 1/2$ and $m = -2$. Both values are present in the options.