Question

The joint equation of pair of lines through the origin and having slopes $(1+\sqrt{2})$ and $\frac{1}{(1+\sqrt{2})}$ is

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

Hint:

Whenever the two slopes are given as reciprocals of each other ($m_2 = \frac{1}{m_1}$), their product $m_1m_2$ is always equal to 1. This means the coefficients of $x^2$ and $y^2$ in the final combined equation must be completely identical, letting you rule out options (B) and (D) instantly!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the joint (homogenous second-degree) equation representing a pair of straight lines passing through the origin with given explicit slopes $m_1$ and $m_2$.

Step 2: Key Formula or Approach:
The equation of any straight line passing through the origin is given by $y = mx \implies mx - y = 0$.
The joint equation of two lines with slopes $m_1$ and $m_2$ is found by multiplying their separate linear factor equations:
$$(m_1x - y)(m_2x - y) = 0 \implies m_1m_2x^2 - (m_1 + m_2)xy + y^2 = 0$$

Step 3: Detailed Explanation:
Let's first simplify the given values for the two slopes:
$m_1 = \sqrt{2} + 1$
$m_2 = \frac{1}{\sqrt{2} + 1}$
We can rationalize the denominator of $m_2$ by multiplying the numerator and denominator by the conjugate $(\sqrt{2} - 1)$:
$$m_2 = \frac{1}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{\sqrt{2} - 1}{(\sqrt{2})^2 - 1^2} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1$$ Now, calculate the sum of the slopes ($m_1 + m_2$) and the product of the slopes ($m_1 m_2$):
$$m_1 + m_2 = (\sqrt{2} + 1) + (\sqrt{2} - 1) = 2\sqrt{2}$$ $$m_1 m_2 = (\sqrt{2} + 1)(\sqrt{2} - 1) = (\sqrt{2})^2 - 1^2 = 2 - 1 = 1$$ Substitute these two simplified symmetric expressions into our structural joint equation:
$$(1)x^2 - (2\sqrt{2})xy + y^2 = 0$$ $$x^2 - 2\sqrt{2}xy + y^2 = 0$$ This matches option (C).

Step 4: Final Answer:
The joint equation of the pair of lines is $x^2 - 2\sqrt{2}xy + y^2 = 0$, which corresponds to option (C).