Question

The joint equation of the pair of lines through the origin and making an angle of \(\frac{π}{4}\) with the line \(3x+y-6 = 0\) is

• A. \(2x^2+3xy-2y^2 = 0\)
• B. \(2x^2+3xy+2y^2 = 0\)
• C. \(x^2+3xy-y^2 = 0\)
• D. \(x^2-3xy+y^2 = 0\)

Hint:

Find the slopes of lines making 45 degrees with a given slope using the angle formula, then multiply the line equations.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Lines through the origin that make a given angle with a given line are found from the angle formula between two lines, which yields two possible slopes.

Step 2: Key Formula or Approach:
\[ \tan\theta = \left|\frac{m - m_1}{1 + m m_1}\right| \]
Here \(\theta = \pi/4\), so \(\tan\theta = 1\). The slope of \(3x + y - 6 = 0\) is \(m_1 = -3\).

Step 3: Solve for the slopes.
\[ \left|\frac{m+3}{1-3m}\right| = 1 \]
Case 1: \(m + 3 = 1 - 3m\), so \(4m = -2\) and \(m = -\tfrac{1}{2}\).
Case 2: \(m + 3 = -(1 - 3m) = 3m - 1\), so \(2m = 4\) and \(m = 2\).

Step 4: Write the lines and combine.
The lines through the origin are \(y = 2x\) and \(y = -\tfrac{x}{2}\), that is \(2x - y = 0\) and \(x + 2y = 0\).
\[ (2x - y)(x + 2y) = 2x^2 + 4xy - xy - 2y^2 = 2x^2 + 3xy - 2y^2 = 0 \]

Step 5: Check the options.
Only option (A) matches. Option (B) has \(+2y^2\) and options (C) and (D) have no coefficient 2 in front of \(x^2\) and \(y^2\).

Final Answer:
The pair of lines is \(2x^2 + 3xy - 2y^2 = 0\), option (A). \[ \boxed{2x^2 + 3xy - 2y^2 = 0} \]