Question

The equation of one of the straight lines which passes through the point (1, 3) and makes an angle tan⁻¹(√2) with the straight line, y + 1 = 3√2 x is :

• A. 4√2 x + 5y - (15 + 4√2) = 0
• B. 4√2 x - 5y - (5 + 4√2) = 0
• C. 5√2 x + 4y - (15 + 4√2) = 0
• D. 4√2 x + 5y - 4√2 = 0

Hint:

Remember to solve for both cases of the absolute value when finding slopes of lines at an angle.

Answer 1

The Correct Option is A

Step 1: Slope of given line $m_1 = 3\sqrt{2}$. Given $\tan \theta = \sqrt{2}$.
Step 2: Use the angle formula: $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Step 3: $\sqrt{2} = \left| \frac{3\sqrt{2} - m_2}{1 + 3\sqrt{2} m_2} \right|$.
Step 4: Solving for $m_2$: Case 1: $\sqrt{2} + 6m_2 = 3\sqrt{2} - m_2 \implies 7m_2 = 2\sqrt{2} \implies m_2 = \frac{2\sqrt{2}}{7}$. Case 2: $-\sqrt{2} - 6m_2 = 3\sqrt{2} - m_2 \implies -5m_2 = 4\sqrt{2} \implies m_2 = \frac{-4\sqrt{2}}{5}$.
Step 5: Use point-slope form with $m_2 = \frac{-4\sqrt{2}}{5}$ and $(1, 3)$: $y - 3 = \frac{-4\sqrt{2}}{5}(x - 1) \implies 5y - 15 = -4\sqrt{2}x + 4\sqrt{2}$.
Step 6: $4\sqrt{2}x + 5y - (15 + 4\sqrt{2}) = 0$.