To solve this problem, we need to determine the equation of a straight line that passes through the point \((1, 3)\) and makes an angle of \(\tan^{-1}(\sqrt{2})\) with the given line \(y + 1 = 3\sqrt{2} x\).
First, identify the slope of the given line. The equation \(y + 1 = 3\sqrt{2} x\) can be rewritten in the slope-intercept form \(y = mx + c\), where \(m\) is the slope.
The equation becomes:
y = 3\sqrt{2} x - 1Thus, the slope m_1 = 3\sqrt{2}.
The required line makes an angle of \(\theta = \tan^{-1}(\sqrt{2})\) with this line. Therefore, if m_2 is the slope of the required line, then:
\frac{m_2 - m_1}{1 + m_2 \cdot m_1} = \sqrt{2}Substitute m_1 = 3\sqrt{2} and solve for \(m_2\):
\frac{m_2 - 3\sqrt{2}}{1 + 3\sqrt{2} \cdot m_2} = \sqrt{2}Cross-multiply to get:
m_2 - 3\sqrt{2} = \sqrt{2}(1 + 3\sqrt{2} \cdot m_2)Expanding and simplifying:
m_2 - 3\sqrt{2} = \sqrt{2} + 6m_2 m_2 - 6m_2 = \sqrt{2} + 3\sqrt{2} -5m_2 = 4\sqrt{2} m_2 = -\frac{4\sqrt{2}}{5}Now, we have the slope of the required line as m_2 = -\frac{4\sqrt{2}}{5} and the line passes through the point (1, 3).
Therefore, using the point-slope form:
y - 3 = -\frac{4\sqrt{2}}{5}(x - 1)Expanding this equation gives:
y - 3 = -\frac{4\sqrt{2}}{5}x + \frac{4\sqrt{2}}{5}Multiplying everything by 5 to eliminate fractions:
5y - 15 = -4\sqrt{2}x + 4\sqrt{2}Rearranging terms, we get:
4\sqrt{2}x + 5y - (15 + 4\sqrt{2}) = 0Thus, the equation of the required line is:
4\sqrt{2}x + 5y - (15 + 4\sqrt{2}) = 0This matches the given option:
This confirms that the option 4√2 x + 5y - (15 + 4√2) = 0 is the correct answer.