The equation of one of the straight lines whi | vedclass

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(A) Let the slope of the required line be $m$. The line passes through $(1, 3)$,so its equation is $y - 3 = m(x - 1)$,or $y = mx + (3 - m)$.The given

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$4\sqrt{2}x + 5y - (15 + 4\sqrt{2}) = 0$

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(A) Let the slope of the required line be $m$. The line passes through $(1, 3)$,so its equation is $y - 3 = m(x - 1)$,or $y = mx + (3 - m)$.The given line is $y = 3\sqrt{2}x - 1$,which has a slope $m_1 = 3\sqrt{2}$.The angle $	heta$ between the two lines is given by $	an 	heta = \left| \frac{m - m_1}{1 + m \cdot m_1} ight|$.Given $	an 	heta = \sqrt{2}$,we have $\sqrt{2} = \left| \frac{m - 3\sqrt{2}}{1 + 3\sqrt{2}m} ight|$.Case $1$: $\frac{m - 3\sqrt{2}}{1 + 3\sqrt{2}m} = \sqrt{2}$$m - 3\sqrt{2} = \sqrt{2} + 6m$$-5m = 4\sqrt{2} \implies m = -\frac{4\sqrt{2}}{5}$.Case $2$: $\frac{m - 3\sqrt{2}}{1 + 3\sqrt{2}m} = -\sqrt{2}$$m - 3\sqrt{2} = -\sqrt{2} - 6m$$7m = 2\sqrt{2} \implies m = \frac{2\sqrt{2}}{7}$.Using $m = -\frac{4\sqrt{2}}{5}$ in $y - 3 = m(x - 1)$:$y - 3 = -\frac{4\sqrt{2}}{5}(x - 1)$$5y - 15 = -4\sqrt{2}x + 4\sqrt{2}$$4\sqrt{2}x + 5y - (15 + 4\sqrt{2}) = 0$.

The equation of one of the straight lines which passes through the point $(1, 3)$ and makes an angle of $\tan^{-1}(\sqrt{2})$ with the straight line $y + 1 = 3\sqrt{2}x$ is:

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