1. Given line and its slope

The given line is $y+1=3\sqrt{2}\,x$ So, $y=3\sqrt{2}\,x-1$ Hence its slope is $m_1=3\sqrt{2}$

2. Required angle condition

We need a line through $(1,3)$ making angle $\theta=\tan^{-1}(\sqrt{2})$ with the given line.

If the required line has slope $m$, then angle between two lines satisfies $\tan\theta=\left|\frac{m-m_1}{1+mm_1}\right|$ Thus, $\left|\frac{m-3\sqrt{2}}{1+3\sqrt{2}m}\right|=\sqrt{2}$

So we solve: $\frac{m-3\sqrt{2}}{1+3\sqrt{2}m}=\pm\sqrt{2}$

3. Find possible slopes

Case 1:

$\frac{m-3\sqrt{2}}{1+3\sqrt{2}m}=\sqrt{2}$

Then, $m-3\sqrt{2}=\sqrt{2}(1+3\sqrt{2}m)$ $m-3\sqrt{2}=\sqrt{2}+6m$ $-5m=4\sqrt{2}$ $m=-\frac{4\sqrt{2}}{5}$

Case 2:

$\frac{m-3\sqrt{2}}{1+3\sqrt{2}m}=-\sqrt{2}$

Then, $m-3\sqrt{2}=-\sqrt{2}(1+3\sqrt{2}m)$ $m-3\sqrt{2}=-\sqrt{2}-6m$ $7m=2\sqrt{2}$ $m=\frac{2\sqrt{2}}{7}$

So the two possible slopes are $m=-\frac{4\sqrt{2}}{5},\qquad m=\frac{2\sqrt{2}}{7}$

4. Equation of lines through $(1,3)$

Using point-slope form: $y-3=m(x-1)$

For $m=-\dfrac{4\sqrt{2}}{5}$:

$y-3=-\frac{4\sqrt{2}}{5}(x-1)$ $5y-15=-4\sqrt{2}x+4\sqrt{2}$ $4\sqrt{2}x+5y-(15+4\sqrt{2})=0$ This matches Option A.

For $m=\dfrac{2\sqrt{2}}{7}$:

$y-3=\frac{2\sqrt{2}}{7}(x-1)$ $7y-21=2\sqrt{2}x-2\sqrt{2}$ $2\sqrt{2}x-7y+(21-2\sqrt{2})=0$ This is not among the options.

5. Check options

• A: $4\sqrt{2}x+5y-(15+4\sqrt{2})=0$ ✓
• B: slope $=-\dfrac{5\sqrt{2}}{4}$, not valid.
• C: does not pass through $(1,3)$.
• D: slope $=\dfrac{4\sqrt{2}}{5}$, not valid.

Therefore, the correct option is: $\boxed{A}$