Intuition behind getting two straight lines as result Question:
Find the equation of the straight line that passes through $(6,7)$ and makes an angle $45^{\circ}$ with the straight line $3x+4y=11$.
My solution (if you want, you can skip to the bottom):
Manipulating the given equation to get it to the slope-intercept form,
$$3x+4y=11...(i)$$
$$\implies 4y=-3x+11$$
$$\implies y=\frac{-3}{4}x+\frac{11}{4}$$
Let, the slope of (i) is $m_1=\frac{-3}{4}$, and the slope of our desired equation is $m_2$. Now, according to the question,
$$\tan(45^{\circ})=\pm\frac{-\frac{3}{4}-m_2}{1-\frac{3}{4}m_2}...(1)$$
$$\implies 1=\pm\frac{-\frac{3}{4}-m_2}{1-\frac{3}{4}m_2}$$
$$\implies \pm \frac{3}{4}+m_2=1-\frac{3}{4}m_2...(ii)$$
Picking positive value from (ii),
$$\frac{3}{4}+m_2=1-\frac{3}{4}m_2$$
$$\implies m_2(1+\frac{3}{4})=1-\frac{3}{4}$$
$$\implies m_2=\frac{1-\frac{3}{4}}{1+\frac{3}{4}}$$
$$\implies m_2=\frac{1}{7}$$
Picking negative value from (ii),
$$-\frac{3}{4}-m_2=1-\frac{3}{4}m_2$$
$$\implies -\frac{3}{4}-m_2=1-\frac{3}{4}m_2$$
$$\implies -m_2(1-\frac{3}{4})=1+\frac{3}{4}$$
$$\implies m_2=-\frac{1+\frac{3}{4}}{1-\frac{3}{4}}$$
$$\implies m_2=-7$$
Picking $m_2=\frac{1}{7}$, the equation of the straight line that passes through $(6,7)$,
$$\frac{y-7}{x-6}=\frac{1}{7}$$
$$\implies 7y-49=x-6$$
$$\implies -x+7y-43=0$$
$$\implies x-7y+43=0...(iii)$$
Picking $m_2=-7$, the equation of the straight line that passes through $(6,7)$,
$$\frac{y-7}{x-6}=-7$$
$$\implies -7x+42=y-7$$
$$\implies 7x+y-49=0...(iv)$$
The general form of equation (1) is,
$$\tan\theta=\pm \frac{m_1-m_2}{1+m_1m_2}$$
Here, $\pm$ has been included to include both the acute and the obtuse angles that are formed when two lines with slopes $m_1$ and $m_2$ intersect each other.
Now, I used this equation to find the straight line that makes $45^{\circ}$ with (i). Why am I getting 2 values of $m_2$ when there is only one value of $m_2$ in the general form of the equation? How can I reconcile between my getting of 2 values of $m_2$ with the $\pm$ sign arising due to the acute and obtuse angles?

We have to be mindful about sign of slope
To the given inclination $ \tan^{-1} (-3/4) = - 36. 87 ^{\circ}$ add, resulting in $ 45 ^{\circ}- 36.87 ^{\circ} = 8.13^{\circ}$ positive ( counter clockwise) to x-axis and you have only one straight line solution AB. Other solution is spurious, needs to be discarded.

