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?
 A: I think your problem is regarding this equation:
$$\tan(45^{\circ})=\pm\frac{-\frac{3}{4}-m_2}{1-\frac{3}{4}m_2} \tag{1} 
 \label{1}$$
(You are using equation tags (i) and (1) both  creating confusion among the readers)
Your misconception: You think the two values of $\theta$ one can get by solving $\tan{\theta}= |\frac{m_1-m_2}{1+m_1m_2}|$ represent acute angle and the obtuse angle respectively between the lines with slopes $m_1$ and $m_2$.
Fact: In the above formula $\theta$ is always the acute angle between the lines with slopes $m_1$ and $m_2$. $\theta$ is simply equal to $|\tan^{-1}{m_1}-\tan^{-1}{m_2}|$ from which one derives the above formula.(In your question, of course $\theta=45°$ is acute and for that acute angle only, you are getting two values of $m_2$)
Further information: The equation of lines passing through point $(x_1,y_1)$ making an angle $\alpha$ with the line $y=mx+c$ are given by:
$y-y_1=\tan(\theta-\alpha)(x-x_1)$ and $y-y_1= \tan(\theta-\alpha)(x-x_1)$, where $\tan{\theta}=m$

Here the two lines $CD$ and $CV$ passing through $C(x_1,y_1)$ make an angle $\alpha$ with line $AB$ having slope $\tan{\theta}$. You'll always get exactly two such lines and their slopes will always be $\tan{(\theta+\alpha)}$ and $\tan{(\theta+180°-\alpha)}=\tan{(\theta-\alpha)}$. These are the two slopes you get while solving for $m_2$ in $(\ref{1})$. Notice that the angles $\theta + \alpha$ and $\theta - \alpha$ are not acute obtuse pair.
Note: Acute and obtuse have nothing to do here unless you were given an inclination angle which is obtuse, for which you simply have to find the corresponding acute angle between the lines and use that for $\theta$ in $\tan{\theta}= |\frac{m_1-m_2}{1+m_1m_2}|$  or alternatively, you can use this formula: $\tan{\theta_{obtuse}}= - |\frac{m_1-m_2}{1+m_1m_2}|$ where $\theta_{obtuse}$ is the obtuse angle between the lines with slopes $m_1$ and $m_2$.

Answering some questions after reading your edit:

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.

No. You are completely wrong there. $\pm$ is there because the angle $\theta$ can be measured from two different directions namely clockwise and anticlockwise.This fact is responsible for two different lines satisfying one condition of inclination angle. See the figure attached in my answer.

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?

You might be thinking $(\ref{1})$ is one equation, but that equation is actually a simpler way of writing two equations connected with "or". This is pretty obvious. Why can't you expect two different solutions of two different equations?
A: You have a right triangle with the right angle at (6, 7) and the line 3x+ 4y= 11 as hypotenuse.  The "two lines" are the two legs of the triangle.
A: 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.

