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 $(6,7)$,
$$\frac{y-7}{x-6}=-7$$
$$\implies -7x+42=y-7$$
$$\implies 7x+y-49=0…(iv)$$
Now, we are getting two straight lines as a result. One makes $45^{\circ}$ with $3x+4y=11$ in the counterclockwise direction ($x-7y+43=0$), and the other makes $45^{\circ}$ with $3x+4y=11$ in the clockwise direction ($7x+y-49=0$). So, is $7x+y-49=0$ a valid solution to the above question as it makes $-45^{\circ}$ not $45^{\circ}$?
Best Answer
You previously asked why your same solution to the same exercise is returning two separate gradients, to which I commented:
“Algebraically, since $\tan\theta=\pm \frac{m_1-m_2}{1+m_1m_2}$ is effectively two separate linear equations in $m_2,$ why wouldn't you get two answers for $m_2?$ BTW, a shorter alternative to the second to fourth chunks of your presentation is $$\arctan(m)-\arctan\left(-\frac34\right)=\pm\frac\pi4\\m=\tan\left(\pm\frac\pi4-\arctan\left(\frac34\right)\right)\\=\frac17\;\text{ or }\;-7,$$ where $\arctan(m)$ is the anticlockwise angle between the line of gradient $m$ and the positive $x$-axis.”
This indirectly addresses your current question of why your solution returns two separate straight lines.
Notice that when two straight lines intersect, the angle between them is uniquely specified by some angle in $[0,90^\circ].$ This is in fact the standard way to specify the angle between two straight lines.
Also, notice that when you say “one makes $45^\circ$ with the given line in the counterclockwise direction”, you could well have equivalently thought of it as a clockwise angle, since there is no inherent directionality when measuring the angle between two lines (i.e., clockwise or counterclockwise depends merely on which you have chosen as the reference line). As such, $45^\circ$ (counterclockwise) and $-45^\circ$ (clockwise) are equivalent in this context. Based on the previous paragraph, they are all just standardly specified as $45^\circ.$
Related: while it's unambiguous to speak of the angle between two lines (and this angle is uniquely specified by $[0^\circ,90^\circ]$) and the angle between two vectors (and this angle is uniquely specified by $[0^\circ,180^\circ]$), a vertex angle might refer to either the reflex or non-reflex one.