<s>[INST] Finding a straight line that makes a given angle with a given straight line 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 $45^{\circ}$?
 [/INST] I don't see any reason why the clockwise angle would be invalid.
In some contexts, such as trigonometry, we, by convention, write the angle as the counterclockwise angle from the positive $x$-axis. In others, such as true compass bearings, the angle is, again by convention, measured clockwise (from North).
So for this question, with no specified convention, a $45$° angle is a $45$° angle, regardless of whether it is clockwise or counterclockwise. Unless this is from a class or textbook in which you only consider counterclockwise angles, or if the question specifies it, both lines would be a valid solution to the given problem; they pass through the given point and they make a $45°$ angle with the given line.
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