How do you find the equations of the lines that intersect a certain line at a $45^\circ$ given that it passes through a certain point? Let's say there is a line $L$, with an equation $y=mx+b$. We also know that there is a point with coordinates $(p,q)$, which we can call as point $Q$. What are the equations of the lines that intersect line $L$ at a $45^\circ$ and pass through point $Q$(any form is fine, but slope-intercept form would be preferable)?
I was thinking to first find the slope of these lines, and then use the point-slope equation for each slope to get the final equations.
MY APPROACH:
I first found the slope of the final lines. I did so by first finding the y-intercept of $L$, which $(0,b)$. Let's denote this point as $M$. I then drew a circle with center $M$ and radius $1$. I then drew a line perpendicular to $L$ that passed threw $M$. Let's say this line is $P$. I then found the intersection of circle $M$ and the line $P$, which I denoted as $I$. I then drew a circle with center $I$ and radius $1$. I also drew a line parallel to $L$ that passed through $I$. I then found the intersection between this line and circle $I$, which I denoted as $A$ and $B$. I then found the slope of $\overline {AM}$ and $\overline{BM}$, which I will call $m_a$ and $m_b$, respectively. I finally used the point-slope equation with slope $m_a$ and $m_b$ and point $Q$ to find the final equations. Below is an example graph:

Graph Link
Is this the correct approach? Are there other ways to do this? Thanks in advance.

That certainly works, but for this problem I think it’s much more convenient to work with the point-normal form of equation of a line: the line through $P=(x_p,y_p)$ with normal $(a,b)$ has the equation $a(x-x_p)+b(y-y_p)=0$.  
The lines that you are tasked with finding are the angle bisectors of the lines through $Q$ that are parallel and perpendicular to $L$. Writing the equation of $L$ as $mx-y+b=0$, we can see that its normal is $(m,-1)$, so an equation of the parallel line through $Q$ is $m(x-p)-(y-q)=0$ and an equation of the perpendicular is $(x-p)+m(y-q)=0$. Using a fairly well-known formula, the equations of their angle bisectors are then $m(x-p)-(y-q)=\pm((x-p)+m(y-q))$, which you can rearrange as desired.  
If you don’t happen to know the formula for angle bisectors of a pair of lines, you can proceed by first finding the normals to the two bisectors: they will be angle bisectors of the normals of the parallel and perpendicular lines through $Q$. For this, we don’t need to know their complete equations. We already know that the normal to the parallel line is $(m,-1)$, so the perpendicular normal is $(1,m)$. These vectors have equal lengths, so their angle bisectors are simply their sum and difference, namely, $(m+1,m-1)$ and $(1-m,1+m)$. Using the point-slope form from above, the equations of the two 45-degree lines are $(m+1)(x-p)+(m-1)(y-q)=0$ and $(1-m)(x-p)+(1+m)(y-q)=0$. I’ll leave verifying that the equations derived using these two methods are equivalent to you.
