wasn't exactly sure how to word this in a quick question title, but say I have the following:
Meaning, I have two lines (or line segments) with known slopes and start / end verticies, and one interception point of the two known lines. I now want to calculate the slope of a third line which should protude from the point, but exactly "away" from both of the two lines together.
I'm not exactly sure how to word this, as it's tecnically not perpendicular, since there are two lines involved, but I suppose it would be perpendicular to a theoretical line drawn across the point which would somehow "equify" the two other lines... I don't know if I'm making any sense, but I just don't know exactly how to word this, but basically:
How do i get the slope of a line protruding from the intersection point of two lines, such that the third line's slope will be completely "equal" to both of the other lines, in the sense that it won't lean towards either one; rather, be "going away" from both of them equally?
You can think of the protruding line as an extension of the angle bisector. Wolfram Alpha gives a detailed explanation on how to calculate the slope of an angle bisector here.
In general, the slope of angle bisector $m_a$ is given by:
$$ m_a = \frac{m_1 m_2 - 1 + \sqrt{(m_1^2 + 1)(m_2^2 + 1)}}{m_1 + m_2} $$
where $m_1$ and $m_2$ are slopes of the two lines that form the angle.