How to find the equation of the bisector
Let there be given two intersecting lines, given their equations. Required to find the equation of the line which passes through the intersection point of these two lines would be divided exactly in half the angle between them, then there would be a bisector.
Instruction how to find the equation of the bisector
Suppose that the lines are defined by their canonical equations. Then A1x + B1y + C1 = 0 and A2x + B2y + C2 = 0. In this case, A1 / B1 ≠ A2 / B2, otherwise the lines are parallel and the problem does not make sense.
Since it is obvious that the two intersecting lines form between them the four pairs of equal angles, there must be exactly two lines that meet the condition of the problem.
These lines are perpendicular to each other. The proof is quite simple. Amount of the four corners formed by intersecting lines, is always equal to 360 °. Since the angles are pairwise equal, then this sum can be expressed as: 2a + 2b = 360 ° or, obviously, a + b = 180 °. Because the first of the required bisectors bisects the angle a, and the second - the angle b, the angle between the bisectors themselves always equal to a / 2 + b / 2 = (a + b) / 2 = 90 °.
Bisected by definition divides the angle between the lines in half, which means that for every point on it, the distance from the two lines are the same.
If the line is given canonical equation, the distance from it to some point (x0, y0), not lying on that line: d = | (Ax0 + By0 + C) / (√ (A ^ 2 + B ^ 2)) |. Consequently, for any point along the bisector required: | (A1 * x + B1 * y + C1) / √ (A1 ^ 2 + B1 ^ 2) | = | (A2 * x + B2 * y + C2) / √ (A2 ^ 2 + B2 ^ 2) |.
Due to the fact that on both sides of the module are signs once it describes both direct desired. Or with a sign to turn it into the equation only one of the bisectors, to open a module or with a + -. Thus, the equation of the first bisector: (A1 * x + B1 * y + C1) / √ (A1 ^ 2 + B1 ^ 2) = (A2 * x + B2 * y + C2) / √ (A2 ^ 2 + B2 ^ 2). The equation of the second bisector: (A1 * x + B1 * y + C1) / √ (A1 ^ 2 + B1 ^ 2) = - (A2 * x + B2 * y + C2) / √ (A2 ^ 2 + B2 ^ 2) .
For example, let there be given direct, certain canonical equations: 2x + y -1 = 0, x + 4y = 0. The equation of the bisector of the first is obtained from the equation: (2x + y -1) / √ (2 + 1 ^ 2 ^ 2) = (x + 4y + 0) / √ (1 + 4 ^ 2 ^ 2), that is (2x + y - 1) / √5 = (x + 4y) / √15. Expanding the brackets and translating the equation in canonical form: (2 * √3 - 1) * x + (√3 - 4) * y - √3 = 0.