Angle Bisector of Parallel Lines

Question

Today I came across a question in which equations of two lines (Which were parallel) were given and it was asked to find their angle bisector.

My answer for this was :

Since there is no point of intersection of Parallel lines, there is no origin of angle bisector. So, answer should be Doesn't Exist, obviously.

But when I checked the answer it was the line equidistant (and parallel) from both of these two i.e. if lines are $ax+by+c_1=0$ and $ax+by+c_2=0$ than angle bisector will be $ax+by+\frac{c_1+c_2}{2}=0$

I am asking this question since I feel that I am not wrong here, bisector shouldn't exist. Can someone please confirm.

Someone may say this is the extension of the property of angle bisector that each point of angle bisector is equidistant from the original lines but I am asking what is defined to be angle bisector ?

For example : $\binom{n}{r}=0$ when $n<r$ is an extension of property of binomial coefficients. But originally $n<r$ isn't in domain of this function.

Answer 1

In Projective Geometry two parallel lines intersect at the infinity point. If you then define the angle bisector as a line through this intersection point, that has the same angle to both of the other lines, every parallel line will be a such (so not only the equidistant line, but it is a possible angle bisector). So principially the answer given by your test is a angle bisector, but it not a unique one.

Hope this helps!

Answer 2

Your objection is valid.

Unless there's some special definition in force (which is why I asked for the textbook), there's no vertex, hence no angle, hence no angle bisector.

Thus, assuming the standard definition, the answer you quoted is simply wrong.

Answer 3

In a similar question, Equation of angle bisector, given the equations of two lines in 2D, someone was given a task similar to yours, except that in their case it was explicitly specified that if the lines were parallel the line halfway between them should be given as the answer.

In an answer to the same question, a formula was given for the angle bisectors of any two intersecting lines with equations \begin{align} a_1x + b_1y + c_1 &= 0, \\ a_2x + b_2y + c_2 &= 0. \end{align}

To write the answer in a more compact format, let \begin{align} q_1 = \sqrt{a_1^2 + b_1^2}, \\ q_2 = \sqrt{a_2^2 + b_2^2}. \end{align} Then the equations of the two angle bisectors are \begin{align} (a_1 q_2 + a_2 q_1)x + (b_1 q_2 + b_2 q_1)y + c_1 q_2 + c_2 q_1 &=0, \tag1\\ (a_1 q_2 - a_2 q_1)x + (b_1 q_2 - b_2 q_1)y + c_1 q_2 - c_2 q_1 &=0. \tag2 \end{align} There are two such lines because the original two lines form two pairs of vertical angles and each of the bisectors bisects just one pair of angles. The two angle bisectors are perpendicular to each other.

In the case of two parallel lines, \begin{align} ax + by + c_1 &=0, \\ ax + by + c_2 &=0, \end{align} Equation $2$ has zero coefficients for both $x$ and $y$ (and therefore no longer describes a line), while Equation $1$ becomes $$ 2ax + 2by + c_1 + c_2 = 0, \tag3 $$ which is the equation for a line midway between the two given lines.

I find this to be an interesting "limit" property, but not a justification for the answer given for your practice problem.

In my opinion it is misleading to call the answer for parallel lines an "angle bisector," and the problem should have been posed in the manner of Equation of angle bisector, given the equations of two lines in 2D instead. But it may be customary on the exam you're preparing for that the "midline if the lines are parallel" clause is implicitly understood to be part of any angle bisector question. I would regard this as a quirk of the exam--a very bad quirk in my opinion, adding a completely unnecessary reason why one would need coaching for such exams, but that's a complaint for another forum, perhaps. (You probably have little choice at this time other than to accept the existence of such quirks and learn to deal with them.) I would not regard this quirk as an application of analytic geometry as most practitioners understand it.