# A triangle has corners at (2 , 5 ), ( 1, 3 ), and ( 8, 1 )#. What are the endpoints and lengths of the triangle's perpendicular bisectors?

Apr 22, 2018

I didn't finish, but here's some interesting stuff.

#### Explanation:

Why does this come up on "just asked" if it's two years old?

I recently gave a long rambling partial answer to one like this. I'll try to step up my game.

Let's first solve the following problem. We're given a triangle ABC where the midpoint of AB is the origin. We can label the points $A \left(a , b\right) , B \left(- a , - b\right) , C \left(c , d\right)$. Determine the endpoints and length of the perpendicular bisector of AB, OD. D is the point on AC or BC (or possibly the vertex C itself) where the perpendicular bisector meets the other sides of the triangle.

I don't know how to do matrices here. Let's just consider the mapping

$R \left(x , y\right) = \left(a x + b y , a y - b x\right)$

That's a dot product and a cross product, FYI. Let's look at our triangle under R:

$A ' = R \left(a , b\right) = \left({a}^{2} + {b}^{2} , 0\right)$
$B ' = R \left(- a , - b\right) = \left(- {a}^{2} - {b}^{2} , 0\right)$
$C ' = R \left(c , d\right) = \left(a c + b d , a d - b c\right)$

In the transformed space, the perpendicular bisector of AB is
the y axis so the question has an obvious answer. The sign of ac+bd determines if the bisector hits A'C' (negative) or B'C' (positive).

The length of the bisector is the y intercept of the AC or BC as chosen. Let's work them out. The general line through $\left(p , q\right)$ and $\left(r , s\right)$ is $\left(y - q\right) \left(r - p\right) = \left(x - p\right) \left(s - q\right)$.

$A ' C ' : p = {a}^{2} + {b}^{2} , q = 0 , r = a c + b d , s = a d - b c$

Line: $y \left(a c + b d - {a}^{2} - {b}^{2}\right) = \left(x - {a}^{2} - {b}^{2}\right) \left(a d - b c\right)$

The y intercept is when $x = 0$.

$y = \frac{\left({a}^{2} + {b}^{2}\right) \left(a d - b c\right)}{{a}^{2} + {b}^{2} - \left(a c + b d\right)}$

For $B ' C '$ it's clearly

$y = - \frac{\left({a}^{2} + {b}^{2}\right) \left(a d - b c\right)}{{a}^{2} + {b}^{2} + \left(a c + b d\right)}$

One of those is the length (or the signed length). Let's call it $Y$.

The inverse transformation to R is

$S \left(x , y\right) = \frac{\left(a x - b y , a y + b x\right)}{{a}^{2} + {b}^{2}}$

$S \left(R \left(c , d\right)\right) = S \left(a c + b d , a d - b c\right)$

$= \frac{\left(a \left(a c + b d\right) - b \left(a d - b c\right) , a \left(a d - b c\right) + b \left(a c + b d\right)\right)}{{a}^{2} + {b}^{2}}$

$= \frac{\left(a \left(a c + b d\right) - b \left(a d - b c\right) , a \left(a d - b c\right) + b \left(a c + b d\right)\right)}{{a}^{2} + {b}^{2}}$

$= \frac{\left({a}^{2} c + {b}^{2} c , {a}^{2} d + {b}^{2} d\right)}{{a}^{2} + {b}^{2}}$

$= \left(c , d\right)$

Let's map our y intercept back:

$S \left(0 , Y\right) = \frac{\left(- b Y , a Y\right)}{{a}^{2} + {b}^{2}}$

The denominator cancels the factor in the numerator of $Y$.

That's pretty cool. I'm getting warnings that the answer is too long, so I'm just going to post this without finishing. Some problems are too long to do with a short answer.