Question

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

Answer 1

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(a,b), B(-a,-b), C(c,d)`. 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(x,y)=(ax + by, ay - b x)`

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

`A'=R(a,b) = (a^2 + b^2, 0)`
`B'=R(-a,-b)=(-a^2-b^2,0)`
`C'=R(c,d)=(ac+bd,ad-bc)`

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 `(p,q)` and `(r,s)` is `(y-q)(r-p)=(x-p)(s-q)`.

`A'C': p=a^2+b^2,q=0, r=ac+bd, s=ad-bc`

Line: `y(ac+bd-a^2-b^2)=(x-a^2-b^2)(ad-bc)`

The y intercept is when `x = 0`.

`y = frac{(a^2+b^2)(ad-bc)} {a^2+b^2-(ac+bd)}`

For `B'C'` it's clearly

`y = -frac{(a^2+b^2)(ad-bc)} {a^2+b^2 + (ac+bd)}`

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

The inverse transformation to R is

`S(x,y)=frac{ (ax-by, ay+bx) }{a^2+b^2 }`

`S(R(c,d))= S(ac+bd,ad-bc)`

`=frac{ (a(ac+bd)-b(ad-bc),a(ad-bc)+b(ac+bd) ) }{a^2+b^2}`

`=frac{ (a(ac+bd)-b(ad-bc),a(ad-bc)+b(ac+bd) ) }{a^2+b^2}`

`=frac{ (a^2 c+b^2 c, a^2 d + b^2 d) }{a^2+b^2}`

`=(c,d)`

Let's map our y intercept back:

`S(0, Y) = frac{( -bY, aY )}{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.