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.

Given that it is two years old, why does this appear on "just asked"?

I will attempt to improve upon my recent lengthy, meandering partial response to one similar to this one.

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.

Since I'm not familiar with matrices, let's just focus on the mapping.

#R(x,y)=(ax + by, ay - b x)#

Just so you know, that's a cross product and a dot product. Now let's examine 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)#

The question has an obvious answer because in the transformed space, the y axis is the perpendicular bisector of AB. Depending on the sign of ac+bd, the bisector hits either B'C' (positive) or A'C' (negative).

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 opposite conversion 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)#

Now let's map our intercept back to y:

# S(0, Y) = frac{( -bY, aY )}{a^2+b^2 } #
The denominator cancels the factor in the numerator of #Y#.

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

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Answer 2

The endpoints of the perpendicular bisectors are as follows:

  1. For the line passing through the midpoint of the segment between (2, 5) and (1, 3): Endpoint 1: (3.5, 4) Endpoint 2: (4, 4)

  2. For the line passing through the midpoint of the segment between (2, 5) and (8, 1): Endpoint 1: (5, 3) Endpoint 2: (5, 3)

  3. For the line passing through the midpoint of the segment between (1, 3) and (8, 1): Endpoint 1: (4.5, 2) Endpoint 2: (4.5, 2)

The lengths of the perpendicular bisectors can be calculated using the distance formula between the endpoints.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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