A triangle has vertices at #A(a,b )#, #C(c,d)#, and #O(0,0)#. What are the endpoints and length of the perpendicular bisector of AC ?

Question

A triangle has vertices at #A(a,b )#, #C(c,d)#, and #O(0,0)#.  What are the endpoints and length of the perpendicular bisector of AC ?

Evelyn Arboleda · Accepted Answer

#t=(a-c)^2+(b-d)^2#

#p = 1/t ( b(b+d)-a(a+c))#

#q = 1/t (ad-bc)#

If #ad-bc = 2t,# i.e #q=1/2,# the bisector goes through #O#.

If #q<1/2# the bisector intersects OA at# ((x),(y))=((c-a,b-d),(d-b,c-a)) (( 1/2 - p),( q/{2p} - q )) #

If #q>1/2# the bisector intersects OB at # ((x),(y))=((c-a,b-d),(d-b,c-a)) (( 1/2 - p),( q/{2-2p} - q )) # I'm looking for a simpler method that I can actually finish because my other answer was featured before I finished it. Let's start with a simpler problem. Triangle OPI, coordinates #O(0,0), P(p,q), I(1,0)#. What are the endpoints and lengths of the perpendicular bisector of # OI?# Solution: The foot of the bisector is the midpoint of #OI,# #F(1/2, 0).# There are three possibilities: If #q=1/2# the bisector goes through #P#. If #q <1/2# the bisector intersects OP at #Q(1/2 , q/{2p}).# That's the intersection of #x=1/2# with #py=qx.# If #q>1/2# the bisector intersects IP #(p-1)y=q(x-1)# at #(1/2, q/{2-2p})# That seems simple enough. We want to map #AOC# to # OPI#. That's #A(a,b) to (0,0),# #C(c,d) to (1,0)# and #O(0,0) to P(p,q)# The standard translation, scaling, and rotation is #x' = rx + sy + p# #y' =-sx + r y + q # We mapped O to P. Unknowns #p,q,r,s# The remaining equations are #0 = ar + bs + p# #0 = -as + br + q # #1 = cr + ds + p# #0 = -sc + rd + q # Let #t=(a-c)^2+(b-d)^2.# We know that squared distance is one in the translated space, so that's the denominator. I'll skip the tedium. #p = 1/t ( b(b+d)-a(a+c))# #q = 1/t (ad-bc)# # r = 1/t(c-a)# # s = 1/t(d-b)# The remaining difficult issue is the inverse mapping so we can get the other endpoints #Q(1/2 , q/{2p})# and #Q'(1/2 , q/{2-2p}).# Sorry to devolve into matrices for a minute. #((x' - p),( y' - q )) = ((r,s),(-s,r))((x),(y))# The nearly opposite is #((r,s),(-s,r)) ((r,-s),(s,r))=((r^2+s^2, 0),(0, r^2+s^2))=1/t((1,0),(0,1))# # ((x),(y)) = t ((r,-s),(s,r)) ((x' - p),( y' - q )) # # ((x),(y)) = ((c-a,b-d),(d-b,c-a)) ((x' - p),( y' - q )) # #Q(1/2 , q/{2p})# maps to # ((c-a,b-d),(d-b,c-a)) (( 1/2 - p),( q/{2p} - q )) # #Q(1/2 , q/{2-2p})# maps to # ((c-a,b-d),(d-b,c-a)) (( 1/2 - p),( q/{2-2p} - q )) # Modifying our initial response, #t=(a-c)^2+(b-d)^2# #p = 1/t ( b(b+d)-a(a+c))# #q = 1/t (ad-bc)# If #ad-bc = 2t,# i.e #q=1/2,# the bisector goes through #O#. If #q<1/2# the bisector intersects OA at# ((x),(y))=((c-a,b-d),(d-b,c-a)) (( 1/2 - p),( q/{2p} - q )) # If #q>1/2# the bisector intersects OB at # ((x),(y))=((c-a,b-d),(d-b,c-a)) (( 1/2 - p),( q/{2-2p} - q )) #

Violet Aldrich · Answer

Hey, this got featured before it was really done. It's still not done.

This is the second in a series of questions I've been asking that I generalize from earlier. The phrase "the triangle's perpendicular bisectors" is not one that is used very often. The perpendicular bisector through the midpoint is the one that intersects each side; it is not always the altitude of an isosceles triangle, but it will occasionally intersect the opposite vertex. The question has been rewritten to place one vertex at the origin and only request the opposite side's perpendicular bisector. The triangle's sides are provided parametrically as #OA: quad quad # #(x,y) = v(a,b) quad quad quad quad 0 le v le 1# #OC: quad quad # #(x,y) = w(c,d) quad quad quad quad 0 le w le 1# #AC: quad quad # #(x,y) = (a,b)+u(c-a,d-b) quad quad quad quad 0 le u le 1# The perpendicular family to #AC# has direction vector given by swapping coordinates and negating one; the bisector is through the midpoint of AC: #(x,y) = 1/2 (a+c, b+d) + t(d-b,a-c) quad quad quad # for real #t# That meets #OA # when #v(a,b)= 1/2 (a+c, b+d) + t(d-b,a-c) # There are two unknowns in two equations. #va = (a+c)/2+t(d-b)# #vb = (b+d)/2+t(a-c)# We're more interested in #v# than #t#: # 2va(a-c) + 2t(b-d)(a-c) = (a+c)(a-c)# #2vb(b-d) - 2t(a-c)(b-d) = ( b+d)(b-d)# # v = 1/2 cdot { (a^2+b^2) - (c^2 + d^2) }/{a^2+b^2 -(ac+bd)} # The roles aren't quite symmetrical so we repeat the process for the meet of the bisector and #OC#. #w(c,d) = 1/ 2(a+c, b+d) + t(d-b,a-c)# #2wc(a-c) = (a+c)(a-c) - 2t(b-d)(a-c) # #2wd(b-d) = (b+d)(b-d) + 2t(a-c)(b-d)# # w = 1/2 cdot {(a^2+b^2) - (c^2 +d ^2) }/{(ac+bd)-(c^2+d^2)}# Yes, Socratic, I realize this is getting a bit long. This is a complicated issue. The numerators are the same in #v# and #w#. We need to show that either #v=w=0# (an isosceles triangle where the perpendicular bisector is an altitude, here to the origin) or exactly one of #0 < v < 1# or #0 < w <1.# ~~~~~~~ # # We want to solve #v>0# # { (a^2+b^2) - (c^2 + d^2) }/{a^2+b^2 -(ac+bd)} > 0# Let's first assume # a^2+b^2 > c^2+d^2,# positive numerator, so we need a positive denominator. # a^2+b^2 > ac+bd # # |A|^2 > A cdot C = |A| \ | C| \cos AOC # # |A| > |C| \ cos AOC # We're assuming #|A|>|C|# and the cosine is never bigger than one so we've shown #|A|>|C| implies v>0.# What about when # a^2+b^2 < c^2+d^2,# or #|A|<|C|#? For #v>0# we need a negative denominator. # a^2+b^2 < ac+bd # # |A|^2 < A cdot C = |A| \ | C| \cos AOC # # |A| < |C| \ cos AOC # # cos AOC > |A|/|C| # How about #w>0#? # w = 1/2 cdot {|A|^2-|C|^2}/{ A cdot C - |C|^2 }# For #|A|>|C|# we need #A cdot C > |C|^2# or #|A| \ |C| \ cos AOC > |C|^2 # # cos AOC > |C|/|A| # Unless #|A|=|C|# exactly one of those fractions #|C| /|A| or |A|/|C|# is in cosine range, between -1 and 1. Not there yet. ~~~~~~~~~ The denominators must be demonstrated to have opposite signs. # ( (a^2+b^2) -(ac+bd) ) ( (ac+bd)-(c^2+d^2) ) # #= (a^2+b^2 + c^2 + d^2)(ac+bd) + (ac+bd)^2 + (a^2+b^2-c^2-d^2)(ac+bd) # #= (ac+bd)( a^2+b^2 + c^2 + d^2 + (ac+bd) + a^2+b^2-c^2-d^2) # =(ac+bd)( 2a^2 + 2b^2 + (ac+bd) ) ~ ~~~~~~ According to the triangle inequality #|AC| > |OA| + |OC|# #|AC|^2 > |OA|^2 + |OC|^2 + 2|OA| \ |OC|# # (a-c)^2+(b-d)^2 > a^2+b^2+c^2+d^2+ 2sqrt{(a^2+b^2)(c^2 + d^2)} # # (a-c)^2+(b-d)^2 -( a^2+b^2+c^2+d^2) > 2sqrt{(a^2+b^2)(c^2 + d^2)} # #-2 ac -2bd > 2sqrt{(a^2+b^2)(c^2 + d^2)} # #ac-bd > 2sqrt{(a^2+b^2)(c^2 + d^2)} # The dot product and the sum of squares have a close relationship. We expand #|AC|^2:# # |AC|^2 = (a-c)^2+(b-d)^2 = a^2+b^2 + c^2 + d^2 - 2(ac+bd) # # (ac+bd) - (c^2+d^2) = (a^2+b^2)- (ac+bd) - |AC|^2 # # w = 1/2 cdot {|A|^2+|C|^2}/{ |A|^2 - Acdot C - |AC|^2 }# # v = 1/2 cdot {|A|^2+|C|^2 }/{|A|^2 -A cdot C) # We still haven't shown we pass through at most one side or the vertex. It remains to show #-1 < v < 1# which ensures that the bisector passes through exactly one side. There is still work to be done.

Parker Allen · Answer

undefined The midpoint of AC is M, where $ M\left(\frac{a+c}{2}, \frac{b+d}{2}\right) $.  
The slope of AC is $ \frac{d-b}{c-a} $, so the slope of the perpendicular bisector of AC is $ -\frac{c-a}{d-b} $.  
Since the perpendicular bisector passes through M, we use the point-slope form of a line to find the equation of the perpendicular bisector:  
$ y - \frac{b+d}{2} = -\frac{c-a}{d-b} (x - \frac{a+c}{2}) $.  
To find the endpoints of the perpendicular bisector, substitute $ x = 0 $ and $ y = 0 $ into the equation.  
The length of the perpendicular bisector is twice the distance from its midpoint (M) to either of its endpoints.