A triangle has corners at #(2 , 4 )#, ( 3, 1 )#, and #( 8, 3 )#. What are the endpoints and lengths of the triangle's perpendicular bisectors?

Question

1845 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-20T18:00:27

One end-pt. of all perp. bsctrs. is #O(169/34,113/34)# & other end. pts. are #D(11/2,2),E(5,7/2), & F(5/2,5/2).#

Length of one perp.bsctr. #OD# is #sqrt(2349)/34.#

Let us name the vertices of #Delta# as #A(2,4),B(3,1),C(8,3)# & let the mid-pts. of sides #BC,CA,AB# be #D,E,F# resp.

Clearly, the mid-pts. are #D(11/2,2),E(5,7/2), & F(5/2,5/2).#

We know that three perp. bsctrs. of sides of a #Delta# are concurrent at a pt., known as the Circumcentre of #Delta ABC.# Let us call it #O.#

To find #O#, we find the eqns. of two perp.bstrs., namely, #OD & OE.#

Eqn. of #OD#:-

#OD# is perp. to #BC#, & slope of #BC# is #(3-1)/(8-3)=2/5,# so, slope of #OD# must be #-5/2#. In addition, #D in OD.#

#:.# eqn. of #OD# is, #y-2=-5/2(x-11/2),# or, #4y-8=-5(2x-11)=-10x+55,# i.e., #10x+4y=63...........(1).#

On the same line, we can work out the Eqn. of #OE# as #12x-2y=53......(2)#

Solving (1) & (2), we get #O(169/34,113/34)#

Length #OD =sqrt{(169/34-11/2)^2+(113/34-2)^2}=sqrt(2349)/34#