A triangle has corners A, B, and C located at #(1 ,1 )#, #(3 ,4 )#, and #(2 , 2 )#, respectively. What are the endpoints and length of the altitude going through corner C?

1 Answer
May 12, 2017

Endpoints of perpendicular bisector are #D(23/13,28/13)# and #C(2,2)# and length of altitude is #sqrt13/13#

Explanation:

One of the endpoint is clearly #C(2,2)#, while other say #D# will lie on #AB#. It is apparent that #CD# and #AB# are perpendicular to each other.

Hence product of their slopes #m_(AB)# and #m_(CD)# would be #-1# and solving for equations of line #AB# and #CD# will give us coordinates of #D#.

Let us first find the equation of #AB#. As it's slope is given by #m_(AB)=(4-1)/(3-1)=3/2# and equation of #AB# using point slope form of equation, would be

#y-1=3/2(x-1)# i.e. #2y-2=3x-3# or #3x-2y=1# .............(1)

Now slope of #CD# is given by #m_(CD)=(-1)/m_(AB)=(-1)/(3/2)=-2/3#

and as #CD# passes through #C(2,2)#, its equation is

#y-2=-2/3(x-2)# or #3y-6=-2x+4# or

#2x+3y=10# ...........(2)

For Solving equations (1) and (2), let us multiply equation (1) by #3# and (2) by #2# and adding them we get

#9x-6y+4x+6y=3+20# or #13x=23# and hence #x=23/13#

Putting this in (2), we get #2xx23/13+3y=10# or #3y=10-46/13=84/13#

i.e. #y=28/13#

and hence coordinates of #D# are #(23/13,28/13)#

So endpoints of perpendicular bisector are #D(23/13,28/13)# and #C(2,2)# and length of altitude is

#sqrt((2-23/13)^2+(2-28/13)^2)=sqrt((3/13)^2+(-2/13)^2)#

= #sqrt(9/169+4/169)=sqrt13/13#

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