A triangle has corners A, B, and C located at #(2 ,7 )#, #(7 ,4 )#, and #(1 , 2 )#, respectively. What are the endpoints and length of the altitude going through corner C?

Answer 1

Endpoints of altitude are #(59/17,104/17), (1,2)#
Length of altitude #=sqrt(392/17)~~4.802#

As shown in the diagram, line #CD# is the altitude perpendicular to line #AB# from point #C#

Given #A(2,7), B(7,4), C(1,2)#,
slope of line #AB = (4-7)/(7-2)=-3/5#
equation of line #AB# is : #y-4=-3/5(x-7)#
#=> 5y=-3x+41# ..... (1)

Let slope of line #AB# be #s_1#, and slope of line #CD# be #s_2#
As line #CD# is perpendicular to line #AB#, #=> s_1*s_2=-1#
#=> -3/5*s_2=-1, => s_2=5/3#
So slope of line #CD= 5/3#
equation of line #CD# is : #(y-2)=5/3(x-1)#
#=> 3y=5x+1# ..... (2)

Solving (1) and (2) we get #x=59/17, y=104/17#
Hence, endpoints of the altitude are #(59/17, 104/17) and (1,2)#

length of altitude #=sqrt((59/17-1)^2+(104/17-2)^2#

#= sqrt(392/17)~~4.802#

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