A triangle has corners A, B, and C located at #(6 ,1 )#, #(2 ,3 )#, and #(3 , 2 )#, respectively. What are the endpoints and length of the altitude going through corner C?

1 Answer
Jul 4, 2016

#"End points "(3,2) & (16/5,12/5)#

#"Altitude"=1/sqrt5#

Explanation:

Given the coordinates of three corners of #DeltaABC#

#A->(6,1)#

#B->(2,3)#

#C->(3,2)#

Let CD be the perpendicular dropped from C on AB at D.

We are to find out the length of CD and coordinate of D.

Let the coordinate of D be (h,k)

#"Now slope of AB"=(1-3)/(6-2)=-1/2#

So slope of CD perpendicular to AB is 2

But the slope CD considering its end points as (h,k) and (3,2) is

#(k-2)/(h-3)#

So

#(k-2)/(h-3)=2#

#=>2h-6=k-2#

#=>2h-k=4.....(1)#

Now equation of AB is

#y-3=-1/2(x-2)#

#2y-6=-x+2#

#x+2y=8#

This equation will be satisfied by (h,k)

So#" "h+2k=8....(2)#

Now multiplying (1) by 2 and then adding with (2) we get

#4h-2k+h+2k=2*4+8#

#5h=16=>h=16/5#

Putting the value of h in (1)

#2*16/5-k=4#

#k=32/5-4=12/5#

So coordinate of D is

#D->(16/5,12/5)#

Length of the altitude CD

#=sqrt((16/5-3)^2+(12/5-2)^2)#

#=sqrt((1/5)^2+(2/5)^2)#

#=sqrt(1/25+4/25)#

#=sqrt(5/25)#

#=1/sqrt5#