A triangle has corners A, B, and C located at #(3 ,5 )#, #(2 ,9 )#, and #(4 , 8 )#, respectively. What are the endpoints and length of the altitude going through corner C?

1 Answer
Apr 21, 2018

Endpoints #(4,8)# and #( 40/17, 129/17) # and length #7/sqrt{17} #.

Explanation:

I am apparently an expert in answering two year old questions. Let's continue.

The altitude through C is the perpendicular to AB through C.

There are a few ways to do this one. We can calculate the slope of AB as #-4,# then the slope of the perpendicular is #1/4# and we can find the meet of the perpendicular through C and the line through A and B. Let's try another way.

Let's call the foot of the perpendicular #F(x,y)#. We know the dot product of the direction vector CF with the direction vector AB is zero if they're perpendicular:

#(B-A) cdot (F - C) = 0#

#(1-,4) cdot (x-4,y-8) = 0#

# x - 4 - 4y + 32 = 0 #

# x - 4y = -28 #

That's one equation. The other equation says #F(x,y)# is on the line through A and B:

#(y - 5)(2-3)=(x-3)(9-5)#

#5 - y = 4(x-3)#

#y = 17 - 4x#

They meet when

#x - 4(17 - 4x) = -28#

# x - 68 + 16 x = -28 #

# 17 x = 40#

# x = 40/17 #

# y = 17 - 4 (40/17) = 129/17 #

The length CF of the altitude is

#h = \sqrt{ (40/17-4)^2 + (129/17 - 8)^2} = 7 /sqrt{17}#

Let's check this by calculating the area using the shoelace formula and then solving for the altitude. A(3,5),B(2,9),C(4,8)

#a = \frac 1 2 | 3(9)-2(5) + 2(8)-9(4) + 4(5)-3(8)| = 7/2#

# AB=sqrt{ (3-2)^2+(9-5)^2 } = sqrt{17}#

#a = \frac 1 2 b h #

# 7/2 = 1/2 h sqrt{17} #

# h = 7/sqrt{17} quad quad quad sqrt #