A triangle has corners A, B, and C located at #(5 ,5 )#, #(3 ,9 )#, and #(4 , 1 )#, respectively. What are the endpoints and length of the altitude going through corner C?

#A(5,5) , B(3,9) , C(4,1)#

Let #CD# be the altitude going through #C# touches #D# on line

#AB#. #C# and #D# are the endpoints of altitude #CD; CD# is

perpendicular on #AB#. Slope of #AB= m_1= (y_2-y_1)/(x_2-x_1)#

#=(9-5)/(3-5) = 4/(-2) =-2 :. # Slope of

#CD=m_2= -1/m_1= 1/2 #

Equation of line #AB# is # y - y_1 = m_1(x-x_1) #or

# y- 5 = -2(x-5) or 2 x+y = 15 ; (1) #

Equation of line #CD# is # y - y_3 = m_2(x-x_3)# or

#y- 1 = 1/2(x-4) or 2 y-2= x-4 or 2 y - x =-2 ;(2)#

the co-ordinates of #D(x_4,y_4)#. Mutiplying equation (2) by #2#

we get #-2x+4y=-4 ; (3)# , adding equation (1) from

equation (3) we get #5 y=11 or y=11/5=2.2#

# :. x=(15-y)/2=(15-2.2)/2=6.4 :. D# is # (6.4,2.2,)#.

The end points of altitude #CD# is #(4,1) and (6.4,2.2)#

Length of altitude #CD# is

#CD = sqrt((x_3-x_4)^2+(y_3-y_4)^2) # or

#CD = sqrt((4-6.4)^2+(1-2.2)^2)= sqrt 7.2 ~~ 2.68# unit [Ans]

Let #triangleABC " be the triangle with corners at"#

#A(5,5), B(3,9) and C(4,1)#

Let #bar(AL) , bar(BM) and bar(CN)# be the altitudes of sides #bar(BC) ,bar(AC) and bar(AB)# respectively.

Let #(x,y)# be the intersection of three altitudes

Slope of #bar(AB) =(5-9)/(5-3)=-4/2=-2#

#bar(AB)_|_bar(CN)=>#slope of # bar(CN)=1/2# ,

# bar(CN)# passes through #C(4,1)#

#:.#The equn. of #bar(CN)# is #:y-1=1/2(x-4)#

#=>2y-2=x-4#

#=>x-2y=2#

#i.e. color(red)(x=2y+2.....to (1)#

Now, Slope of #bar(AB) =-2# and #bar(AB)# passes through
#A(5,5)#

So, eqn. of #bar(AB)# is: #y-5=-2(x-5)#

#=>y-5=-2x+10#

#=>color(red)(y=15-2x...to(2)#

From #(1)and (2)#

#y=15-2(2y+2)=15-4y-4=11-4y#

#=>y+4y=11=>5y=11=>color(blue)(y=11/5#

From #(1) ,#

#x=2(11/5)+2=>color(blue)(x=32/5#

#=>N(32/5,11/5) and C(4,1)#

Using Distance formula,

#CN=sqrt((32/5-4)^2+(11/5-1)^2)=sqrt(144/25+36/25)#
#:.CN=sqrt(180/25)#

#:.CN=sqrt(36/5)~~2.68#