#A(4,3) , B(7,4) , C(6,5)#
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)#
#=(4-3)/(7-4) = 1/3 :. # Slope of #CD=m_2= -1/m_1= -3 #
Equation of line #AB# is # y - y_1 = m_1(x-x_1) #or
# y- 3 = 1/3(x-4) or 3y-9 = x-4 or x-3y = -5 ; (1) #
Equation of line #CD# is # y - y_3 = m_2(x-x_3)# or
#y- 5 = -3(x-6) or 3x+y=23 ; (2) # )# Sollving equation
(1) and (2) we get the co-ordinates of #D(x_4,y_4)#.
Mutiplying equation (1) by #3# we get #3x-9y=-15 ; (3)#
Subtracting equation (3) from equation (2) we get
#10y=38 or y=3.8 :. x= -5+3y=-5+11.4# or
#x= 6.4 :. (x_4,y_4) = (6.4,3.8)# The end points of altitude are
#CD# is #(6,5) and (6.4,3.8)# . Length of altitude #CD# is
#CD = sqrt((x_3-x_4)^2+(y_3-y_4)^2) # or
#CD = sqrt((6-6.4)^2+(5-3.8)^2)= sqrt1.6 ~~ 1.26# unit [Ans]
