#A(4,7) , B(3,2) , C(2,1)#
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)#
#=(2-7)/(3-4) = 5 :. # Slope of #CD=m_2= -1/m_1= - 1/5#
Equation of line #AB# is # y - y_1 = m_1(x-x_1) # or
# y- 7 = 5(x-4) or 5x-y = 13 ; (1) #
Equation of line #CD# is # y - y_3 = m_2(x-x_3)# or
#y- 1 = -1/5(x-2) or x+5y=7 (2)# Mutiplying equation (2)
by #5# we get # 5x+25y=35 (3)# Subtracting equation (1)
from equation(3) we get , #26y=22 or y =22/26=11/13 #
#:. x= 7-5y=7- 55/13=36/13 :. # The co-ordinates of point
#D# is#(36/13,11/13)# and the coordinates of end point of
altitude is #(2,1) and (36/13,11/13)#
Length of altitude #CD# is
#CD = sqrt((x_3-x_4)^2+(y_3-y_4)^2) # or
#CD = sqrt((2-36/13)^2+(1-11/13)^2)~~0.784# unit [Ans]
