#A(6,8) , B(7,4) , C(2,9)#
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)#
#=(4-8)/(7-6) = -4 :. # Slope of #CD=m_2= -1/m_1= 1/4 #
Equation of line #AB# is # y - y_1 = m_1(x-x_1) #or
# y- 8 = -4(x-6) or 4 x+y = 32 ; (1) #
Equation of line #CD# is # y - y_3 = m_2(x-x_3)# or
#y- 9 = 1/4(x-2) or 4 y-36= x-2 or x -4 y= -34 ; (2) #
Solving equation (1) and (2) we get the co-ordinates of
#D(x_4,y_4)#. Multiplying equation (2) by #4# we get
#4 x - 16 y= -136 ; (3)# Subtracting equation (3) from
equation (1) we get # 17 y= 168 or y=168/17#
Putting #y=168/17# in equation (2) we get,
# :. x= -34 + 4*168/17 = 94/17:. D# is # (94/17, 168/17)#.
The end points of altitude are
#CD# is #(2,9) and (94/17, 168/17)# . Length of altitude #CD# is
#CD = sqrt((x_3-x_4)^2+(y_3-y_4)^2) # or
#CD = sqrt((2-94/17)^2+(9-168/17)^2) ~~ 3.638# unit [Ans]
