Let #O(a, b)# be the ortho-center of #\Delta ABC# with the vertices #A(4,3)#, #B(9, 1)# & #C(8, 3)#
Ortho-center of a triangle is the point of intersection of its altitudes drawn from the vertices to opposite sides.
By property of ortho-center, the line #AO# will be perpendicular to the opposite side #BC# i.e. the product of the slopes of lines #AO# & #BC# will be #-1# hence we have
#(\frac{b-3}{a-4})(\frac{3-1}{8-9})=-1#
#a-2b=-2\ ..........(1)#
Similarly, the line #BO# will be perpendicular to the opposite side #AC# i.e. the product of the slopes of lines #BO# & #AC# will be #-1# hence we have
#(\frac{b-1}{a-9})(\frac{3-3}{8-4})=-1#
#a-9=0#
#a=9#
setting #a=9# in (1), we get
#9-2b=-2#
#b=5.5#
hence, the ortho-center of given triangle is
#(9, 5.5)#
