The vertices of #\Delta ABC# are #A(1, 8)#, #B(2, 3)# & #C(5, 9)#
The area #\Delta# of #\Delta ABC# is given by following formula
#\Delta=1/2|1(3-9)+2(9-8)+5(8-3)|#
#=21/2#
Now, the length of side #AB# is given as
#AB=\sqrt{(1-2)^2+(8-3)^2}=\sqrt26#
If #CN# is the altitude drawn from vertex C to the side AB then the area of #\Delta ABC# is given as
#\Delta =1/2(CN)(AB)#
#21/2=1/2(CN)(\sqrt26)#
#CN=21/\sqrt26#
Let #N(a, b)# be the foot of altitude CN drawn from vertex #C(5, 9)# to the side AB then side #AB# & altitude #CN# will be normal to each other i.e. the product of slopes of AB & CN must be #-1# as follows
#\frac{b-9}{a-5}\times \frac{3-8}{2-1}=-1#
#a=5b-40\ ............(1)#
Now, the length of altitude CN is given by distance formula
#\sqrt{(a-5)^2+(b-9)^2}=21/\sqrt26#
#(5b-40-5)^2+(b-9)^2=(21/\sqrt26)^2#
#(b-9)^2=21^2/26^2#
#b=255/26, 213/26#
Setting above values of #b# in (1), we get the corresponding values of #a#
#a=235/26, 25/26#
hence, the end points of altitudes are
#(235/26, 255/26), (25/26, 213/26)#
