The vertices of #\Delta ABC# are #A(7, 6)#, #B(8, 3)# & #C(2, 1)#
The area #\Delta# of #\Delta ABC# is given by following formula
#\Delta=1/2|7(3-2)+8(1-6)+2(6-3)|#
#=27/2#
Now, the length of side #AB# is given as
#AB=\sqrt{(7-8)^2+(6-3)^2}=\sqrt10#
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)#
#27/2=1/2(CN)(\sqrt10)#
#CN=27/\sqrt10#
Let #N(a, b)# be the foot of altitude CN drawn from vertex #C(2, 1)# 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-1}{a-2}\times \frac{6-3}{7-8}=-1#
#a=3b-1\ ............(1)#
Now, the length of altitude CN is given by distance formula
#\sqrt{(a-2)^2+(b-1)^2}=27/\sqrt10#
#(3b-1-2)^2+(b-1)^2=(27/\sqrt10)^2#
#(b-1)^2=27^2/10^2#
#b=37/10, -17/10#
Setting these values of #b# in (1), we get the corresponding values of #a#
#a=101/10, -61/10#
hence, the end points of altitudes are
#(101/10, 37/10), (-61/10, -17/10)#
