The vertices of #\Delta ABC# are #A(3, 5)#, #B(2, 1)# & #C(5, 8)#
The area #\Delta# of #\Delta ABC# is given by following formula
#\Delta=1/2|3(1-8)+2(8-5)+5(5-1)|=2.5#
Now, the length of side #AB# is given as
#AB=\sqrt{(3-2)^2+(5-1)^2}=\sqrt17#
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)#
#2.5=1/2(CN)(\sqrt17)#
#CN=5/\sqrt17#
Let #N(a, b)# be the foot of altitude CN drawn from vertex #C(5, 8)# 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-8}{a-5}\times \frac{5-1}{3-2}=-1#
#b=\frac{37-a}{4}\ ............(1)#
Now, the length of altitude CN is given by distance formula
#\sqrt{(a-5)^2+(b-8)^2}=5/\sqrt17#
#(a-5)^2+(\frac{37-a}{4}-8)^2=(5/\sqrt17)^2#
#a=105/17, 65/17#
Setting above values of #a# in (1), the corresponding values of #b# respectively are
#b=131/17, 141/17#
hence, the end points of altitude #CN# are
#(105/17, 131/17), (65/17, 141/17)#
