The vertices of #\Delta ABC# are #A(3, 4)#, #B(2, 5)# & #C(2, 1)#
The area #\Delta# of #\Delta ABC# is given by following formula
#\Delta=1/2|3(5-1)+2(1-4)+2(4-5)|#
#=2#
Now, the length of side #AB# is given as
#AB=\sqrt{(3-2)^2+(4-5)^2}=\sqrt2#
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=1/2(CN)(\sqrt2)#
#CN=2\sqrt2#
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{5-4}{2-3}=-1#
#a=b+1\ ............(1)#
Now, the length of altitude CN is given by distance formula
#\sqrt{(a-2)^2+(b-1)^2}=2\sqrt2#
#(b+1-2)^2+(b-1)^2=(2\sqrt2)^2#
#(b-1)^2=4#
#b=3, -1#
Setting these values of #b# in (1), we get the corresponding values of #a# as follows
#a=3+1=4\ \ & \ \ \ a=-1+1=0#
#a=4, 0#
The endpoints of altitude from vertex #C(2, 1)# are #(4, 3)#
& #(0, -1)# But #(0, -1)# is not the end point of altitude.
hence, the end points of altitude from vertex #C# are
#(2, 1), (4, 3)#
