The vertices of #\Delta ABC# are #A(2, 3)#, #B(5, 8)# & #C(3, 4)#
The area #\Delta# of #\Delta ABC# is given by following formula
#\Delta=1/2|2(8-4)+5(4-3)+3(3-8)|=1#
Now, the length of side #AB# is given as
#AB=\sqrt{(2-5)^2+(3-8)^2}=\sqrt34#
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)#
#1=1/2(CN)(\sqrt34)#
#CN=2/\sqrt34#
Let #N(a, b)# be the foot of altitude CN drawn from vertex #C(3, 4)# 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-4}{a-3}\times \frac{8-3}{5-2}=-1#
#b=\frac{29-3a}{5}\ ............(1)#
Now, the length of altitude CN is given by distance formula
#\sqrt{(a-3)^2+(b-4)^2}=2/\sqrt34#
#(a-3)^2+(\frac{29-3a}{5}-4)^2=(2/\sqrt34)^2#
#a=46/17, 56/17#
Setting the above values of #a# in (1), the corresponding values of #b# respectively are
#b=71/17, 65/17#
hence, the end points of altitude #CN# are
#(46/17, 71/17), (56/17, 65/17)#
