An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from #(5 ,1 )# to #(3 ,2 )# and the triangle's area is #12 #, what are the possible coordinates of the triangle's third corner?

Refer to the figure below

It's known that:
#P_2(5,1)#, #P_3(3,2)#
#S=12#
#B=C#
It's asked #P_1#

#A=sqrt((5-3)^2+(2-1)^2)=sqrt(5)#
#S=("base"*"height")/2 => 12=(sqrt5*h)/2 => h=24/sqrt5#
#B^2=(A/2)^2+h^2=(sqrt5/2)^2+(24/sqrt5)^2=5/4+576/5=2329/20 => B~=10.791# (this is the distance #P_1P_2# or #P_1P_3#)

Now we could use the equations of the distance between 2 points to determine #P_1#, but since #P_1# is directly above (or below) #M(4,1.5)# we can find #P_1# in the following easier way:

#k_(P_2P_3)=(Deltay)/(Deltax)=(1-2)/(5-3)=-1/2#
line #P_2P_3#: #(y-1)=(-1/2)(x-5) => 2y-2=-x+5 => x+2y-7=0# [1]
Distance between a point and a line
#d=|ax_0+by_0+c|/sqrt(a^2+b^2)#
So, making #d=h#
#24/cancel(sqrt5)=|x+2y-7|/cancel(sqrt5)#
We get two equations:
(I) #24=x+2y-7 => 2y=-x+31 => y=-x/2+15.5#[2]
(II) #-24=x+2y-7 => 2y= -x-17 => y=-x/2-8.5#[3]

Since #P_1# is also in the bisector line of side A, let's find its equation
#p=-1/k_(P_2P_3)=2#
line #P_1M#: #(y-1.5)=2(x-4) => y=2x-8+1.5 => y=2x-6.5#[4]

Finding the two possible answers (by combining [2] and [4] or [3] and [4]):

(I) #2x-6.5=-x/2+15.5 => 2.5x=22=> x=8.8#
#->y=2*8.8-6.5 => y=11.1#
#=> (8.8,11.1)#

(II) #2x-6.5=-x/2-8.5 => 2.5x=-2 => x=-0.8#
#->y=2*(-0.8)-6.5 => y=-8.1#
#=> (-0.8,-8.1)#

Checking the results:
Notice that #(8.8-0.8)/2=4=x_M#
and #(11.1-8.1)/2=1.5=y_M#
Notice also that
(testing result I)#P_1P_2=sqrt(3.8^2+10.1^2)=sqrt116.45~=10.791#
(testing result I)#P_1P_3=sqrt(5.8^2+9.1^2)=sqrt116.45~=10.791#
(Try result II)
As it should be: all checked