SOLUTION: Triangle ABC has vertices A(-2,3), B(1,-3) and C(p,1). For how many values of p is the triangle isosceles?

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Question 1187102: Triangle ABC has vertices A(-2,3), B(1,-3) and C(p,1). For how many values of p is the triangle isosceles?
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

Isosceles Triangle has:
Two equal sides
Two equal angles


A(-2,3), B(1,-3) and C(p,1)
the triangle will be isosceles if sides AC+and BC+ are equal in length

so, use distance formula:

d=sqrt%28%28x%5B2%5D-x%5B1%5D%29%5E2%2B%28y%5B2%5D-y%5B1%5D%29%5E2%29
find the length of the side AC
AB=sqrt%28%28p-%28-2%29%29%5E2%2B%281-3%29%5E2%29
AB=sqrt%28%28p%2B2%29%5E2%2B%28-2%29%5E2%29
AB=sqrt%28p%5E2+%2B+4+p+%2B+4%2B4%29
AB=sqrt%28p%5E2+%2B+4+p+%2B+8%29

find the length of the side +BC
BC=sqrt%28%28p-1%29%5E2%2B%281%2B3%29%5E2%29
BC=sqrt%28p%5E2+-+2+p+%2B+1%2B16%29
BC=sqrt%28p%5E2+-+2+p+%2B+17%29

the triangle will be isosceles if AC=BC

sqrt%28p%5E2%2B4+p+%2B+8%29=sqrt%28p%5E2+-2+p+%2B+17%29..........square both sides
p%5E2+%2B4+p+%2B+8=p%5E2+-+2+p+%2B+17...........solve for+p
p%5E2+%2B4+p+%2B+8-p%5E2+%2B+2+p+-+17=0.............simplify
6p++-+9+=0
6p++=9
p=9%2F6
p=3%2F2
the triangle will be isosceles if p=3%2F2
then vertices C(3%2F2,1)



Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.
Triangle ABC has vertices A(-2,3), B(1,-3) and C(p,1).
For how many values of p is the triangle isosceles?
~~~~~~~~~~~~~~~~~~~~~


            The solution by @MathLover1 is  FATALLY  INCOMPLETE.

            There are several other,  different isosceles triangles,  that @MathLover1 missed/failed to find.

            I came to bring the correct solution.



First,  find the length of the side  AB.  It is  (apply the distance formula)

        d = sqrt%28%281-%28-2%29%29%5E2+%2B+%28-3-3%29%5E2%29 = sqrt%283%5E2+%2B+6%5E2%29 = sqrt%2845%29 = 3%2Asqrt%285%29.

Next,  consider different cases.


Case AB = BC.


   In this case, BC = sqrt%2845%29 = sqrt%28%28p-1%29%5E2+%2B+%281-%28-3%29%29%5E2%29 = sqrt%28%28p-1%29%5E2%2B4%5E2%29.

   Squaring both sides, you get

      45 = %28p-1%29%5E2+%2B+16,

      45 = p%5E2+-+2p+%2B+1+%2B+16

      p%5E2+-+2p+-+28 = 0

      p%5B1%2C2%5D = %282+%2B-+sqrt%282%5E2+-+4%2A%28-28%29%29%29%2F2 = %282+%2B-+sqrt%284+%2B+4%2A28%29%29%2F2 = 1+%2B-+sqrt%2829%29.


   Thus, it gives two different real values for p, and, hence, two different isosceles triangles with AB = BC.


   These two different isosceles triangles are shown in the Figure 1 below (blue lines).



   


                      Figure 1.  Two different triangles for case 1.



Case AB = AC.


   In this case, AC = sqrt%2845%29 = sqrt%28%28p-%28-2%29%29%5E2+%2B+%281-3%29%5E2%29 = sqrt%28%28p%2B2%29%5E2%2B2%5E2%29.

   Squaring both sides, you get

      45 = %28p%2B2%29%5E2+%2B+4,

      45 = p%5E2+%2B+4p+%2B+4+%2B+4

      p%5E2+%2B+4p+-+37 = 0

      p%5B1%2C2%5D = %28-4+%2B-+sqrt%284%5E2+-+4%2A%28-37%29%29%29%2F2 = %28-4+%2B-+sqrt%2816+%2B+4%2A37%29%29%2F2 = -2+%2B-+sqrt%2841%29.


   Thus, it gives two different real values for p, and, hence, two different isosceles triangles with AB = AC.


   These two different isosceles triangles are shown in the Figure 2 below (blue lines).



   


                      Figure 2.  Two different triangles for case 2.


So,  in my post  I  shown  4  possible isosceles triangles,  missed by @MathLover1.

The fifth triangle is the one found in her post.

The  ANSWER  to the problem's question  IS :   5  triangles are possible.