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?
~~~~~~~~~~~~~~~~~~~~~



            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((1-(-2))^2 + (-3-3)^2)}}} = {{{sqrt(3^2 + 6^2)}}} = {{{sqrt(45)}}} = {{{3*sqrt(5)}}}.


Next,  consider different cases.



<pre>
<U>Case AB = BC</U>.


   In this case, BC = {{{sqrt(45)}}} = {{{sqrt((p-1)^2 + (1-(-3))^2)}}} = {{{sqrt((p-1)^2+4^2)}}}.

   Squaring both sides, you get

      45 = {{{(p-1)^2 + 16}}},

      45 = {{{p^2 - 2p + 1 + 16}}}

      {{{p^2 - 2p - 28}}} = 0

      {{{p[1,2]}}} = {{{(2 +- sqrt(2^2 - 4*(-28)))/2}}} = {{{(2 +- sqrt(4 + 4*28))/2}}} = {{{1 +- sqrt(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).



   {{{ drawing( 600, 300, -10, 10, -5, 5,
       locate(-2.7, 3.6, A(-2,3)),
       locate( 1,  -3,   B(1,-3)),

       green(line(-7,1, 9,1)),
       blue(line(-2,3, 1,-3)),

       blue(line(-2, 3, 6.385,1)),
       blue(line( 1,-3, 6.385,1)),
       locate(6.385,1, C),

       blue(line(-2, 3, -4.385,1)),
       blue(line( 1,-3, -4.385,1)),
       locate(-4.6,1, C),

       graph( 600, 300, -10, 10, -5, 5, 0)) }}}


                      Figure 1.  Two different triangles for case 1.



<U>Case AB = AC</U>.


   In this case, AC = {{{sqrt(45)}}} = {{{sqrt((p-(-2))^2 + (1-3)^2)}}} = {{{sqrt((p+2)^2+2^2)}}}.

   Squaring both sides, you get

      45 = {{{(p+2)^2 + 4}}},

      45 = {{{p^2 + 4p + 4 + 4}}}

      {{{p^2 + 4p - 37}}} = 0

      {{{p[1,2]}}} = {{{(-4 +- sqrt(4^2 - 4*(-37)))/2}}} = {{{(-4 +- sqrt(16 + 4*37))/2}}} = {{{-2 +- sqrt(41)}}}.


   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).



   {{{ drawing( 600, 300, -10, 10, -5, 5,
       locate(-2.7, 3.6, A(-2,3)),
       locate( 1,  -3,   B(1,-3)),

       green(line(-9.5,1, 9,1)),
       blue(line(-2,3,    1,-3)),

       blue(line(-2, 3, 4.403,1)),
       blue(line( 1,-3, 4.403,1)),
       locate(4.5,1, C),

       blue(line(-2, 3, -8.403,1)),
       blue(line( 1,-3, -8.403,1)),
       locate(-8.8,1, C),

       graph( 600, 300, -10, 10, -5, 5, 0)) }}}


                      Figure 2.  Two different triangles for case 2.

</pre>

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


The fifth triangle is the one found in her post.


The &nbsp;<U>ANSWER</U> &nbsp;to the problem's question &nbsp;IS :  &nbsp;&nbsp;5 &nbsp;triangles are possible.