Question 1187049

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((x[2]-x[1])^2+(y[2]-y[1])^2)}}}

find the length of the side  {{{AC}}}

{{{AB=sqrt((p-(-2))^2+(1-3)^2)}}}

{{{AB=sqrt((p+2)^2+(-2)^2)}}}

{{{AB=sqrt(p^2 + 4 p + 4+4)}}}

{{{AB=sqrt(p^2 + 4 p + 8)}}}


find the length of the side {{{ BC}}}

{{{BC=sqrt((p-1)^2+(1+3)^2)}}}

{{{BC=sqrt(p^2 - 2 p + 1+16)}}}

{{{BC=sqrt(p^2 - 2 p + 17)}}}


 the triangle will be isosceles if {{{AC=BC}}}


{{{sqrt(p^2+4 p + 8)=sqrt(p^2 -2 p + 17)}}}..........square both sides

{{{p^2 +4 p + 8=p^2 - 2 p + 17}}}...........solve for{{{ p}}}

{{{p^2 +4 p + 8-p^2 + 2 p - 17=0}}}.............simplify

{{{6p  - 9 =0}}}

{{{6p  =9}}}

 {{{p=9/6}}}

{{{p=3/2}}}

 the triangle will be isosceles if  {{{p=3/2}}}

then vertices  C({{{3/2}}},{{{1}}})


{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(-2,3,.12), locate(-2,3,A(-2,3)),
circle(1,-3,.12), locate(1,-3,B(1,-3)),
circle(3/2,1,.12), locate(3/2,1,C(3/2,1)),
green(line(-2,3,3/2,1)),green(line(1,-3,3/2,1)),
blue(line(-2,3,1,-3)),
graph( 600, 600, -10, 10, -10, 10, 0)) }}}