Question 970876
a ({{{5}}}, {{{5}}}); b ({{{-7}}},{{{1}}}); and c ({{{1}}},{{{-7}}}) are the vertices of  triangle ABC

to show that ABC is an {{{isosceles}}} triangle, we need to show that the distance between two vertices is {{{same}}}; so, use the distance formula

the distance between {{{a}}} and {{{b}}}
a ({{{5}}}, {{{5}}}); 
b ({{{-7}}},{{{1}}})

{{{d=sqrt((5-(-7))^2+(5-1)^2)}}}

{{{d=sqrt((5+7)^2+(5-1)^2)}}}

{{{d=sqrt(12^2+4^2)}}}

{{{d=sqrt(144+16)}}}

{{{d=sqrt(160)}}}

{{{d=4sqrt(10)}}}

or

{{{d[ab]=12.65}}}


the distance between {{{a}}} and {{{c}}}
a ({{{5}}}, {{{5}}}); 
c ({{{1}}},{{{-7}}})

{{{d=sqrt((5-1))^2+(5-(-7))^2)}}}

{{{d=sqrt(4^2+(5+7)^2)}}}

{{{d=sqrt(16+12^2)}}}

{{{d=sqrt(16+144)}}}

{{{d=sqrt(160)}}}

{{{d=4sqrt(10)}}}

or

{{{d[ac]=12.65}}}

as you can see, {{{d[ab]=d[ac]=12.65}}} => so, ABC is an {{{isosceles}}} triangle


let's draw it too

first find equation of a line through each two points:


a line through points {{{a}}} and {{{b}}}

*[invoke change_this_name10094 5, 5, -7, 1]



a line through points {{{a}}} and {{{c}}}

*[invoke change_this_name10094 5, 5, 1, -7]


a line through points {{{b}}} and {{{c}}}

*[invoke change_this_name10094 -7, 1, 1, -7]


so, equations are:

{{{y=0.333333333333333x + 3.33333333333333}}}

{{{y=3x -10}}}

{{{y=-x -6}}}

graph all together:


{{{drawing( 600, 600, -10, 10, -10,10,
circle(5,5,.13),locate(5,5,a(5,5)),
circle(-7,1,.13),circle(1,-7,.13),locate(-7,1,b(-7,1)),locate(1,-7,c(1,-7)),
 graph( 600, 600, -10, 10, -10, 10,0.333333333333333x + 3.33333333333333, 3x -10,-x -6)) }}}