Question 1045501

 the triangle formed by the points {{{A}}}=({{{-2}}},{{{5}}}), {{{B}}}=({{{1}}},{{{3}}}) and {{{C}}}=({{{5}}},{{{9}}})

we need to find sides length using distance formula

{{{AB}}}
({{{-2}}},{{{5}}}), ({{{1}}},{{{3}}})
 {{{d=sqrt((x[1]-x[2])^2+(y[1]-y[2])^2)}}}

 {{{d=sqrt((-2-1)^2+(5-3)^2)}}}

 {{{d=sqrt((-3)^2+(2)^2)}}}

 {{{d=sqrt(9+4)}}}

 {{{d=sqrt(13)}}}

->{{{AB=sqrt(13)}}}

{{{AC}}}
({{{-2}}},{{{5}}}),({{{5}}},{{{9}}})

 {{{d=sqrt((x[1]-x[2])^2+(y[1]-y[2])^2)}}}

 {{{d=sqrt((-2-5)^2+(5-9)^2)}}}

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

 {{{d=sqrt(49+16)}}}

 {{{d=sqrt(65)}}}

->{{{AC=sqrt(65)}}}

{{{BC}}}

({{{1}}},{{{3}}},({{{5}}},{{{9}}})

 {{{d=sqrt((x[1]-x[2])^2+(y[1]-y[2])^2)}}}

 {{{d=sqrt((1-5)^2+(3-9)^2)}}}

 {{{d=sqrt((-4)^2+(-6)^2)}}}

 {{{d=sqrt(16+36)}}}

 {{{d=sqrt(52)}}}
->{{{BC=sqrt(52)}}}

so,
->{{{AB=sqrt(13)}}}
->{{{AC=sqrt(65)}}}->hypotenuse
->{{{BC=sqrt(52)}}}

if {{{(AC)^2=(AB)^2+(BC)^2}}} the triangle formed by the points is right angled

{{{(sqrt(65))^2=(sqrt(13))^2+(sqrt(52))^2}}}

{{{65=13+52}}}

{{{65=65}}} which is true

so, the triangle formed by the points {{{is}}} right angled