Answer:
see explanation
Step-by-step explanation:
For the triangle to be isosceles it must have 2 equal sides.
Calculate the length of the sides using the distance formula
d = [tex]\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }[/tex]
with (x₁, y₁ ) = A (- 1, 2) and (x₂, y₂ ) = B (3, 5)
AB = [tex]\sqrt{(3-(-1))^2+(5-2)^2}[/tex]
= [tex]\sqrt{(3+1)^2+3^2}[/tex]
= [tex]\sqrt{4^2+9}[/tex]
= [tex]\sqrt{16+9}[/tex] = [tex]\sqrt{25}[/tex] = 5
Repeat with (x₁, y₁ ) = A (- 1, 2) and (x₂, y₂ ) = C (3, - 1)
AC = [tex]\sqrt{(3+1)^2+(-1-2)^2}[/tex]
= [tex]\sqrt{4^2+(-3)^2}[/tex]
= [tex]\sqrt{16+9}[/tex] = [tex]\sqrt{25}[/tex] = 5
Consider B (3, 5) and C (3, - 1)
Since both x- coordinates are 3, this is a vertical line and its length is the difference of the y- coordinates, that is
BC = | - 1 - 5 | = | - 6 | = 6
Then
AB = AC = 5 , BC = 6
Since AB = AC = 5 then the triangle is isosceles
