Answer:
Triangle ABC is isosceles.
Step-by-step explanation:
An isosceles triangle means two sides of the triangle should be equal.
Given vertices of the triangle are A(0, 2), B(2, 5) and C(-1, 7)
We will use the formula to find the length
[tex]=\sqrt{(x-x')^2+(y+y')^2}[/tex]
Now length AB = [tex]\sqrt{(0-2)^2+(2-5)^2}[/tex]
= [tex]\sqrt{2^2+3^2}[/tex]
= [tex]\sqrt{4+9}[/tex]
= [tex]\sqrt{13}[/tex]
Length AV = [tex]\sqrt{(0+1)^2+(2-7)^2}[/tex]
= [tex]\sqrt{1^2+(-5)^2}[/tex]
= [tex]\sqrt{1+25}[/tex]
= [tex]\sqrt{26}[/tex]
Length BC = [tex]\sqrt{(2+1)^2+(5-7)^2}[/tex]
= [tex]\sqrt{3^2+2^2}[/tex]
= [tex]\sqrt{9+4}[/tex]
= [tex]\sqrt{13}[/tex]
mAB = mBC = [tex]\sqrt{13}[/tex],
Therefore, triangle ABC is isosceles.
