Answer
$AB = BC = \sqrt{13}$,
So triangle ABC is isosceles
Work Step by Step
Given three points
A = (3, 4), B = (1, 1), and C = (- 2, 3)
To show they are vertices of isosceles triangle
Finding lengths AB, BC, AC
$AB = \sqrt{(3-1)^2 + (4-1)^2} = \sqrt{13}$
$AC = \sqrt{(3-(-2))^2 + (4-3)^2} = \sqrt{26}$
$BC = \sqrt{(1-(-2))^2 +(1-3)^2} = \sqrt{13}$
Since $AB = BC = \sqrt{13}$, Triangle ABC is isosceles