Question

(3,0),(6,4),(-1,3)are vertices of a right isoceless triangle

Answer 1

A(3, 0)    B(6, 4),   C(-1, 3)

AB² = (6-3)² + (4-0)² = 25
AC² = (3-0)² + (-1-3)² = 25
BC² = (3-4)² + (-1-6)² = 50

We can see that   AB² = AC²  and  AB² + AC² = BC²
SOΔ ABC is isosceles and right angled.

Answer 2

let A(3,0) B(6,4) and c(-1,3)

now , use distance formulae ,
AB =√{(6-3)^2+(4-0)^2}=√(3^2+4^2)
=5 unit
BC =√{(-1-6)^2+(3-4)^2} =√(7^2+1^2)=5√2

CA=√{(-1-3)^2+(3-0)^2}=√(3^2+4^2)=5

here we see ,
AB = CA
so triangle is isosceles traingle

also we see that ,
AC^2 +AB^2 = BC^2

hence ,
ABC is right angle triangle .

so, ∆ABC is right isosceles triangle