Question

prove that the points 3, 0 and 6, 4 and -1, 3 are the vertices of right angled isosceles triangle

Answer 1

let A (3, 0), B (6, 4), C (-1, 3).
AB is 5
BC is 50^1/2
AC is 5
here AC is equal to AB,
AC^2+AB^2=BC^2
5^2+5^2=50
25+25=50
Hence proved it is a right isosceles triangle

Answer 2

Given A(3, 0), B(6, 4) and C(–1, 3)

AB^2 = (3 - 6)^2 + (0 - 4)^2

= 9 + 16 = 25

BC^2 = (6 + 1)^2 + (4 - 3)^2

= 49 + 1 = 50

CA^2 = (-1 - 3)^2 + (3 - 0)^2

= 16 + 9 = 25  

AB^2 = CA^2 ⇒ AB = CA

Triangle is isosceles

Also, 25 + 25 = 50

⇒ AB^2 + CA^2 = BC^2

Therefore  triangle is a right angled triangle