Question

Prove that the points (3,0) (6,4) (-1,3) are the vertices of a right angle isosceles triangle.

Answer 1

let A (3,0) B (6,4) and C (-1,3)
now use distance formula ,
AB=root {(6-3)^2+(4-0)^2}=root (25)=5
BC=root {(-1-6)^2+(3-4)^2}=root (50)
CA =root {((3+1)^2+(-3)^2}=5.

you see ,
BC^2 =AC^2 + AB^2
which is related to Pythagoras theorem .
hence , ABC is the vertices of right angle isoceles triangle .in which A is right angle .

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

Since Pythagoras theorem is verified, therefore

Triangle is a right angled triangle