Question

Prove that the points (3, 0), (6, 4) and (-1, 3) are verticies of a right-angled isoceles triangle.

Answer 1

Let A (3,0) B (6,4) and C (-1,3)
now use distance formula ,
$AB=\sqrt{(6-3{)}^2+(4-0{)}^2}=\sqrt{(}25)=5$
$BC=\sqrt{(-1-6{)}^2+(3-4{)}^2}=\sqrt{(}50)$
$CA=\sqrt{\left(\right(3+1{)}^2+(-3{)}^2}=5.$

you see ,
$B{C}^2=A{C}^2+A{B}^2$
which is related to Pythagoras theorem .
hence , ABC is the vertices of right angle isoceles triangle .in which A is right angle .