prove that the point (3,0),(6,4),and (-1,3) are the vertices of a right angled triangle?
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Let the points be A(3,0), B(6,4) and C(-1,3)
Now,By distance formula
AB^2=(6-3)^2 + (4-0)^2
=3^2 + 4^2
=9+16
-=25
AB=5
Now,
BC^2= (-1-6)^2 + (3-4)^2
=(-7)^2 + (-1)^2
= 49 + 1
=50
BC= root50
Now,
AC^2= (-1-3)^2 + (3-0)^2
=(-4)^2 + 3^2
=16+9
=25
AC=5
Clearly AB=AC
So, The Points A, B, C are vertices of a right angled triangle
Now,By distance formula
AB^2=(6-3)^2 + (4-0)^2
=3^2 + 4^2
=9+16
-=25
AB=5
Now,
BC^2= (-1-6)^2 + (3-4)^2
=(-7)^2 + (-1)^2
= 49 + 1
=50
BC= root50
Now,
AC^2= (-1-3)^2 + (3-0)^2
=(-4)^2 + 3^2
=16+9
=25
AC=5
Clearly AB=AC
So, The Points A, B, C are vertices of a right angled triangle
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