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Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.

Q ; 6    Without using the Pythagoras theorem, show that the points  (4,4),(3,5)  and  (-1,-1),  are  the vertices of a right angled triangle.

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It is given that  point A(4,4) , B(3,5) and C(-1,-1) are the vertices of a right-angled triangle
Now,
We know that the distance between two points is given by
D = |\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}|
Length of AB  = |\sqrt{(4-3)^2+(4-5)^2}|= |\sqrt{1+1}|= \sqrt2
Length of BC = |\sqrt{(3+1)^2+(5+1)^2}|= |\sqrt{16+36}|= \sqrt{52}
Length of AC = |\sqrt{(4+1)^2+(4+1)^2}|= |\sqrt{25+25}|= \sqrt{50}
Now, we know that  Pythagoras theorem is
H^2= B^2+L^2
Is clear that
(\sqrt{52})^2=(\sqrt{50})^2+(\sqrt 2)^2\\ 52 = 52\\ i.e\\ BC^2= AB^2+AC^2
Hence proved

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