Show that the points (-2, 3), (8, 3) and (6, 7) are the vertices of a right triangle.

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Solution

Let A (-2, 3), B(8,3), C(6, 7).

$(A B)^{2} = (8 + 2)^{2} + (3 - 3)^{2} = 10^{2} + 0^{2} = 100$

$(B C)^{2} = (6 - 8)^{2} + (7 - 3)^{2} = (- 2)^{2} + 4^{2} = 20$

$(A C)^{2} = (6 + 2)^{2} + (7 - 3)^{2} = 8^{2} + 4^{2} = 80$

Now, $(B C)^{2} + (A C)^{2} = 20 + 80 = 100 = (A B)^{2}$

By converse of Pythagoras’ theorem

Therefore, Points A, B, C are the vertices of a right triangle.

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