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Triangle ABC is defined by the points A(-2, 4), B(6,2), and C(1,-1). Using what you know about the distance formula, what type of triangle would ABC be?

Sagot :

sreedevi102

Answer:

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Step-by-step explanation:

[tex]distance = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}[/tex]

[tex]AB = \sqrt{(6--2)^2 + (2-4)^2} = \sqrt{8^2 + 2^2} = \sqrt{68}\\\\BC = \sqrt{(1-6)^2 + (-1-2)^2} = \sqrt{5^2 + 3^2} = \sqrt{34}\\\\AC= \sqrt{(1--2)^2 + (-1-4)^2}} = \sqrt{3^2 + 5^2 } = \sqrt{34}[/tex]

[tex]Clearly, this\ satisfies \ the\ Pythagoras\ theorem : AC^2 = AB^2 + BC^2[/tex]

                                                                      [tex](\sqrt{68})^2 = (\sqrt{34} )^2 + (\sqrt{34} )^2\\\\68 = 34 + 34 \\68 = 68\\Hence\ satisfies .[/tex]

Therefore,  the triangle ABC is a right angle triangle.

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