Question

IF YOU HELP ME ILL give U 2$ AFTER UR DONE

Answer 1

Answer:

LOL i dont need money jus mark me brainliest :P

Step-by-step explanation:

$distance = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

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

$Clearly, this\ satisfies \ the\ Pythagoras\ theorem : AC^2 = AB^2 + BC^2$

                                                                      $(\sqrt{68})^2 = (\sqrt{34} )^2 + (\sqrt{34} )^2\\\\68 = 34 + 34 \\68 = 68\\Hence\ satisfies .$

Therefore,  the triangle ABC is a right angle triangle.