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Math Help - confused with these trigono....

  1. #1
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    Post confused with these trigono....

    Show that the given points are the vertices of an isosceles triangle. Use the

    distance formula. (Show that two sides of the triangle are equal in length).


    1.) A (-1, -6), B (2, -7), C (-2, -3)

    2.) A (-9, 3), B (-4, 8), C (0, -1)

    3.) A (2, -3), B (3, 4), C (-5, -2)

    4.) A (-3, 0), B (-6, 7), C (1, 4)
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  2. #2
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    Quote Originally Posted by orochimaru700 View Post
    Show that the given points are the vertices of an isosceles triangle. Use the

    distance formula. (Show that two sides of the triangle are equal in length).


    1.) A (-1, -6), B (2, -7), C (-2, -3)

    2.) A (-9, 3), B (-4, 8), C (0, -1)

    3.) A (2, -3), B (3, 4), C (-5, -2)

    4.) A (-3, 0), B (-6, 7), C (1, 4)
    This seems pretty straightforward. What part don't you understand? Do you know the distance formula?

    d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.
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    don't know how to solve for the following, our instructor said, that we plot the following and solve for each problem... don't know what to do
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    Quote Originally Posted by orochimaru700 View Post
    don't know how to solve for the following, our instructor said, that we plot the following and solve for each problem... don't know what to do
    The question tells you exactly what to do.

    You work out the distance between the vertices using the distance formula.

    Once you have all three distances of the triangles, you should be able to see that in each triangle, two distances are equal. So the triangles are isosceles.


    E.g. for the first one, you have

    A = (-1, -6), B = (2, -7), C= (-2, -3).


    AC = \sqrt{[-2 - (-1)]^2 + [-3 - (-6)]^2}

     = \sqrt{(-2 + 1)^2 + (-3 + 6)^2}

     = \sqrt{(-1)^2 + 3^2}

     = \sqrt{1 + 9}

     = \sqrt{10}.


    Now work out the distance of AB and BC.
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  5. #5
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    ok thanks for the first one, i'll try to solve the other problem...
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