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Math Help - Analytical Geometry

  1. #1
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    Analytical Geometry

    Hey guys.
    I need lots of help.
    I need to find out how to do these questions: Answers (showing your work, like how to get the answer) would be really great. I'm really stuck with this stuff. And its drving me insane because I don't undertsand it.

    1. The vertices of a right triangle are A(3,3), B(-1,4), and C(1,-5). Find the area of the triangle.

    2. For the line segment DE, one endpoint is D(3,1) and the midpoint is M(0,-4). Find the coordinates of endpoint E.

    3.Triangle DEF is a right triangle with vertices D(-2,5), E(-4,1), and (2,3). Verify that the midpoint of the hypotenuse is equidistant from the three vertices of the triangle.

    4. Verify that the vertices A(-6,1), B(2,-5), C(6,1), and D(2,4)are the vertices of a trapezoid.

    5. Quadrilaterial RSTU has vertices R(3,2), S(0,4), T(-2,1), and U(1,-1). Verify that
    a) quadrilateral RSTU is a square
    b) the diagonals of quadrilateral RSTU perpendicularly bisect each other and are equal in length.

    I know these are alot of questions....but if you guys could try and anser them all i'd really appercaite it.
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  2. #2
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    1. problem only

    Quote Originally Posted by SaRah<3
    Hey guys.
    I need lots of help.
    I need to find out how to do these questions: Answers (showing your work, like how to get the answer) would be really great. I'm really stuck with this stuff. And its drving me insane because I don't undertsand it.
    1. The vertices of a right triangle are A(3,3), B(-1,4), and C(1,-5). Find the area of the triangle. ...
    Hello,

    first you have to calculate the length of both legs of this triangle. Use the Pythagorian Theorem:

    \overline{AC}=\sqrt{(3-1)^2+(3-(-5))^2} = \sqrt{68}

    \overline{BC}=\sqrt{(-1-1)^2+(4-(-5))^2} = \sqrt{85}

    \overline{AB}=\sqrt{(3-(-1))^2+(3-4))^2} = \sqrt{17}

    As you can see at once (BC) is the hypotenuse.

    Because a right triangle is a half rectangle the area is calculated here by:

    a=\frac{1}{2}\cdot \overline{AC}\cdot \overline{AB}. So:

    a=\frac{1}{2}\cdot \sqrt{68}\cdot \sqrt{17}= 17.

    Greetings

    EB
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  3. #3
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    hey thanks alot.

    ummm do you think you could help me with the other questions I put in the first post?
    Even just a few of the questions would be real great.
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  4. #4
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    Hello, SaRah<3!

    Here's #1 . . . in baby steps . . .


    1. The vertices of a right triangle are A(3,3), B(-1,4), and C(1,-5).
    Find the area of the triangle.

    A sketch is always a good idea . . .
    Code:
         (-1,4)
          B o   |
              * |
                *
             *  | *
                |   * 
              * |     *
                |       *
               *|         *
                |           o A
                *         * (3,3)
          - - - + - - - * - - - 
                |*    *
                |   *
                | o
                | C (1,-5)

    Even a rough sketch makes us suspect that:  AB \perp AC

    We find that: . m_{AB}\:=\:\frac{4-3}{-1-3}\:=\:\frac{1}{-4}\:=\:-\frac{1}{4}

    . . and that: . m_{AC}\:=\:\frac{-5-3}{1-3}\:=\:\frac{-8}{-2}\:=\:+4

    Since the slopes are negative reciprocals of each other: \angle A = 90^o


    The base of the triangle is: . AC\;=\;\sqrt{(1-3)^2 + (-5-3)^2}\:=\:\sqrt{68}\:=\:2\sqrt{17}

    The height of the triangle is: . AB\;=\;\sqrt{-1-3)^2 + (4-3)^2}\:=\;\sqrt{17}


    The area of a triangle is: . A \;=\;\frac{1}{2}(base)(height)

    . . so we have: . A\;=\;\frac{1}{2}\left(2\sqrt{17}\right)\left( \sqrt{17}\right) \;= \;17 square units.


    Edit: EB's too fast for me . . . and with a shorter solution!
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  5. #5
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    Hello again, SaRah<3!

    Here's #4 . . .


    4. Verify that the vertices A(-6,1), B(2,-5), C(6,1), and D(2,4)are the vertices of a trapezoid.

    A trapezoid is a quadrilateral with one pair of parallel sides.


    From a quick sketch (which I'll omit here), we suspect that: AB \parallel CD


    We find that: . m_{AB} \:=\:\frac{\text{-}5-1}{2-(\text{-}6)}\:=\;\frac{\text{-}6}{8}\;=\;-\frac{3}{4}

    . . . and that: . m_{CD}\:=\:\frac{4 - 1}{2 - 6}\:=\:\frac{3}{\text{-}4}\:=\:-\frac{3}{4}


    We were right! . . . AB \parallel CD

    Therefore, ABCD is a trapezoid.

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  6. #6
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    Quote Originally Posted by Soroban
    [size=3]...
    Edit: EB's too fast for me . . . and with a shorter solution!
    Hello,

    I'm sorry - but I live at least 6 hours east of you. So I am slightly in the lead. (I found this expression in my dictionary and I'm not certain if it is appropriate here).

    Best wishes

    EB
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  7. #7
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    Thanks so much for these guys!
    Try and keep them coming if you can
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  8. #8
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    Hello, SaRah<3!

    Don't get discouraged . . .
    Think of all the stuff you do know: distance formula, slope formula, etc.
    These problems make you apply them in various ways.
    You must learn to Think your way through them . . .


    5. Quadrilaterial RSTU has vertices R(3,2), S(0,4), T(-2,1), and U(1,-1).
    Verify that:
    a) Quadrilateral RSTU is a square
    b) The diagonals of quadrilateral RSTU perpendicularly bisect each other and are equal in length.

    a) How do we prove that a quadrilateral is a square?
    We must show that the four sides are equal (making it a rhombus)
    . . then show that it has one right angle.

    To show equals sides, we need the distance formula for find the lengths, right?

    \overline{RS}\:=\:\sqrt{0-3)^2 + (4-2)^2} \:= \:\sqrt{4 + 9}\:=\:\sqrt{13}
    \overline{ST} \:=\:\sqrt{(\text{-}2-0)^2 + (1-4)^2} \:=\:\sqrt{4 + 9} \:=\:\sqrt{13}
    \overline{TU} \:=\:\sqrt{(1+2)^2 + (\text{-}1-1)^2} \:=\:\sqrt{9+4}\;=\;\sqrt{13}
    \overline{UR}\:=\:\sqrt{(3-1)^2 + (2+1)^2} \:=\:\sqrt{4+9}\:=\:\sqrt{13}
    . . All four sides are equal; we have a rhombus - opposite sides are parallel.

    The slope of RS is: . m_{RS}\:=\:4 - 2}{0 - 3} \:=\:-\frac{2}{3}
    The slope of ST is: . m_{ST} \:=\:\frac{1 - 4}{\text{-}2 - 0} \:=\:+\frac{3}{2}

    . . Hence: RS \perp ST,\;\angle S = 90^o

    Therefore, quadrilateral RSTU is a square.



    b) We must show that the diagonals are perpendicular,
    bisect each other, and are equal in length.


    The diagonal from R(3,2) to T(-2,1)

    . . It has slope: . m_{RT}\:=\:\frac{1-2}{\text{-}2-3}\:=\:\frac{1}{5}

    . . Its midpoint is: . M_{RT}\:=\:\left(\frac{3-2}{2},\;\frac{2+1}{2}\right)\:=\:\left(\frac{1}{2}  ,\,\frac{3}{2}\right)

    . . Its length is: . \overline{RT}\:=\:\sqrt{(\text{-}2-3)^2+(1-2)^2}\:=\:\sqrt{25+1}\:=\:\sqrt{26}


    The diagonal from S(0,4) to U(1,-1)

    . . It has slope: . m_{SU}\:=\:\frac{\text{-}1-4}{1-0}\:=\:-5

    . . Its midpoint is: . M_{SU}\:=\:\left(\frac{0+1}{2},\,\frac{4-1}{2}\right)\:=\:\left(\frac{1}{2},\,\frac{3}{2} \right)

    . . Its length is: . \overline{SU}\:=\:\sqrt{(1-0)^2+ (\text{-}1-4)^2}\:=\:\sqrt{1 = 25}\:=\:\sqrt{26}


    Therefore:
    . . Their slopes are \frac{1}{5} and -5 . . . they are perpendicular.

    . . They have the same midpoint . . . they bisect each other.

    . . They both have length \sqrt{26} . . . they are equal.

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  9. #9
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    Wow I wish I was as smart as you. Math is so difficult for me. Im currently failing my grade 10 course.
    The exam is tommorow.
    Ill be so happy if I pass.
    Please god
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