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Math Help - Proving points with position vectors form a square

  1. #1
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    Proving points with position vectors form a square

    Prove that the points with the position vectors
    rA = (2,2), rB = (-1,6), rC = (-5,3) and rD = (-2,-1) in the (xy)-plane are the vertices of a square.

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  2. #2
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    Hello, knuckles1234!

    Prove that the points with the position vectors:
    r_A = (2, 2),\; r_B = (-1,6),\;r_C = (-5, 3),\; r_D = (-2,-1) in the xy-plane
    are the vertices of a square.
    It helps if you plotted the points.

    Code:
                    B |
              (-1,6)* |
                      |
            C         |
     (-5,3) *         |  A
                      |   *(2,2)
                      |
      - - - - - - - - + - - - - - -
          ( -2,-1)*   |
                  D   |

    \begin{array}{ccccccc}\overrightarrow{AB} \:=\:\langle\text{-}3,4\rangle & \Rightarrow & |\overrightarrow{AB}| \:=\:\sqrt{\text{-}3)^2 + 4^2} \:=\:5 \\ \overrightarrow{DC} \:=\:\langle\text{-}3,4\rangle & \Rightarrow& |\overrightarrow{DC} \:=\:\sqrt{\text{-}3)^2 + 4^2} \:=\:5 \end{array}

    \begin{array}{ccccccc}\overrightarrow{BC} \:=\:\langle\text{-}4,\text{-}3\rangle & \Rightarrow & |\overrightarrow{BC}| \:=\:\sqrt{(\text{-}4)^2 + (\text{-}3)^2} \:=\:5 \\<br />
\overrightarrow{AD} \:=\:\langle\text{-}4,\text{-}3\rangle & \Rightarrow & |\overrightarrow{AD}| \:=\:\sqrt{(\text{-}4)^2 + (\text{-}3)^2} \:=\:5<br />
\end{array}

    We have: . AB = BC = CD = DA.
    . . Quadrilateral ABCD is a rhombus.


    \overrightarrow{AB} \cdot \overrightarrow{AD} \:=\:\langle\text{-}3,4\rangle \cdot \langle \text{-}4,\text{-}3\rangle \:=\:12 - 12 \:=\:0

    . . Hence: . \overrightarrow{AB} \perp \overrightarrow{AD}


    Therefore, ABCD is a square.

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