Proving points with position vectors form a square

**knuckles1234** · November 13th 2009, 09:11 AM

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.

**Soroban** · November 13th 2009, 11:19 AM

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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