Prove that the points with the position vectorsrA = (2,2), rB = (-1,6), rC = (-5,3) and rD = (-2,-1) in the (xy)-plane are the vertices of a square.
Hello, knuckles1234!
It helps if you plotted the points.Prove that the points with the position vectors:
$\displaystyle r_A = (2, 2),\; r_B = (-1,6),\;r_C = (-5, 3),\; r_D = (-2,-1)$ in the xy-plane
are the vertices of a square.
Code:B | (-1,6)* | | C | (-5,3) * | A | *(2,2) | - - - - - - - - + - - - - - - ( -2,-1)* | D |
$\displaystyle \begin{array}{ccccccc}\overrightarrow{AB} \:=\:\langle\text{-}3,4\rangle & \Rightarrow & |\overrightarrow{AB}| \:=\:\sqrt{\text{-}3)^2 + 4^2} \:=\:5 \\ \overrightarrow{DC} \:=\:\langle\text{-}3,4\rangle & \Rightarrow& |\overrightarrow{DC} \:=\:\sqrt{\text{-}3)^2 + 4^2} \:=\:5 \end{array}$
$\displaystyle \begin{array}{ccccccc}\overrightarrow{BC} \:=\:\langle\text{-}4,\text{-}3\rangle & \Rightarrow & |\overrightarrow{BC}| \:=\:\sqrt{(\text{-}4)^2 + (\text{-}3)^2} \:=\:5 \\
\overrightarrow{AD} \:=\:\langle\text{-}4,\text{-}3\rangle & \Rightarrow & |\overrightarrow{AD}| \:=\:\sqrt{(\text{-}4)^2 + (\text{-}3)^2} \:=\:5
\end{array}$
We have: .$\displaystyle AB = BC = CD = DA.$
. . Quadrilateral $\displaystyle ABCD$ is a rhombus.
$\displaystyle \overrightarrow{AB} \cdot \overrightarrow{AD} \:=\:\langle\text{-}3,4\rangle \cdot \langle \text{-}4,\text{-}3\rangle \:=\:12 - 12 \:=\:0$
. . Hence: .$\displaystyle \overrightarrow{AB} \perp \overrightarrow{AD}$
Therefore, $\displaystyle ABCD$ is a square.