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if three points of a parallelogram are located at (1,0), (4,3), and (5,-2) what is the fourth coordinate

Guest May 5, 2015

Best Answer 

 #3
avatar+19495 
+10

If three points of a parallelogram are located at (1,0), (4,3), and (5,-2) what is the fourth coordinate ?

$$\small{\text{$
\begin{array}{rclclclcl}
\vec{P_1}=\begin{pmatrix} 1 \\ 0 \end{pmatrix} \qquad
\vec{P_2}=\begin{pmatrix} 4 \\ 3 \end{pmatrix} \qquad
\vec{P_3}=\begin{pmatrix} 5 \\ -2 \end{pmatrix} \\\\\\
\vec{P_4}&=& &-&\vec{P_1} &+& \vec{P_2} &+& \vec{P_3}
=\begin{pmatrix} -1+4+5 \\ 0+3-2 \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \end{pmatrix}\\\\
\vec{P_4}&=& &&\vec{P_1} &-& \vec{P_2} &+& \vec{P_3}
=\begin{pmatrix} 1-4+5 \\ 0-3-2 \end{pmatrix} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}\\\\
\vec{P_4}&=& &&\vec{P_1} &+& \vec{P_2} &-& \vec{P_3}
=\begin{pmatrix} 1+4-5 \\ 0+3+2 \end{pmatrix} = \begin{pmatrix} 0 \\ 5 \end{pmatrix}\\\\
\end{array}
$}}$$

heureka  May 6, 2015
 #1
avatar+86889 
+5

Here are at least three possibilities......https://www.desmos.com/calculator/lkiacmskg7

The  possible points are  (8,1) , (2, -5) and (0, 5)

 

 

  

CPhill  May 5, 2015
 #2
avatar+1068 
+5

These coordinates are in a clockwise (logic) order.

 

( 1, 0 )

 

( 4, 3 )

 

( 5, -2 )                         

 

( 2, -5 )

civonamzuk  May 5, 2015
 #3
avatar+19495 
+10
Best Answer

If three points of a parallelogram are located at (1,0), (4,3), and (5,-2) what is the fourth coordinate ?

$$\small{\text{$
\begin{array}{rclclclcl}
\vec{P_1}=\begin{pmatrix} 1 \\ 0 \end{pmatrix} \qquad
\vec{P_2}=\begin{pmatrix} 4 \\ 3 \end{pmatrix} \qquad
\vec{P_3}=\begin{pmatrix} 5 \\ -2 \end{pmatrix} \\\\\\
\vec{P_4}&=& &-&\vec{P_1} &+& \vec{P_2} &+& \vec{P_3}
=\begin{pmatrix} -1+4+5 \\ 0+3-2 \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \end{pmatrix}\\\\
\vec{P_4}&=& &&\vec{P_1} &-& \vec{P_2} &+& \vec{P_3}
=\begin{pmatrix} 1-4+5 \\ 0-3-2 \end{pmatrix} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}\\\\
\vec{P_4}&=& &&\vec{P_1} &+& \vec{P_2} &-& \vec{P_3}
=\begin{pmatrix} 1+4-5 \\ 0+3+2 \end{pmatrix} = \begin{pmatrix} 0 \\ 5 \end{pmatrix}\\\\
\end{array}
$}}$$

heureka  May 6, 2015
 #4
avatar+86889 
+5

Thanks, heureka.....I see what you did here with vectors.......could you explain WHY this works  ????

 

 

  

CPhill  May 6, 2015
 #5
avatar+19495 
+5

could you explain WHY this works ?

$$\vec{P_1}=\begin{pmatrix} 1 \\ 0 \end{pmatrix} \qquad\vec{P_2}=\begin{pmatrix} 4 \\ 3 \end{pmatrix} \qquad\vec{P_3}=\begin{pmatrix} 5 \\ -2 \end{pmatrix} \\\\
\small{\text{direction vectors: }}\\
\begin{array}{crcl}
& \vec{d_1} &=& \vec{P_2} - \vec{P_1}\\
& \vec{d_2} &=& \vec{P_3} - \vec{P_2} \\
& \vec{d_3} &=& \vec{P_1} - \vec{P_3}\\
\end{array}\\\\\\
\small{\text{$\begin{array}{lclclclcl}
\vec{P_4} = \vec{P_1}+\vec{d_1}-\vec{d_3}
&=& &-&\vec{P_1} &+& \vec{P_2} &+& \vec{P_3} =\begin{pmatrix} -1+4+5 \\ 0+3-2 \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \end{pmatrix}\\\\
\vec{P_4} = \vec{P_2}+\vec{d_2}-\vec{d_1}
&=& &&\vec{P_1} &-& \vec{P_2} &+& \vec{P_3}=\begin{pmatrix} 1-4+5 \\ 0-3-2 \end{pmatrix} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}\\\\
\vec{P_4} = \vec{P_3}+\vec{d_3}-\vec{d_2}
&=& &&\vec{P_1} &+& \vec{P_2} &-& \vec{P_3}=\begin{pmatrix} 1+4-5 \\ 0+3+2 \end{pmatrix} = \begin{pmatrix} 0 \\ 5 \end{pmatrix}\\\\\end{array}$}}$$

heureka  May 7, 2015

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