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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

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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

#3
+18827
+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
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#1
+80785
+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
+1068
+5

# ( 2, -5 )

civonamzuk  May 5, 2015
#3
+18827
+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
#4
+80785
+5

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

CPhill  May 6, 2015
#5
+18827
+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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