if three points of a parallelogram are located at (1,0), (4,3), and (5,-2) what is the fourth coordinate
If three points of a parallelogram are located at (1,0), (4,3), and (5,-2) what is the fourth coordinate ?
$$\small{\text{$
\begin{array}{rclclclcl}
\vec{P_1}=\begin{pmatrix} 1 \\ 0 \end{pmatrix} \qquad
\vec{P_2}=\begin{pmatrix} 4 \\ 3 \end{pmatrix} \qquad
\vec{P_3}=\begin{pmatrix} 5 \\ -2 \end{pmatrix} \\\\\\
\vec{P_4}&=& &-&\vec{P_1} &+& \vec{P_2} &+& \vec{P_3}
=\begin{pmatrix} -1+4+5 \\ 0+3-2 \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \end{pmatrix}\\\\
\vec{P_4}&=& &&\vec{P_1} &-& \vec{P_2} &+& \vec{P_3}
=\begin{pmatrix} 1-4+5 \\ 0-3-2 \end{pmatrix} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}\\\\
\vec{P_4}&=& &&\vec{P_1} &+& \vec{P_2} &-& \vec{P_3}
=\begin{pmatrix} 1+4-5 \\ 0+3+2 \end{pmatrix} = \begin{pmatrix} 0 \\ 5 \end{pmatrix}\\\\
\end{array}
$}}$$
Here are at least three possibilities......https://www.desmos.com/calculator/lkiacmskg7
The possible points are (8,1) , (2, -5) and (0, 5)
If three points of a parallelogram are located at (1,0), (4,3), and (5,-2) what is the fourth coordinate ?
$$\small{\text{$
\begin{array}{rclclclcl}
\vec{P_1}=\begin{pmatrix} 1 \\ 0 \end{pmatrix} \qquad
\vec{P_2}=\begin{pmatrix} 4 \\ 3 \end{pmatrix} \qquad
\vec{P_3}=\begin{pmatrix} 5 \\ -2 \end{pmatrix} \\\\\\
\vec{P_4}&=& &-&\vec{P_1} &+& \vec{P_2} &+& \vec{P_3}
=\begin{pmatrix} -1+4+5 \\ 0+3-2 \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \end{pmatrix}\\\\
\vec{P_4}&=& &&\vec{P_1} &-& \vec{P_2} &+& \vec{P_3}
=\begin{pmatrix} 1-4+5 \\ 0-3-2 \end{pmatrix} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}\\\\
\vec{P_4}&=& &&\vec{P_1} &+& \vec{P_2} &-& \vec{P_3}
=\begin{pmatrix} 1+4-5 \\ 0+3+2 \end{pmatrix} = \begin{pmatrix} 0 \\ 5 \end{pmatrix}\\\\
\end{array}
$}}$$
Thanks, heureka.....I see what you did here with vectors.......could you explain WHY this works ????
could you explain WHY this works ?
$$\vec{P_1}=\begin{pmatrix} 1 \\ 0 \end{pmatrix} \qquad\vec{P_2}=\begin{pmatrix} 4 \\ 3 \end{pmatrix} \qquad\vec{P_3}=\begin{pmatrix} 5 \\ -2 \end{pmatrix} \\\\
\small{\text{direction vectors: }}\\
\begin{array}{crcl}
& \vec{d_1} &=& \vec{P_2} - \vec{P_1}\\
& \vec{d_2} &=& \vec{P_3} - \vec{P_2} \\
& \vec{d_3} &=& \vec{P_1} - \vec{P_3}\\
\end{array}\\\\\\
\small{\text{$\begin{array}{lclclclcl}
\vec{P_4} = \vec{P_1}+\vec{d_1}-\vec{d_3}
&=& &-&\vec{P_1} &+& \vec{P_2} &+& \vec{P_3} =\begin{pmatrix} -1+4+5 \\ 0+3-2 \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \end{pmatrix}\\\\
\vec{P_4} = \vec{P_2}+\vec{d_2}-\vec{d_1}
&=& &&\vec{P_1} &-& \vec{P_2} &+& \vec{P_3}=\begin{pmatrix} 1-4+5 \\ 0-3-2 \end{pmatrix} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}\\\\
\vec{P_4} = \vec{P_3}+\vec{d_3}-\vec{d_2}
&=& &&\vec{P_1} &+& \vec{P_2} &-& \vec{P_3}=\begin{pmatrix} 1+4-5 \\ 0+3+2 \end{pmatrix} = \begin{pmatrix} 0 \\ 5 \end{pmatrix}\\\\\end{array}$}}$$