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I need help with: #1, b

 

 Sep 27, 2018

Best Answer 

 #1
avatar+25344 
+10

b)
Find the coordinates of point N, such that the midpoint of \(\mathbf{\overline{MN}}\) is the origin.

 

\(M = \left( 3a,-\dfrac{b}{2} \right) \\ N = \left( N_x, N_y \right) \)

 

1. Solution:

\(\begin{array}{|rclrcl|} \hline \left( \dfrac{3a+N_x}{2}, \dfrac{-\dfrac{b}{2}+N_y}{2} \right) &=& \left( 0, 0 \right) \\\\ \dfrac{3a+N_x}{2} &=& 0 & \dfrac{-\dfrac{b}{2}+N_y}{2} &=& 0 \\ 3a+N_x &=& 0 & -\dfrac{b}{2}+N_y &=& 0 \\ N_x &=& -3a & N_y &=& \dfrac{b}{2} \\\\ \mathbf{N} & \mathbf{=} & \mathbf{ \left( -3a , \dfrac{b}{2} \right) } \\ \hline \end{array}\)

 

2. Solution:

\(\begin{array}{|rcll|} \hline N &=& -M \\ &=& - \left( 3a,-\dfrac{b}{2} \right) \\ \mathbf{N} & \mathbf{=} & \mathbf{ \left( -3a , \dfrac{b}{2} \right) } \\ \hline \end{array} \)

 

laugh

 Sep 27, 2018
edited by heureka  Sep 27, 2018
 #1
avatar+25344 
+10
Best Answer

b)
Find the coordinates of point N, such that the midpoint of \(\mathbf{\overline{MN}}\) is the origin.

 

\(M = \left( 3a,-\dfrac{b}{2} \right) \\ N = \left( N_x, N_y \right) \)

 

1. Solution:

\(\begin{array}{|rclrcl|} \hline \left( \dfrac{3a+N_x}{2}, \dfrac{-\dfrac{b}{2}+N_y}{2} \right) &=& \left( 0, 0 \right) \\\\ \dfrac{3a+N_x}{2} &=& 0 & \dfrac{-\dfrac{b}{2}+N_y}{2} &=& 0 \\ 3a+N_x &=& 0 & -\dfrac{b}{2}+N_y &=& 0 \\ N_x &=& -3a & N_y &=& \dfrac{b}{2} \\\\ \mathbf{N} & \mathbf{=} & \mathbf{ \left( -3a , \dfrac{b}{2} \right) } \\ \hline \end{array}\)

 

2. Solution:

\(\begin{array}{|rcll|} \hline N &=& -M \\ &=& - \left( 3a,-\dfrac{b}{2} \right) \\ \mathbf{N} & \mathbf{=} & \mathbf{ \left( -3a , \dfrac{b}{2} \right) } \\ \hline \end{array} \)

 

laugh

heureka Sep 27, 2018
edited by heureka  Sep 27, 2018

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