+26393 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:
\(\begin{array}{|rcll|} \hline N &=& -M \\ &=& - \left( 3a,-\dfrac{b}{2} \right) \\ \mathbf{N} & \mathbf{=} & \mathbf{ \left( -3a , \dfrac{b}{2} \right) } \\ \hline \end{array} \)
+26393 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:
\(\begin{array}{|rcll|} \hline N &=& -M \\ &=& - \left( 3a,-\dfrac{b}{2} \right) \\ \mathbf{N} & \mathbf{=} & \mathbf{ \left( -3a , \dfrac{b}{2} \right) } \\ \hline \end{array} \)
