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The points $B(1, 1)$, $I(2, 4)$ and $G(5, 1)$ are plotted in the standard rectangular coordinate system to form triangle $BIG$. Triangle $BIG$ is translated five units to the left and two units upward to triangle $B'I'G'$, in such a way that $B'$ is the image of $B$, $I'$ is the image of $I$, and $G'$ is the image of $G$. What is the midpoint of segment $B'G'$? Express your answer as an ordered pair.

Guest Jun 19, 2018
 #1
avatar+12560 
0

The STARTING midpoint of BG  is at   3, 1   then it is translated 5 to the left

  so the 'x' coordinate becomes   3-5 = -2  

    and translated UP 2 units

       so the 'y' coordinate becomes 1+2 = 3

 

(-2,3)

ElectricPavlov  Jun 19, 2018
 #2
avatar+19624 
0

The points $B(1, 1)$, $I(2, 4)$ and $G(5, 1)$ are plotted in the standard rectangular coordinate system to form triangle $BIG$.

Triangle $BIG$ is translated five units to the left and two units upward to triangle $B'I'G'$,

in such a way that $B'$ is the image of $B$, $I'$ is the image of $I$, and $G'$ is the image of $G$.

What is the midpoint of segment $B'G'$?

Express your answer as an ordered pair.

 

\(\begin{array}{|l|lrr|} \hline & \text{translation} \\ & x' =x-5\\ & y' = y +2 & \\ \hline \text{B} =(1,1) & \text{B'} =(1-5,1+2) = (-4,3) \\ \text{I} = (2,4) & \text{I'} = (2-5,4+2) = (-3,6) \\ \text{G} = (5,1) & \text{G'} = (5-5,1+2) = (0,3) \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \text{midpoint of } B'G' &=& \dfrac{B'+G'}{2} \\\\ &=& \dfrac{\dbinom{-4}{3}+\dbinom{0}{3}}{2} \\\\ &=& \dfrac{\dbinom{-4+0}{3+3}}{2} \\\\ &=& \dfrac{\dbinom{-4}{6}}{2} \\\\ &\mathbf{=}& \mathbf{ \dbinom{-2}{3} } \\ \hline \end{array}\)

 

laugh

heureka  Jun 20, 2018

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