+0

geometry

0
278
1

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 $BIG'$, 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.

Jun 12, 2018

#1
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 $BIG'$,
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.

1.

$$\text{The midpoint of BG is \frac{B+G}{2} }$$

$$\begin{array}{|rcll|} \hline && \dfrac{B+G}{2} \\\\ &=& \dfrac{\dbinom{1}{1}+\dbinom{5}{1}}{2} \\\\ &=& \dfrac{\dbinom{1+5}{1+1}}{2} \\\\ &=& \dfrac{\dbinom{6}{2}}{2} \\\\ &=& \dbinom{\frac{6}{2}}{\frac{2}{2}}\\\\ &=& \dbinom{3}{1}\\ \hline \end{array}$$

The midpoint of BG is $$\mathbf{(3,\ 1)}$$

2.

$$\text{The translation of the midpoint of BG is the midpoint of B'G' }$$

$$\begin{array}{|ll|} \hline \text{Translation}\\ x' = x-5 \quad & \text{five units to the left} \\ y' = y+2 \quad & \text{two units upward} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \underbrace{\dbinom{3}{1}}_{\text{midpoint of BG }} \rightarrow \quad \underbrace{\dbinom{3-5}{1+2}}_{\text{translation}} &=& \underbrace{ \mathbf{ \dbinom{-2}{3}} }_{ \text{midpoint of B'G'} } \\ \hline \end{array}$$

The midpoint of B'G' is $$\mathbf{(-2,\ 3)}$$ Jun 12, 2018
edited by heureka  Jun 13, 2018