Answer:
A’(5 , -3)
B’(1 , 5)
C’(-9 , 1)
Step-by-step explanation:
Let A be the point of coordinates (-4, 3)
Let B be the point of coordinates (-2, -1)
Let C be the point of coordinates (3, 1)
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In this case we can form three parallelograms ABA’C, BAB’C , CAC’B
How to find the coordinates of the fourth point A’ of the parallelogram ABA’C
Since we know that the diagonals AA’ and BC of the parallelogram ABA’C intersect in their midpoint then we need to solve these two equations in order to get the coordinates of the point A’ :
[tex]\begin{aligned}\left\{\begin{array}{l}\frac{x_{A}+x_{A^{\prime}}}{2}=\frac{x_{B}+x_{C}}{2} \\\frac{y_{A}+y_{A^{\prime}}}{2}=\frac{y_{B}+y_{C}}{2}\end{array}\right.& \Leftrightarrow\left\{\begin{array}{l}-4+x_{A^{\prime}}=-2+3 \\3+y_{A^{\prime}}=-1+1\end{array}\right.\\\\& \Leftrightarrow\left\{\begin{array}{l}x_{A^{\prime}}=1+4=5 \\y_{A^{\prime}}=0-3=-3\end{array}\right.\\\\& \Leftrightarrow A^{\prime}(5,-3)\end{aligned}[/tex]
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We can follow the same method for the other two parallelograms In order to get the coordinates of the points B’ and C’.
B’(1 , 5)
C’(-9 , 1)
