Question

identify any congruent figures in the coordinate plane.
congruent figures

Explanation:

Step1: Calculate side lengths using distance formula

The distance formula is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
For figure \(EFGH\):

• \(EF=\sqrt{(- 4+3)^2+(3 - 2)^2}=\sqrt{1 + 1}=\sqrt{2}\)
• \(FG = 1\) (horizontal side, change in \(x\) is \(1\), change in \(y\) is \(0\))
• \(GH=\sqrt{( - 2+1)^2+(3 - 2)^2}=\sqrt{1+1}=\sqrt{2}\)
• \(HE = 3\) (horizontal side, change in \(x\) is \(3\), change in \(y\) is \(0\))

For figure \(BCDA\):

• \(AB=\sqrt{(1 - 1)^2+(4 - 1)^2}=3\) (vertical side, change in \(x\) is \(0\), change in \(y\) is \(3\))
• \(BC=\sqrt{(2 - 1)^2+(3 - 4)^2}=\sqrt{1 + 1}=\sqrt{2}\)
• \(CD=\sqrt{(2 - 2)^2+(2 - 3)^2}=1\) (vertical side, change in \(x\) is \(0\), change in \(y\) is \(1\))
• \(DA=\sqrt{(1 - 2)^2+(1 - 2)^2}=\sqrt{1+1}=\sqrt{2}\)

For figure \(JKLM\):

• \(JK=\sqrt{(-2 + 3)^2+(-3+4)^2}=\sqrt{1 + 1}=\sqrt{2}\)
• \(KL=\sqrt{(0 + 2)^2+(-3 + 3)^2}=2\) (horizontal side, change in \(x\) is \(2\), change in \(y\) is \(0\))
• \(LM=\sqrt{(1 - 0)^2+(-4+3)^2}=\sqrt{1+1}=\sqrt{2}\)
• \(MJ=\sqrt{(-3 - 1)^2+(-3 + 4)^2}=\sqrt{16 + 1}=\sqrt{17}\)

Step2: Compare side - lengths

Since the side - lengths of \(EFGH\) and \(BCDA\) are equal (\(EF = CD = \sqrt{2}\), \(FG=DA = 1\), \(GH=BC=\sqrt{2}\), \(HE = AB = 3\)).

Answer:

Figure \(EFGH\) and Figure \(BCDA\)