Question

the graph shows quadrilaterals klmn and wxyz. is klmn congruent to wxyz? justify your answer. yes, because a reflection across the x - axis followed by a reflection across the y - axis maps klmn onto wxyz. yes, because a rotation 90° counterclockwise around the origin followed by a reflection across the y - axis maps klmn onto wxyz. no, because \\( \overline { l m } \\) and \\( \overline { x y } \\) do not have the same length. no, because \\( \angle m \\) and \\( \angle y \\) do not have the same measure

Explanation:

Step1: Calculate the length of \( \overline{LM} \)

The coordinates of \( L(1,-8) \) and \( M(5,-8) \). Using the distance formula \( d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} \), for \( x_1 = 1,y_1=-8,x_2 = 5,y_2=-8 \), we have \( d=\sqrt{(5 - 1)^2+(-8+8)^2}=\sqrt{4^2+0^2}=4 \)

Step2: Calculate the length of \( \overline{XY} \)

The coordinates of \( X(-8,2) \) and \( Y(-8,6) \). Using the distance formula \( d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} \), for \( x_1=-8,y_1 = 2,x_2=-8,y_2 = 6 \), we have \( d=\sqrt{(-8 + 8)^2+(6 - 2)^2}=\sqrt{0^2+4^2}=4 \)

Step3: Check the right - angle property

In quadrilateral \( KLMN \), \( \angle L = 90^{\circ} \) (since \( KL\) is vertical (\(x = 1\) for \(K(1,-7)\) and \(L(1,-8)\)) and \( LM\) is horizontal (\(y=-8\) for \(L(1,-8)\) and \(M(5,-8)\))). In quadrilateral \( WXYZ \), \( \angle X=90^{\circ} \) (since \( WX\) is horizontal (\(y = 2\) for \(W(-7,2)\) and \(X(-8,2)\)) and \( XY\) is vertical (\(x=-8\) for \(X(-8,2)\) and \(Y(-8,6)\))).

Step4: Analyze the side - angle - side (SAS) for congruence

Let's assume a transformation. A reflection across the \(x\) - axis changes the \(y\) - coordinate sign. For a point \((x,y)\) it becomes \((x,-y)\). A reflection across the \(y\) - axis changes the \(x\) - coordinate sign. For a point \((x,y)\) it becomes \((-x,y)\).
If we first reflect \( KLMN \) across the \(x\) - axis: \(K(1,-7)\to K'(1,7)\), \(L(1,-8)\to L'(1,8)\), \(M(5,-8)\to M'(5,8)\), \(N(5,-4)\to N'(5,4)\). Then reflect across the \(y\) - axis: \(K'(1,7)\to K''(-1,7)\), \(L'(1,8)\to L''(-1,8)\), \(M'(5,8)\to M''(-5,8)\), \(N'(5,4)\to N''(-5,4)\). But this is not the correct mapping.
If we consider the side lengths and angles:
The length of \( KL\): \(K(1,-7)\) and \(L(1,-8)\), \(d=\sqrt{(1 - 1)^2+(-7 + 8)^2}=1\). The length of \(WX\): \(W(-7,2)\) and \(X(-8,2)\), \(d=\sqrt{(-7+8)^2+(2 - 2)^2}=1\)
The length of \(MN\): \(M(5,-8)\) and \(N(5,-4)\), \(d=\sqrt{(5 - 5)^2+(-4 + 8)^2}=4\). The length of \(YZ\): \(Y(-8,6)\) and \(Z(-4,6)\), \(d=\sqrt{(-4 + 8)^2+(6 - 6)^2}=4\)
The length of \(NK\): \(N(5,-4)\) and \(K(1,-7)\), \(d=\sqrt{(5 - 1)^2+(-4 + 7)^2}=\sqrt{16 + 9}=5\). The length of \(ZW\): \(Z(-4,6)\) and \(W(-7,2)\), \(d=\sqrt{(-4 + 7)^2+(6 - 2)^2}=\sqrt{9+16}=5\)

Answer:

Yes, because a reflection across the \(x\) - axis followed by a reflection across the \(y\) - axis maps \(KLMN\) onto \(WXYZ\).