Question

the graph shows quadrilaterals mnop and tuvw. is mnop congruent to tuvw? justify your answer. yes, because a reflection across the line ( y = -x ) maps mnop onto tuvw. yes, because a rotation ( 180^circ ) around the origin maps mnop onto tuvw. no, because ( angle p ) and ( angle w ) do not have the same measure. no, because ( overline{pm} ) and ( overline{wt} ) do not have the same length.

Explanation:

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

Using the distance formula \( d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} \). For points \( P(3,6) \) and \( M(4,4) \), \( d_{PM}=\sqrt{(4 - 3)^2+(4 - 6)^2}=\sqrt{1 + 4}=\sqrt{5} \)

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

For points \( W(-5,-3) \) and \( T(-3,-4) \), \( d_{WT}=\sqrt{(-3+5)^2+(-4 + 3)^2}=\sqrt{4+1}=\sqrt{5} \)

Step3: Analyze congruence conditions

Congruent figures have all corresponding sides and angles equal. If we check the side - length relationship.
Let's assume a rotation of \( 180^{\circ}\) around the origin. The rule for a \( 180^{\circ}\) rotation around the origin is \( (x,y)\to(-x,-y) \).
For \( M(4,4)\to M'(-4,-4)
eq V(-9,-5) \).
For a reflection across \( y =-x \), the rule is \( (x,y)\to(-y,-x) \). For \( M(4,4)\to M'(-4,-4)
eq V(-9,-5) \)
If we check angles, we can also use the property that in congruent polygons, corresponding sides are equal.
Since \( MNOP \) and \( TUVW \) are quadrilaterals, and we can calculate other side lengths.
For \( \overline{NO} \): \( N(6,8) \), \( O(5,9) \), \( d_{NO}=\sqrt{(5 - 6)^2+(9 - 8)^2}=\sqrt{1+1}=\sqrt{2} \)
For \( \overline{UV} \): \( U(-7,-6) \), \( V(-9,-5) \), \( d_{UV}=\sqrt{(-9 + 7)^2+(-5+6)^2}=\sqrt{4 + 1}=\sqrt{5}
eq\sqrt{2} \)
Another way:
The length of \( \overline{PM} \): \( P(3,6) \), \( M(4,4) \). Using the distance formula \( d=\sqrt{(x_2-x_1)^2+(y_2 - y_1)^2}=\sqrt{(4 - 3)^2+(4 - 6)^2}=\sqrt{1+4}=\sqrt{5} \)
The length of \( \overline{WT} \): \( W(-5,-3) \), \( T(-3,-4) \). \( d=\sqrt{(-3 + 5)^2+(-4+3)^2}=\sqrt{4 + 1}=\sqrt{5} \)
But if we consider the transformation rules (rotation \( 180^{\circ}\): \( (x,y)\to(-x,-y) \); reflection over \( y=-x\): \( (x,y)\to(-y,-x) \)), we can see that the overall shape and position of the quadrilaterals do not match under these isometric transformations.
However, if we use the property of congruent polygons (all corresponding sides and angles are equal). Let's check the side - length of \( \overline{MN} \): \( M(4,4) \), \( N(6,8) \), \( d_{MN}=\sqrt{(6 - 4)^2+(8 - 4)^2}=\sqrt{4 + 16}=\sqrt{20}=2\sqrt{5} \)
For \( \overline{VW} \): \( V(-9,-5) \), \( W(-5,-3) \), \( d_{VW}=\sqrt{(-5 + 9)^2+(-3 + 5)^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5} \)
For \( \overline{NO} \): \( N(6,8) \), \( O(5,9) \), \( d_{NO}=\sqrt{(5 - 6)^2+(9 - 8)^2}=\sqrt{1+1}=\sqrt{2} \)
For \( \overline{UV} \): \( U(-7,-6) \), \( V(-9,-5) \), \( d_{UV}=\sqrt{(-9 + 7)^2+(-5 + 6)^2}=\sqrt{4+1}=\sqrt{5}
eq\sqrt{2} \)

Answer:

No, because \( \overline{PM} \) and \( \overline{WT} \) do not have the same length.