Question

show that quadrilateral jklm is a parallelogram.

because ( overline{jk} = overline{lm} = square ), ( overline{jk} cong square ). because the slopes of ( overline{jk} ) and ( overline{lm} ) are ( square ), they are parallel. so, ( overline{jk} ) and ( overline{lm} ) are ( square ) and ( square ), which means that jklm is a parallelogram by the ( square ).

Explanation:

Step1: Identify coordinates of vertices

From the graph: $J(-3,-4)$, $K(-4,1)$, $L(2,3)$, $M(3,-2)$

Step2: Calculate length of $\overline{JK}$

Use distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$\overline{JK}=\sqrt{(-4-(-3))^2+(1-(-4))^2}=\sqrt{(-1)^2+(5)^2}=\sqrt{1+25}=\sqrt{26}$

Step3: Calculate length of $\overline{LM}$

$\overline{LM}=\sqrt{(3-2)^2+(-2-3)^2}=\sqrt{(1)^2+(-5)^2}=\sqrt{1+25}=\sqrt{26}$

Step4: State segment congruence

Since $JK=LM=\sqrt{26}$, $\overline{JK}\cong\overline{LM}$

Step5: Calculate slope of $\overline{JK}$

Slope formula $m=\frac{y_2-y_1}{x_2-x_1}$
$m_{JK}=\frac{1-(-4)}{-4-(-3)}=\frac{5}{-1}=-5$

Step6: Calculate slope of $\overline{LM}$

$m_{LM}=\frac{-2-3}{3-2}=\frac{-5}{1}=-5$

Step7: Analyze slope relationship

Slopes of $\overline{JK}$ and $\overline{LM}$ are equal, so they are parallel.

Step8: Apply parallelogram theorem

$\overline{JK}$ and $\overline{LM}$ are congruent and parallel. By the theorem: if one pair of opposite sides of a quadrilateral is both congruent and parallel, then the quadrilateral is a parallelogram.

Answer:

Because $JK = LM = \boldsymbol{\sqrt{26}}$, $\overline{JK}\cong\boldsymbol{\overline{LM}}$. Because the slopes of $\overline{JK}$ and $\overline{LM}$ are $\boldsymbol{equal}$, they are parallel. So, $\overline{JK}$ and $\overline{LM}$ are $\boldsymbol{congruent}$ and $\boldsymbol{parallel}$, which means that $JKLM$ is a parallelogram by the $\boldsymbol{theorem: if one pair of opposite sides of a quadrilateral is both congruent and parallel, then the quadrilateral is a parallelogram}$.