Question

name:
directions: now graph and label quadrilateral to prove that it is parallelogram

1. graph and label quadrilateral math with vertices m(- 5, 6), a(6, 6), t(8, - 3) and h(- 3, - 3)

b) prove that quadrilateral math is a parallelogram but not a rhombus.*hint look at the examples to help you.

Explanation:

Step1: Plot vertices on grid

Mark points: $M(-5, 6)$, $A(6, 6)$, $T(8, -3)$, $H(-3, -3)$ on the coordinate grid, then connect them in order to form quadrilateral MATH.

Step2: Calculate side slopes (slope formula $m=\frac{y_2-y_1}{x_2-x_1}$)

Slope of $MA$: $\frac{6-6}{6-(-5)}=\frac{0}{11}=0$
Slope of $TH$: $\frac{-3-(-3)}{-3-8}=\frac{0}{-11}=0$
Slope of $AT$: $\frac{-3-6}{8-6}=\frac{-9}{2}=-\frac{9}{2}$
Slope of $HM$: $\frac{6-(-3)}{-5-(-3)}=\frac{9}{-2}=-\frac{9}{2}$

Step3: Verify parallel opposite sides

Since $m_{MA}=m_{TH}$ and $m_{AT}=m_{HM}$, opposite sides are parallel, so MATH is a parallelogram.

Step4: Calculate side lengths (distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$)

Length of $MA$: $\sqrt{(6-(-5))^2+(6-6)^2}=\sqrt{11^2+0}=11$
Length of $AT$: $\sqrt{(8-6)^2+(-3-6)^2}=\sqrt{2^2+(-9)^2}=\sqrt{4+81}=\sqrt{85}$

Step5: Verify not a rhombus

Since $MA
eq AT$, not all sides are congruent, so MATH is not a rhombus.

Answer:

1. The graphed quadrilateral has vertices $M(-5, 6)$, $A(6, 6)$, $T(8, -3)$, $H(-3, -3)$ connected in sequence, with $MA$ and $TH$ as horizontal top/bottom sides, and $AT$ and $HM$ as sloped left/right sides.
2. Quadrilateral MATH is a parallelogram (opposite sides have equal slopes, so they are parallel) and not a rhombus (adjacent sides have unequal lengths, so not all sides are congruent).