Question

directions: prove whether the quadrilateral is a parallelogram using the specified method.

1. q(-10, -2), r(1, -1), s(1, -7), t(-11, -8) (distance formula)

d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}
qr: \sqrt{(-10 - 1)^2+(-2-(-1))^2}=\sqrt{22}
rs: \sqrt{(1 - 1)^2+(-1-(-7))^2}=6
st: \sqrt{(1-(-11))^2+(-7-(-8))^2}=\sqrt{145}
qt: \sqrt{(-10-(-11))^2+(-2-(-8))^2}=\sqrt{37}

1. k(2, 7), l(6, 12), m(13, 13), n(9, 8) (slope formula)
2. d(-5, -6), e(5, 2), f(4, -4), g(-6, -12) (distance & slope formulas)

Explanation:

Step1: Recall the properties of a parallelogram

In a parallelogram, opposite - sides are equal in length (using distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$) and opposite - sides are parallel (using slope formula $m=\frac{y_2 - y_1}{x_2 - x_1}$).

Step2: Solve problem 10

Calculate side - lengths

For side $QR$ with $Q(-10,-2)$ and $R(1,-1)$:
$d_{QR}=\sqrt{(-10 - 1)^2+(-2+1)^2}=\sqrt{(-11)^2+(-1)^2}=\sqrt{121 + 1}=\sqrt{122}$
For side $RS$ with $R(1,-1)$ and $S(1,-7)$:
$d_{RS}=\sqrt{(1 - 1)^2+(-1 + 7)^2}=\sqrt{0+36}=6$
For side $ST$ with $S(1,-7)$ and $T(-11,-8)$:
$d_{ST}=\sqrt{(1 + 11)^2+(-7 + 8)^2}=\sqrt{144+1}=\sqrt{145}$
For side $QT$ with $Q(-10,-2)$ and $T(-11,-8)$:
$d_{QT}=\sqrt{(-10 + 11)^2+(-2 + 8)^2}=\sqrt{1 + 36}=\sqrt{37}$
Since opposite sides are not equal in length, the quadrilateral $QRST$ is not a parallelogram.

Step3: Solve problem 11

Calculate slopes

For side $KL$ with $K(2,7)$ and $L(6,12)$:
$m_{KL}=\frac{12 - 7}{6 - 2}=\frac{5}{4}$
For side $LM$ with $L(6,12)$ and $M(13,13)$:
$m_{LM}=\frac{13 - 12}{13 - 6}=\frac{1}{7}$
For side $MN$ with $M(13,13)$ and $N(9,8)$:
$m_{MN}=\frac{8 - 13}{9 - 13}=\frac{-5}{-4}=\frac{5}{4}$
For side $NK$ with $N(9,8)$ and $K(2,7)$:
$m_{NK}=\frac{7 - 8}{2 - 9}=\frac{-1}{-7}=\frac{1}{7}$
Since $m_{KL}=m_{MN}$ and $m_{LM}=m_{NK}$, the opposite - sides are parallel, so the quadrilateral $KLMN$ is a parallelogram.

Step4: Solve problem 12

Calculate side - lengths

For side $DE$ with $D(-5,-6)$ and $E(5,2)$:
$d_{DE}=\sqrt{(5 + 5)^2+(2 + 6)^2}=\sqrt{100 + 64}=\sqrt{164}=2\sqrt{41}$
For side $EF$ with $E(5,2)$ and $F(4,-4)$:
$d_{EF}=\sqrt{(4 - 5)^2+(-4 - 2)^2}=\sqrt{1 + 36}=\sqrt{37}$
For side $FG$ with $F(4,-4)$ and $G(-6,-12)$:
$d_{FG}=\sqrt{(-6 - 4)^2+(-12 + 4)^2}=\sqrt{100 + 64}=\sqrt{164}=2\sqrt{41}$
For side $DG$ with $D(-5,-6)$ and $G(-6,-12)$:
$d_{DG}=\sqrt{(-6 + 5)^2+(-12 + 6)^2}=\sqrt{1 + 36}=\sqrt{37}$

Calculate slopes

For side $DE$: $m_{DE}=\frac{2+6}{5 + 5}=\frac{8}{10}=\frac{4}{5}$
For side $EF$: $m_{EF}=\frac{-4 - 2}{4 - 5}=\frac{-6}{-1}=6$
For side $FG$: $m_{FG}=\frac{-12 + 4}{-6 - 4}=\frac{-8}{-10}=\frac{4}{5}$
For side $DG$: $m_{DG}=\frac{-12 + 6}{-6 + 5}=\frac{-6}{-1}=6$
Since opposite sides are equal in length and parallel, the quadrilateral $DEFG$ is a parallelogram.

Answer:

1. No
2. Yes
3. Yes