Question

note that g h j k has vertices g(1, 3), h(4, -4), j(-1, -5), and k(-4, 2). answer the following to determine if the parallelogram is a rectangle, rhombus, square, or none of these. (a) find the slope of \\(\overline{gh}\\) and the slope of a side adjacent to \\(\overline{gh}\\). slope of \\(\overline{gh}\\): \\(\square\\) slope of side adjacent to \\(\overline{gh}\\): \\(\square\\) (b) find the length of \\(\overline{gh}\\) and the length of a side adjacent to \\(\overline{gh}\\). give exact answers (not decimal approximations). length of \\(\overline{gh}\\): \\(\square\\) length of side adjacent to \\(\overline{gh}\\): \\(\square\\) (c) from parts (a) and (b), what can we conclude about parallelogram g h j k? check all that apply. \\(\square\\) g h j k is a rectangle. \\(\square\\) g h j k is a rhombus. \\(\square\\) g h j k is a square. \\(\square\\) g h j k is none of these.

Explanation:

Step1: Find slope of $\overline{GH}$

Points \( G(1, 3) \) and \( H(4, -4) \). Slope formula: \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
\( m_{GH} = \frac{-4 - 3}{4 - 1} = \frac{-7}{3} \).

Step2: Find slope of adjacent side (e.g., $\overline{GK}$)

Points \( G(1, 3) \) and \( K(-4, 2) \).
\( m_{GK} = \frac{2 - 3}{-4 - 1} = \frac{-1}{-5} = \frac{1}{5} \)? Wait, no, adjacent side to \( GH \) in parallelogram: \( GH \) and \( GK \)? Wait, parallelogram vertices: \( G, H, J, K \), so sides \( GH \), \( HJ \), \( JK \), \( KG \). Adjacent to \( GH \) is \( GK \) or \( HJ \)? Wait, \( GH \) connects \( G \) to \( H \), next side from \( H \) is \( HJ \), or from \( G \) is \( GK \). Let's take \( GK \): \( G(1,3) \), \( K(-4,2) \). Wait, no, maybe \( GK \) is not adjacent? Wait, parallelogram: \( \overline{GH} \parallel \overline{JK} \), \( \overline{GK} \parallel \overline{HJ} \). So adjacent to \( GH \) is \( GK \) (from \( G \)) or \( HJ \) (from \( H \)). Let's recalculate \( GK \): \( G(1,3) \), \( K(-4,2) \): slope \( \frac{2 - 3}{-4 - 1} = \frac{-1}{-5} = \frac{1}{5} \)? Wait, no, maybe I made a mistake. Wait, \( H(4,-4) \), \( J(-1,-5) \): slope of \( HJ \): \( \frac{-5 - (-4)}{-1 - 4} = \frac{-1}{-5} = \frac{1}{5} \). Wait, no, adjacent to \( GH \) should be \( GK \) (from \( G \)): \( G(1,3) \), \( K(-4,2) \): slope \( \frac{2 - 3}{-4 - 1} = \frac{-1}{-5} = \frac{1}{5} \). Wait, but let's check \( GH \) and \( GK \) slopes: \( \frac{-7}{3} \) and \( \frac{1}{5} \), product is \( \frac{-7}{3} \times \frac{1}{5} = \frac{-7}{15} eq -1 \), so not perpendicular. Wait, maybe I picked wrong adjacent side. Wait, \( GH \) and \( HJ \): \( H(4,-4) \), \( J(-1,-5) \): slope \( \frac{-5 - (-4)}{-1 - 4} = \frac{-1}{-5} = \frac{1}{5} \), same as \( GK \). Wait, no, \( GK \) is from \( G \) to \( K \), \( HJ \) from \( H \) to \( J \). So adjacent to \( GH \) (from \( G \)) is \( GK \), from \( H \) is \( HJ \). So slope of \( GH \): \( \frac{-4 - 3}{4 - 1} = \frac{-7}{3} \). Slope of \( GK \): \( G(1,3) \), \( K(-4,2) \): \( \frac{2 - 3}{-4 - 1} = \frac{-1}{-5} = \frac{1}{5} \). Wait, but maybe I messed up the adjacent side. Wait, let's do part (b) first for length.

Step3: Length of $\overline{GH}$

Distance formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
\( GH \): \( (4 - 1)^2 + (-4 - 3)^2 = 3^2 + (-7)^2 = 9 + 49 = 58 \), so \( \sqrt{58} \).

Step4: Length of adjacent side (e.g., $\overline{GK}$)

\( G(1,3) \), \( K(-4,2) \): \( (-4 - 1)^2 + (2 - 3)^2 = (-5)^2 + (-1)^2 = 25 + 1 = 26 \), so \( \sqrt{26} \). Wait, but if adjacent side is \( HJ \): \( H(4,-4) \), \( J(-1,-5) \): \( (-1 - 4)^2 + (-5 - (-4))^2 = (-5)^2 + (-1)^2 = 25 + 1 = 26 \), same as \( GK \). So length of \( GH \) is \( \sqrt{58} \), adjacent side (e.g., \( GK \)) is \( \sqrt{26} \).

Wait, back to slopes: \( m_{GH} = -7/3 \), \( m_{GK} = 1/5 \). Product: \( (-7/3)(1/5) = -7/15 eq -1 \), so not perpendicular (so not rectangle). Lengths: \( \sqrt{58} eq \sqrt{26} \), so not rhombus (all sides equal). So not square. So (c) answer: none of these? Wait, but let's recheck slopes. Wait, maybe I took the wrong adjacent side. Wait, \( GH \) and \( GK \): are they adjacent? In parallelogram \( GHIJ \) (wait, vertices \( G, H, J, K \)), so edges: \( G-H \), \( H-J \), \( J-K \), \( K-G \). So \( GH \) is adjacent to \( HJ \) and \( KG \). Wait, \( KG \) is from \( K \) to \( G \), same as \( GK \). So slope of \( KG \) is same as \( GK \). Wait, maybe I made a mistake in slope of \( GK \). Wait, \( G(1,3) \), \( K(-4,2) \): \( x_2 - x_1 = -5 \), \( y_2 - y_1 = -1 \), so slope \( (-1)/(-5) = 1/5 \). Slope of \( GH \): \( -7/3 \). Product: \( (-7/3)(1/5) = -7/15 eq -1 \), so not perpendicular. Lengths: \( GH \): \( \sqrt{(4-1)^2 + (-4-3)^2} = \sqrt{9 + 49} = \sqrt{58} \). \( GK \): \( \sqrt{(-4-1)^2 + (2-3)^2} = \sqrt{25 + 1} = \sqrt{26} \). So sides are not equal, slopes not perpendicular. So (c) is "GHJK is none of these".

But let's redo (a) and (b) properly:

(a) Slope of \( \overline{GH} \):
\( G(1,3) \), \( H(4,-4) \): \( m = \frac{-4 - 3}{4 - 1} = \frac{-7}{3} \).

Adjacent side: let's take \( \overline{GK} \) (from \( G \) to \( K \)): \( G(1,3) \), \( K(-4,2) \):
\( m = \frac{2 - 3}{-4 - 1} = \frac{-1}{-5} = \frac{1}{5} \). Wait, but in parallelogram, adjacent sides are \( GH \) and \( GK \) (or \( GH \) and \( HJ \)). Wait, \( HJ \): \( H(4,-4) \), \( J(-1,-5) \):
\( m = \frac{-5 - (-4)}{-1 - 4} = \frac{-1}{-5} = \frac{1}{5} \), same as \( GK \). So slope of adjacent side is \( \frac{1}{5} \).

(b) Length of \( \overline{GH} \):
\( \sqrt{(4 - 1)^2 + (-4 - 3)^2} = \sqrt{3^2 + (-7)^2} = \sqrt{9 + 49} = \sqrt{58} \).

Length of adjacent side (e.g., \( \overline{GK} \)):
\( \sqrt{(-4 - 1)^2 + (2 - 3)^2} = \sqrt{(-5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} \).

(c) Since slopes are not negative reciprocals (not perpendicular, so not rectangle), lengths are not equal (not rhombus), so not square. Thus, "GHJK is none of these".

Answer:

(a) Slope of \( \overline{GH} \): \( -\frac{7}{3} \); Slope of adjacent side: \( \frac{1}{5} \)
(b) Length of \( \overline{GH} \): \( \sqrt{58} \); Length of adjacent side: \( \sqrt{26} \)
(c) \( \boldsymbol{GHJK} \) is none of these (check "GHJK is none of these").