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Hitman42 [59]
9 months ago
6

Determine the most specific name for quadrilateral JKLM if the coordinates of the vertices are:J(-4,6), K(-1,2), L(1,6), M(4,2)J

L ll KM PROOF:J.5JL is parallel to X-axis.bothvertices have y-coordinate at y = 6.KM Parallel to x axis, bothvertices have y-coordinate aty=2.43KM M1Determine Stopes of JK & LM(If slopes ave ithen sides arparallel):4517-42-8-3-10JK: 31-4,6) K(+1,2)x2 Y2xiyo2-6-4

Mathematics
1 answer:
Arturiano [62]9 months ago
5 0

To finish the demonstration that the quadrilateral JKLM is a rhombus we need to prove that side JK is congruent with side LM.

The length of a segment with endpoints (x1, y1) and (x2, y2) is calculated as follows:

\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}

Substituting with points L(1,6) and M(4,2) we get:

\begin{gathered} LM=\sqrt[]{(4-1)^2+(2-6)^2} \\ LM=\sqrt[]{3^2+(-4)^2} \\ LM=\sqrt[]{9+16^{}} \\ LM=5 \end{gathered}

Given that opposite sides are parallel, all sides have the same length, and, from the diagram, the quadrilateral is not a square, we conclude that it is a rhombus.

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Given that WA = 5x – 8 and WC = 3x + 2, find WB. A. WB = 5 B. WB = 8 C. WB = 10 D. WB = 17
rodikova [14]
The rest of the question is the attached figure.
============================================
Δ AYW a right triangle at Y ⇒⇒⇒ ∴ WA² = AY² + YW²
And AY = YB ⇒⇒⇒ ∴ WA² = YB² + YW²   → (1)
Δ BYW a right triangle at Y ⇒⇒⇒ ∴ WB² = BY² + YW²  → (2)
From (1) , (2)  ⇒⇒⇒ ∴ WA = WB  →→ (3)
Δ CXW a right triangle at Y ⇒⇒⇒ ∴ WC² = CX² + XW²
And CX = XB ⇒⇒⇒ ∴ WC² = XB² + XW²   → (4)
Δ BXW a right triangle at Y ⇒⇒⇒ ∴ WB² = XB² + XW²  → (5)
From (4) , (5)  ⇒⇒⇒ ∴ WC = WB  →→ (6)
From (3) , (6)
WA = WB = WC
given ⇒⇒⇒ WA = 5x – 8 and WC = 3x + 2
∴ <span> 5x – 8 = 3x + 2</span>
Solve for x ⇒⇒⇒ ∴ x = 5
∴ WB = WA = WC = 3*5 + 2 = 17

The correct answer is option D. WB = 17






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