Question

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1. a rhombus is an equilateral quadrilateral. prove that abcd is a rhombus given a(2,2), b(1,-2), c(-3,-3), and d(-2,1).

Explanation:

Identify given coordinates

We are given vertices:
\(A(2,2)\), \(B(1,-2)\), \(C(-3,-3)\), \(D(-2,1)\).

State definition of rhombus

A rhombus has four equal sides.
We must show:
\(AB = BC = CD = DA\).

Calculate side lengths

Using the distance formula:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

For side \(AB\):

$$AB = \sqrt{(1 - 2)^2 + (-2 - 2)^2} = \sqrt{(-1)^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17}$$

For side \(BC\):

$$BC = \sqrt{(-3 - 1)^2 + (-3 - (-2))^2} = \sqrt{(-4)^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17}$$

For side \(CD\):

$$CD = \sqrt{(-2 - (-3))^2 + (1 - (-3))^2} = \sqrt{(1)^2 + (4)^2} = \sqrt{1 + 16} = \sqrt{17}$$

For side \(DA\):

$$DA = \sqrt{(2 - (-2))^2 + (2 - 1)^2} = \sqrt{(4)^2 + (1)^2} = \sqrt{16 + 1} = \sqrt{17}$$

Compare calculated lengths

All four side lengths are equal:

$$AB = BC = CD = DA = \sqrt{17}$$

Conclude the proof

Since all sides are equal, \(ABCD\) is a rhombus.

Answer:

To prove that quadrilateral \(ABCD\) is a rhombus, we calculate the lengths of its four sides using the distance formula:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

1. Length of \(AB\):

$$AB = \sqrt{(1 - 2)^2 + (-2 - 2)^2} = \sqrt{(-1)^2 + (-4)^2} = \sqrt{17}$$

1. Length of \(BC\):

$$BC = \sqrt{(-3 - 1)^2 + (-3 - (-2))^2} = \sqrt{(-4)^2 + (-1)^2} = \sqrt{17}$$

1. Length of \(CD\):

$$CD = \sqrt{(-2 - (-3))^2 + (1 - (-3))^2} = \sqrt{(1)^2 + (4)^2} = \sqrt{17}$$

1. Length of \(DA\):

$$DA = \sqrt{(2 - (-2))^2 + (2 - 1)^2} = \sqrt{(4)^2 + (1)^2} = \sqrt{17}$$

Since \(AB = BC = CD = DA = \sqrt{17}\), all four sides of the quadrilateral are equal in length. By definition, an equilateral quadrilateral is a rhombus. Therefore, \(ABCD\) is a rhombus.