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Simora [160]
3 years ago
8

In the coordinate plane, draw quadrilateral ABCD with A(–5, 0), B(2, –6), C(8, 1), and D(1, 7), then demonstrate that ABCD is a

rectangle.
Mathematics
1 answer:
Len [333]3 years ago
7 0

Answer:

ABCD is a rectangle

Step-by-step explanation:

∵ A = (-5 , 0) , B = (2 , -6) , C = (8 , 1) , D = (1 , 7)

∵ The x-coordinate of the mid-point of AC = (-5 + 8)/2 =3/2

∵ The y-coordinate of the mid-point of AC = (0 + 1)/2 = 1/2

∴ The mid-point of AC = (3/2 , 1/2)

∵ The x-coordinate of the mid-point of BD = (2 + 1)/2 =3/2

∵ The y-coordinate of the mid-point of BD = (-6 + 7)/2 = 1/2

∴ The mid-point of BD = (3/2 , 1/2)

∴ The mid-point of AC = The mid-point of BD ⇒ (1)

∵ AC = √[(8 - -5)²+(1 - 0)² = √170

∵ BD = √[(1 - 2)²+(7 - -6)² = √170

∴ AC = BD ⇒ (2)

From (1) and (2)

AC and BD equal each other and bisects each other

∴ ABCD is a rectangle

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We can use the distance formula to calculate the lengths of the line segments.

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}

1. A (1,5), B (4,5) (red)

d = \sqrt{(x_{2} - x_{1}^{2}) + (y_{2} - y_{1})^{2}} = \sqrt{(4 - 1)^{2} + (5 - 5)^{2}}\\= \sqrt{3^{2} + 0^{2}} = \sqrt{9 + 0} = \sqrt{9} = \mathbf{3}

2. A (2,-5), B (2,7) (blue)

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(2 - 2)^{2} + (7 - (-5))^{2}}\\= \sqrt{0^{2} + 12^{2}} = \sqrt{0 + 144} = \sqrt{144} = \mathbf{12}

3. A (3,1), B (-1,4 ) (green)

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(-1 - 3)^{2} + (4 - 1)^{2}}\\= \sqrt{(-4)^{2} + 3^{2}} = \sqrt{16 + 9} = \sqrt{25} = \mathbf{5}

4. A (-2,-5), B (3,7) (orange)

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5. A (5,4), B (-3,-2) (purple)

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