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Ad libitum [116K]
3 years ago
8

Two of the vertices of a rectangle are (1, -6) and ( -8, -6 ) if the rectangle has a perimeter of 26 units what are the coordina

tes of it's other vertices?
Mathematics
1 answer:
Murrr4er [49]3 years ago
5 0

Answer:

(1, -2)

(-8, -2)

Step-by-step explanation:

(1 , -6) and (-8 , -6)

1 - (-8) = 9

we know that the length of the side that we know the vertices is 9

from there we make an equation with the sum of the sides equal to the perimeter

we will have 2 times 9 and 2 times x beacause it is a rectangle

x + x + 9 +9 = 26

2x + 18 = 26

2x = 26 - 18

2x = 8

x = 8/2

x = 4

Now that we know the missing side we just have to add or subtract this value to the coordinate in and of the vertices we have and we will obtain the missing vertices

(1, -6 + 4)

(1, -2)

( -8, -6+4 )

(-8, -2)

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write the standard form of the equation of the circle for which the endpoints of a diameter are (0,0) and (4,-6)
Elenna [48]

Given:

The endpoints of a diameter are (0,0) and (4,-6).

To find:

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Solution:

The endpoints of a diameter are (0,0) and (4,-6). So, the length of the diameter is

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Now, radius is half of the diameter.

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Center of the circle is the midpoint of the endpoints of a diameter.

Center=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)

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Center=\left(\dfrac{4}{2},\dfrac{-6}{2}\right)

Center=\left(2,-3\right)

Standard form of a circle is

(x-h)^2+(y-k)^2=r^2

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The center of the circle is (2,-3) and radius is \sqrt{13}. So,

(x-2)^2+(y-(-3))^2=(\sqrt{13})^2

(x-2)^2+(y+3)^2=13

Therefore, the standard form of the circle is (x-2)^2+(y+3)^2=13.

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Answer:

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