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Ivanshal [37]
3 years ago
14

I need help! Find the equation of a line parallel to 3x-3y=5 that contains the point (-1,-5). Write the equation in slope Interc

ept form.

Mathematics
1 answer:
Readme [11.4K]3 years ago
6 0

the equation of the line is y= x-4

Step-by-step explanation:

so so the first thing that you're going to do is get y alone

-x-5=y

next you're going to find your slope which in this case is negative one.

after that in order to find the equation of a parallel line you're going to find the reciprocal of the slope.

so in this case the reciprocal is one.

next you're going to substitute in the cordinates to find the y intercept

x+b=y. (-1)+b= -5. (. b is the y intercept)

b= -4

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A circular platform is to be built in a playground. The center of the structure is required to be equidistant from three support
castortr0y [4]

Answer:

The coordinates for the location of the center of the platform are (0, 1)

Step-by-step explanation:

The equation of the circle of center (h , k) and radius r is:

(x - h)² + (y - k)² = r²

Now,

- The center is equidistant from any point lies on the circumference of the circle

- There are three points equidistant from the center of the circle

- We have three unknowns in the equation of the circle h , k , r

Thus, let's substitute the coordinates of these point in the equation of the circle to find h , k , r.

The equation of the circle is (x - h)² + (y - k)² = r²

∵ Points A(2,−3), B(4,3), and C(−2,5)

- Substitute the values of x and y the coordinates of these points

Point A (2 , -3)

(2 - h)² + (-3 - k)² = r² - - - (1)

Point B (4 , 3)

(4 - h)² + (3 - k)² = r² - - - - (2)

Point C (-2 , 5)

(-2 - h)² + (5 - k)² = r² - - - - (3)

- To find h , k equate equation (1) and (2) and same for equation (2) and (3) because all of them equal r²

Thus;

(2 - h)² + (-3 - k)² = (4 - h)² + (3 - k)² - - - - - (4)

(4 - h)² + (3 - k)² = (-2 - h)² + (5 - k)² - - - - -(5)

- Simplify (5);

h² - 8h + 16 + k² - 6k + 9 = h² + 4h + 4 + k² - 10k + 25

h² and k² will cancel out to give;

-8h - 6k + 25 = 4h - 10k + 29

Rearranging, we have;

12h - 4k = -4 - - - - (6)

Similarly, for equation 4;

(2 - h)² + (-3 - k)² = (4 - h)² + (3 - k)²

h² - 4h + 4 + k² + 6k + 9 = h² - 8h + 16 + k² - 6k + 9

h², k² and 9 will cancel out to give;

4 - 4h + 6k = 16 - 8h - 6k

Rearranging;

4h + 12k = 12 - - - - (7)

Divide by 4 to give;

h + 3k = 3

Making h the subject;

h = 3 - 3k

Put 3 - 3k for h in eq 6;

12(3 - 3k) - 4k = -4

36 - 36k - 4k = -4

40k = 40

k = 40/40

k = 1

h = 3 - 3(1)

h = 0

The coordinates for the location of the center of the platform are (0, 1)

5 0
3 years ago
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