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IrinaVladis [17]
2 years ago
5

Write in standard form an equation for the line with slope 1/3 through the point (2,-2).

Mathematics
1 answer:
Alexxandr [17]2 years ago
6 0

Answer:

x - 3y = 8.

Step-by-step explanation:

Use the point-slope form of the equation of a line:

y - y1 = m(x - x1)  where m = the slope and (x1, y1) is a point on the line.

So substituting the given values:

y - (-2) = 1/3(x - 2)

y + 2 = 1/3x - 2/3     Multiply through by 3:

3y + 6 = x - 2

x - 3y  =  6 + 2

x - 3y = 8  <---- Standard Form.

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Answer:

(x - 1)²/4² - (y - 2)²/2² = 1 ⇒ The bold labels are the choices

Step-by-step explanation:

* Lets explain how to solve this problem

- The equation of the hyperbola is x² - 4y² - 2x + 16y - 31 = 0

- The standard form of the equation of hyperbola is

  (x - h)²/a² - (y - k)²/b² = 1 where a > b

- So lets collect x in a bracket and make it a completing square and

  also collect y in a bracket and make it a completing square

∵ x² - 4y² - 2x + 16y - 31 = 0

∴ (x² - 2x) + (-4y² + 16y) - 31 = 0

- Take from the second bracket -4 as a common factor

∴ (x² - 2x) + -4(y² - 4y) - 31 = 0

∴ (x² - 2x) - 4(y² - 4y) - 31 = 0

- Lets make (x² - 2x) completing square

∵ √x² = x

∴ The 1st term in the bracket is x

∵ 2x ÷ 2 = x

∴ The product of the 1st term and the 2nd term is x

∵ The 1st term is x

∴ the second term = x ÷ x = 1

∴ The bracket is (x - 1)²

∵  (x - 1)² = (x² - 2x + 1)

∴ To complete the square add 1 to the bracket and subtract 1 out

   the bracket to keep the equation as it

∴ (x² - 2x + 1) - 1

- We will do the same withe bracket of y

- Lets make 4(y² - 4y) completing square

∵ √y² = y

∴ The 1st term in the bracket is x

∵ 4y ÷ 2 = 2y

∴ The product of the 1st term and the 2nd term is 2y

∵ The 1st term is y

∴ the second term = 2y ÷ y = 2

∴ The bracket is 4(y - 2)²

∵ 4(y - 2)² = 4(y² - 4y + 4)

∴ To complete the square add 4 to the bracket and subtract 4 out

   the bracket to keep the equation as it

∴ 4[y² - 4y + 4) - 4]

- Lets put the equation after making the completing square

∴ (x - 1)² - 1 - 4[(y - 2)² - 4] - 31 = 0 ⇒ simplify

∴ (x - 1)² - 1 - 4(y - 2)² + 16 - 31 = 0 ⇒ add the numerical terms

∴ (x - 1)² - 4(y - 2)² - 16 = 0 ⇒ add 14 to both sides

∴ (x - 1)² - 4(y - 2)² = 16 ⇒ divide both sides by 16

∴ (x - 1)²/16 - (y - 2)²/4 = 1

∵ 16 = (4)² and 4 = (2)²

∴ The standard form of the equation of the hyperbola is

   (x - 1)²/4² - (y - 2)²/2² = 1

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