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EleoNora [17]
2 years ago
12

Find an equation of the line parallel to 8x − 3y = 1 and containing the point (−2, 0)

Mathematics
1 answer:
Anna007 [38]2 years ago
5 0

The equation of the line parallel to 8x − 3y = 1 and containing the point (−2, 0) is 8x -3y = -16

<h3>Equation of a straight line</h3>

From the question, we are to determine the equation of the line parallel to the given equation and that passes through the given point

Lines that are parallel to each other will have the same slope.

First, we will determine the slope of the line in the given equation.

The given equation of the line is

8x − 3y = 1

Express the equation in the slope-intercept form of a line, y = mx + b

Where m is the slope and

b is the y-intercept

That is,

8x − 3y = 1

8x -1 = 3y

3y = 8x - 1

y = 8/3(x) - 1/3

By comparing,

∴ The slope of the line is 8/3

Since we are to find the equation of a line parallel to this, the slope of the line will also be 8/3

Now, find the equation of a line containing the point (-2, 0) and that has  a slope of 8/3

Using the point-slope form of a line

y - y₁ = m(x - x₁)

x₁ = -2

and y₁ = 0

∴ y - 0 = 8/3(x - -2)

y = 8/3(x +2)

3y = 8(x +2)

3y = 8x + 16

8x -3y = -16

Hence, the equation of the line parallel to 8x − 3y = 1 and containing the point (−2, 0) is 8x -3y = -16

Learn more on Equation of a line here: brainly.com/question/13763238

#SPJ1

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Let's do

\\ \rm\dashrightarrow y=\dfrac{3}{x-h}+k

Release k for some while

If

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So

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So vertical asymptote is at origin now

It mentioned that it's at x=-5 so we need to change x

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\\ \rm\dashrightarrow y=\dfrac{3}{x-(-5)}

\\ \rm\dashrightarrow y=\dfrac{3}{x+5}

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Now

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But it's given

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Same put y=12 in place of k

\\ \rm\dashrightarrow y=\dfrac{3}{x+5}+12

  • h=-5
  • k=12

Graph attached for verification

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