Karinarinaidn
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Fieryanswererftidn

Answer:

1st = [tex]A. \sf \ y = 1[/tex]

2nd = [tex]D. \ \sf y - 5 = 3(x + 2)[/tex]

PART A

Given: y = -2

This equation has a slope of 0

Equation that pass (-1, 1)

y - 1 = 0(x -(-1))

y = 1

PART B

[tex]\sf Given : y = -\dfrac{1}{3} x + 2[/tex]

Here slope is -1/3 and y-intercept is 2

Perpendicular lines has negatively inverse slope.

→ per. slope = -(slope)⁻¹ = -(-1/3)⁻¹ = 3

Equation in point slope form:

[tex]\sf y - y_1 = m(x - x_1)[/tex]

[tex]\sf y - 5 = 3(x - (-2))[/tex]

[tex]\sf y - 5 = 3(x + 2)[/tex]

Semsee45idn

Answer:

[tex]y = 1[/tex]

[tex]y-5=3(x+2)[/tex]

Step-by-step explanation:

The line [tex]y = -2[/tex] is a horizontal line.

Therefore, a line that is parallel to the given line and passes through point (-1, 1) is another horizontal line with the y-value of the point.

Therefore, the line is [tex]y = 1[/tex]

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Slope-intercept form of a linear equation:

 [tex]y=mx+b[/tex]

where:

  • m is the slope
  • b is the y-intercept

Given linear equation:

[tex]y=-\dfrac{1}{3}x+2[/tex]

Therefore, the slope of this equation is -1/3.

If two lines are perpendicular to each other, the product of their slopes will be -1.   Therefore, the slope (m) of the line perpendicular to the given line is:

[tex]\implies m \cdot -\dfrac{1}{3}=-1[/tex]

[tex]\implies m=3[/tex]

Use the point-slope form of a linear equation, with the found slope and the point (-2, 5) to create the equation:

[tex]\implies y-y_1=m(x-x_1)[/tex]

[tex]\implies y-5=3(x-(-2))[/tex]

[tex]\implies y-5=3(x+2)[/tex]

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