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Line s has an equation of y=8x+7. Line t is perpendicular to line s and passes through (–8,–2). What is the equation of line t?Write the equation in slope-intercept form. Write the numbers in the equation as simplified proper fractions, improper fractions, or integers.

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akposevictor

The simplified equation in slope-intercept form of the line t that passes through (-8, -2) and is perpendicular to line s is: [tex]\mathbf{y = -\frac{1}{8} x - 3}[/tex]

Recall:

  • The slope-intercept form of any line can be written as: [tex]\mathbf{y = mx + b}[/tex]
  • point-slope form:  [tex]\mathbf{y - b = m(x - a)}[/tex]
  • If a line is perpendicular to another line, the slope of one will be the negative reciprocal of the other line it is perpendicular to.

Given:

  • Line s equation: y = 8x+7
  • line t passes through line s
  • line t passes through a point (-8, -2)

Thus:

  • The slope of line s, [tex]y = 8x+7[/tex] is 8.

  • The slope (m) of line t, will be the negative reciprocal of 8, which is [tex]-\frac{1}{8}[/tex]

Write the equation of line t in point-slope form by substituting m = [tex]-\frac{1}{8}[/tex] and (a, b) = (-8, -2) into [tex]\mathbf{y - b = m(x - a)}[/tex].

  • Thus:

[tex]\mathbf{y - (-2) = -\frac{1}{8} (x - (-8))}\\\\\mathbf{y + 2 = -\frac{1}{8}(x + 8)}[/tex]

Rewrite the equation in the form of  [tex]\mathbf{y = m(x - b)}[/tex]

  • Thus:

[tex]\mathbf{y + 2 = -\frac{1}{8} (x + 8)}\\\\y + 2 = -\frac{1}{8} x - 1\\\\y + 2 - 2 = -\frac{1}{8} x - 1 - 2\\\\y = -\frac{1}{8} x - 3[/tex]

Therefore, the simplified equation in slope-intercept form of the line t that passes through (-8, -2) and is perpendicular to line s is: [tex]\mathbf{y = -\frac{1}{8} x - 3}[/tex]

Learn more here:

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