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Write an equation of a line that is perpendicular to y=3x+3 and passes through (-6,3)

User MrSpaar
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Answer:

The equation of a line that is perpendicular to y = 3x + 3 is option A y equals negative one-third times x plus 1(-1/3x + 1)Given,The equation of line, y = 3x + 3The line passes through (-6, 3)We have to find the equation of the line that is perpendicular to y = 3x + 3Slope of the line, m = 3Here, in perpendicular case, slope is negative of its reciprocal.That is, m = -1/3Now, we can find the equationy - y₁ = m(x - x₁)

Here,

y₁ = 3x₁ = -6m = -1/3

So,y - 3 = -1/3(x - (-6))

y - 3 = -1/3(x + 6)

y - 3 = -1/3x + -1/3(6)

y - 3 = -1/3x - 6/3

y - 3 = -1/3

x - 2y = -1/3

x - 2 + 3

y = -1/3x + 1

That is, the equation of a line that is perpendicular to y = 3x + 3 is y equals negative one-third times x plus 1(y = -1/3x + 1)

Explanation:

User Dub Stylee
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