Write the equation of a line that passes through (6,-5) and is perpendicular to y=3x-2 NEED HELP

Answer 1

The line which is perpendicular will have a slope of (-1/3) the negative reciprocal of 3

So equation is y = (-1/3)x +b You can solve for b using the given point (6,-5)

-5= (-1/3)(6) +b

-5 = -2 +b

b=-3

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So equation is y= (-1/3)x -3

Answer 2

A line that is perpendicular to another has a slope that is the negative reciprocal of its slope. The slope of y = 3x - 2 is "3" (when in the form "y = mx + b", the slope is the "m")

the reciprocal of 3 (aka 3/1) is 1/3. The negative reciprocal is -1/3.

So we need a line with a slope of -1/3 that goes through the point (6, -5).

One way to get the equation of that line is to use the point-slope form:

y - y₁ = m(x - x₁) where "m" is the slope and (x₁, y₁) is the point it goes through. So, since x₁ = 6 and y₁ = -5 we get:

y - (-5) = -1/3(x - 6)

y + 5 = -1/3(x - 6)

This is an equation that will work. If for some reason, you want to put it in y = mx + b format, just multiply it out and combine like-terms:

y + 5 = -1/3(x - 6)

y + 5 = -1/3x + 2

y = -1/3x - 3