write the equation of a line that is perpendicular to y= 3x-2 and that passes through -9,5 PLEASE HELP !!! I don’t understand

Answer 1

Perpendicular lines have slopes whose product is -1. The slope given in

y = 3x - 2

is -3.

The slope of the perpendicular line will be m, and m needs to satisfy the equation

3 * m = -1

Hence m = -1/3.

Our perpendicular line is thus

y = (-1/3)x + b

for some constant b.

We are told that the perpendicular line goes through the point (-9,5). If so, then the following must hold:

5 = (-1/3)(-9) + b

= 3 + b

b = 2

Ans: y = (-1/3)x + 2

Answer 2

The key to understand this problem is that perpendicular lines have slopes that are the negative reciprocals of each other. Example, a line with slope 3/4 has a perpendicular with a slope of -4/3. So flip it over and take the negative. Like 7/12 and -12/7 or 2 and -1/2 (because 2 = 2/1)

The slope of y = 3x - 2 is 3 (the number preceding the "x' if the line is in the form y = mx + b)

So the line perpendicular to y = 3x - 2 has a slope of -1/3.

So now you just need to write the equation of the line with slope -1/3 that goes through (-9, 5). Use the point-slope form, y - y₁ = m(x - x₁) where (x₁, y₁) is (-9, 5) and m = -1/3

y - 5 = -1/3(x - -9)

y - 5 = -1/3(x + 9)

That's the answer. If they want you to give the answer in the y= mx + b format (slope-intercept) then multiply it out and combine like terms.

y - 5 = -1/3(x + 9)

y - 5 = -1/3x - 3

y = -1/3x + 2