Find the equation of the line passing through the point (1,−5) and perpendicular to y=18x+2

Hey there! :)

User wrote question wrong, correct question is:

Find the equation of the line passing through the point (1,−5) and perpendicular to y=1/8x+2

Line passes through (1, -5) & perpendicular to y = 1/8x + 2

There are a variety of different ways to start this off but I like to start off by identifying what our slope is. In the slope-intercept form y=mx+b, we know that "m" is our slope. "M" is simply a place mat so if we look at our given line, the "m" value is 1/8. Therefore, the slope of the given line is 1/8.

However, we must note that we're looking for a line that's perpendicular to the given one. This means that our new line needs to have a slope that is the negative reciprocal of it. A negative reciprocal is simply a number that is flipped and either given a negative or gets a negative removed.

So the negative reciprocal of 1/8 is -8. This means that our new slope is -8.

Now, we use point-slope form ( y-y₁=m(x-x₁) ) to complete our task of finding a line that passes through (1, -5). Our new slope is -8 & it passes through (1,-5).

y-y₁=m(x-x₁)

Let's start by plugging in -8 for m (our new slope), 1 for x1 and -5 for y1.

y - (-5) = -8(x - 1)

Simplify.

y + 5 = -8x + 8

Simplify by subtracting 5 from both sides.

y = -8x + 3

~Hope I helped!~

idk but i found this "the graph of y = f(x-2) is the graph of y = f(x) translated two units rightward.  

so, since the point (9,2) lies on the graph of y = f(x), the point (11,2) lies on the graph of y = f(x-2)."

step-by-step explanation:

link is

-6 is the correct option from the above given options