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. <u>Therefore, the slope of the given line is 1/8.</u>
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!~
