Write the equation of a line that is perpendicular to y=-0.3x+6 and passes through the point 3,-8

Answer 1

We want to start with the fact that the line we’re solving for is perpendicular to the line they gave us - that means that the slope of the line we’re solving for is the negative reciprocal of the given slope. The equation they gave us is in slope-intercept form, which is y = mx + b, where m is the slope.

since the slope of the given line is -0.3, our slope will be

+ 1 / 0.3, which gives us 3.33 (with the three repeating). Now that we know our slope, we can use y = mx + b as a formula, plugging in the slope we know and the point it passes through, to solve for the “b” (aka the y-intercept) of the equation. We’ll use the point (3, -8) as the (x,y) in our equation. That gives us:

-8 = 3.33333333(3) + b

Start by multiplying the 3.33333333 and the 3 to get:

-8 = 10 + b

Subtract the 10 to the other side to get b = -18

Now we have our full equation: y = 3.33x - 18

Answer 2

Use the procedure described here:

https://www.wyzant.com/resources/answers/727327/write-the-equation-that-is-parallel-to-y-3-2x-1-and-that-passes-through-4-6

except that--for a perpendicular line, the slope is the negative reciprocal of the original line---so the new line will have a slope of +10/3