Question

@((*-*))
write an equation in slope-intercept form of a line that passes through the given point and is perpendicular to the graph of the given equation.
(0,0); y=-6x+5

Answer 1

your Q?

Answer 2

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Answer 3

^

Answer 4

@((*-*)) @Demonx341 @AnswerMyQuestions

Answer 5

you are tagging people, but question isn't being asked.

Answer 6

^

Answer 7

refresh your page I edited the question with the question posted in the bow.

Answer 8

a line perpendicular to a line with slope m, would have a slope of -1/m.

Answer 9

a line perpendicular to a line with slope -6, would have a slope of -1/(-6).

Answer 10

-1/(-6) is simplified to?

Answer 11

(two negatives=positive)

Answer 12

I cant post fast enough my computer is too slow. sorry.

Answer 13

sure, so do yu understand what I said about a slope of a perpendicular line?

Answer 14

positive 6 would be right?

Answer 15

positive 1/9\6.

Answer 16

oops 1/6

Answer 17

slope -6. perpendicular line has: -1/(-6) >>> 1/6.

Answer 18

for any slope K, of a line that passes through (0,0), you would have a line y=Kx

Answer 19

so the answer would be y=6x+5?

Answer 20

what is your new slope (perp. to a line y=6x+5) ?

Answer 21

I meant to a line y=-6x+5, typo

Answer 22

given a line: \(\large\color{r}{ y=-6x+5 }\)
you have to find a slope that would form a perpendicluar line:

your slope \(\large\color{r}{ -6 }\), slope of a new line (that would form perpendicular line), \(\large\color{r}{ -1/(-6) }\)
(simplifying the new slope, we get....) \(\large\color{r}{ 1/6 }\)

your equation has to go through (0,0), that means it has no non-zero y-intercept, saying that in
\(\large\color{r}{ y=mx+b}\), the constant b, is zero.

Answer 23

so all you need to do is:
\(\large\color{r}{ y={\rm(your~slope~that~forms~a~perpendicular~line)}\cdot x }\)