Question

I need help figuring out how to get the perpendicular line to -2x+3y=9 through (-1,-3)

Answer 1

First put the equation in slope-intercept form

Answer 2

After that, obtain the slope. And perpendicular means the negative reciprocal, so plug that into the point-slope formula, and if needed simplify.

Answer 3

Also\[\large \mathbb{\color{red}{WELCOME~TO~OPENSTUDY}}\]

Answer 4

So the slope intercept form would be y= 2/3x +3 ? But I don't know exactly what to do after, I need a more clear explaination please.

Answer 5

Do you know the slope?

Answer 6

That's what I need help on trying to figure out.

Answer 7

The slope in y=mx+b is m
SO in y=(2/3)x+3 would be 2/3

Answer 8

So when I put it in reciprocal form it will be -2/3 ? Right?

Answer 9

negative reciprocal:
\[\Large \frac{a}{b}=-\frac{b}{a}\]

Answer 10

ohh , now I understand it better. So it would be -3/2 , then I plug in the points that I have , then use the method to find b so y=-3/2x+b? or do I just put b as 0?

Answer 11

Or do I do it the other way around

Answer 12

No you're right, -3/2, plug in and find in slope intercept form :)

Answer 13

It would be \[y=\frac{- 3 }{2 }x+\frac{ 5 }{ 2 }\] then right?

Answer 14

Hmm, I got \[y=-\frac{3}{2}x-\frac{9}{2}\]
Maybe I messed up somewhere

Answer 15

I did mess up, whoops.. but now I got -3/2 instead .-.

Answer 16

I'll show you my process if you want?

Answer 17

Yea that will be kinda better because I think I did something wrong.

Answer 18

Okay, so we got the formula:
\[\LARGE y-y_{1}=m(x-x_{1})\] and the points \[\Large (-1, -3)==>(x_{1}, y_{1})\]
So we plug that in:
\[\huge y+3=-\frac{3}{2}(x+1)\]
Now distribute and get:
\[\huge y+3=-\frac{3}{2}x-\frac{3}{2}\]
Now subtract 3 from both sides
\[\huge y+3-3=-\frac{3}{2}x -\frac{3}{2}-\frac{6}{2}\]
Since 3=6/2 in fraction form, now simplify:
\[\huge y=-\frac{3}{2}x-\frac{9}{2}\]

Try to see where I messed up >.<

Answer 19

You actually got it right, I had got it wrong because I didnt use the right formula ._.

Answer 20

Oh, well then :P

Answer 21

Glad I could help ^-^

Answer 22

Thank you (:

Answer 23

Anytime :)
and not trying to be selfish or anything but can I have a medal? :P

Answer 24

Thank you :D

Answer 25

No Problem