Find the equation in standard form...

[flora33]

I would say this is my last question, but I might be lying! Please forgive my million posts, this unit has been really excruciating!

Find the equation, in standard form, with all integer coefficients, of the line perpendicular to x + 3y = 6 and passing through (-3, 5).
What I did was changed the x + 3y = 6 to the form: y = mx + b:
y = 1/3x + 2 and the problem is asking to find the equation for the line perpindicular to x + 3y = 6 which would be -3, I believe?
Then, use the formula y-y1 = m(x - x1)
y - 5 = -3(x - (-3))
y - 5 = -3(x +3)
y - 5 = 3x + 9
y - 5 + 5 = 3x + 9 + 5
y = 3x + 14 This is the slope-intercept form, not "standard form" so my next guess would be to change it to:
y - 3x = 14

Maybe I'm overthinking these problems, or maybe I'm just totally missing something (maybe from the last unit we covered?!) but I am having such a tough time! As always, I really, really appreciate your help!

 

[pka]

Find the equation, in standard form, with all integer coefficients, of the line perpendicular to x + 3y = 6 and passing through (-3, 5).
What I did was changed the x + 3y = 6 to the form: y = mx + b:
y = 1/3x + 2 SIGN ERROR

Any line perpendicular to \(\displaystyle x + 3y = 6\) will look like \(\displaystyle 3x - y = C\) now substitute –3 for x and 5 for y to determine C.

In general any line perpendicular to \(\displaystyle Ax + By = C\) through \(\displaystyle \left( {p,q} \right)\) is \(\displaystyle Bx - Ay = Bp - Ay.\)

 

[flora33]

Find the equation, in standard form, with all integer coefficients, of the line perpendicular to x + 3y = 6 and passing through (-3, 5).
What I did was changed the x + 3y = 6 to the form: y = mx + b:
y = 1/3x + 2 SIGN ERROR

Any line perpendicular to \(\displaystyle x + 3y = 6\) will look like \(\displaystyle 3x - y = C\) now substitute –3 for x and 5 for y to determine C.

In general any line perpendicular to \(\displaystyle Ax + By = C\) through \(\displaystyle \left( {p,q} \right)\) is \(\displaystyle Bx - Ay = Bp - Ay.\)

So I did:
3(-3) - 5 = C
9-5 = C
4 = C
Then:
3x - y = 4
-1y = -3x + 4
y = 3x - 4
Standard form:
3x + y = -4

 

[pka]

Flora33, It appears that you have an almost complete misunderstanding about how ‘signs’ work. You have multiple sign errors in the above. There is no point in your trying these problems until you master that material.

 

[flora33]

All right, I just wanted to post and put the correct solution up. I was able to find someone to help me out with it today, and I am pretty confident it is right:

3y = -1x + 6
3y/3 = -1x/3 + 6/3
y = -1/3x + 2

y – 5 = -1(x – (-3))
y = 3x + 9 + 5
y = 3x + 14
-3x + y = 14

Wee!

 

[Deleted member 4993]

All right, I just wanted to post and put the correct solution up. I was able to find someone to help me out with it today, and I am pretty confident it is right:

3y = -1x + 6
3y/3 = -1x/3 + 6/3
y = -1/3x + 2

y – 5 = -1(x – (-3))
y = 3x + 9 + 5<<<<< does not follow from above
y = 3x + 14
-3x + y = 14

Wee!

 

[flora33]

All right, I just wanted to post and put the correct solution up. I was able to find someone to help me out with it today, and I am pretty confident it is right:

3y = -1x + 6
3y/3 = -1x/3 + 6/3
y = -1/3x + 2

y – 5 = -1(x – (-3))
y = 3x + 9 + 5<<<<< does not follow from above
y = 3x + 14
-3x + y = 14

Wee!

I'm not sure from where above you are talking about? I used the y -y^1 = m(x - x^1) point-slope form. I put m=3 because it is the negative recriprocal of -1/3 and the problem wants the equation of the line perpendicular... I don't know where the I made an error, but the person I spoke with was a math tutor at my school, and they did not say anything was incorrect.

 

[stapel]

y – 5 = -1(x – (-3))
y = 3x + 9 + 5<<<<< does not follow from above

I'm not sure from where above you are talking about?

He was talking about the bit he was "pointing" to: How did you go from "-1(x)..." in one line to "3x" in the next? :shock:

Eliz.

 

[flora33]

Re:

y – 5 = -1(x – (-3))
y = 3x + 9 + 5<<<<< does not follow from above

I'm not sure from where above you are talking about?

He was talking about the bit he was "pointing" to: How did you go from "-1(x)..." in one line to "3x" in the next? :shock:

Eliz.

All right, I see what I did. This is terribly frustrating for me. I know I keep making these small (but very important) errors, and that's my whole problem. That -1 is meant to be 3. I had the work written down in front of me as I worked with the tutor, and I was looking at something else when I typed that -1... Hopefully, I didn't do the same when I submitted it to my teacher.
I hate algebra and I can not wait to be done with this class. It is absolute torture. I have struggled each week (as some of you already know, I am severely challenged when it comes to math stuff!) and I still have 5 weeks to go... Hopefully I can squeak by. So far I've got almost a 90% average ( I know, amazing, right?!)
Well, thanks for showing me the error in my ways yet again... This week I'm gonna try not to bug you all too much with questions... I'm working with exponents and polynomials, and I'm understanding a lot more then last week...

I'm sure I'll be back though!
Flora