1. ## Sketch the line

Hi,
I have the straight line equation of, 2x+3y-12=0. I need to sketch the line but I'm not really sure how to do this so I require some help please!

Thanks!

2. ## Re: Sketch the line

Originally Posted by andyboy179
Hi,
I have the straight line equation of, 2x+3y-12=0. I need to sketch the line but I'm not really sure how to do this so I require some help please!

Thanks!
Rearrange into the form $\displaystyle y = mx+c$ and then plot it. m is the gradient and (0,c) is a point on the line

To plot a line you need two points. I suggest using $\displaystyle (0,c)$ and $\displaystyle (t,0)$ where t is the value that makes y=0

3. ## Re: Sketch the line

Originally Posted by andyboy179
Hi,
I have the straight line equation of, 2x+3y-12=0. I need to sketch the line but I'm not really sure how to do this so I require some help please!

Thanks!
One way to do this is to rewrite the general equation to a linear equation you are more familiar with y = mx + b, where m is the slope and b is the y intercept.

2x + 3y - 12 = 0
3y = -2x + 12
y = (-2/3)x + 4

can you take it from here? EDIT: -1 was faster than I was.

4. ## Re: Sketch the line

Thanks for the posts!!
I understand up to this point however when it comes to the graph I tknow where I would plot the (-2,3) but I don't understand the slope part. How would I draw this?

5. ## Re: Sketch the line

Originally Posted by andyboy179
Thanks for the posts!!
I understand up to this point however when it comes to the graph I tknow where I would plot the (-2,3) but I don't understand the slope part. How would I draw this?
start at the y-intercept (0,4) ... slope is rise/run, -2/3 ... down 2 , right 3

6. ## Re: Sketch the line

Oh I see! Thanks!
Do you know how I would be able to find the co-ordinates of A and B?

7. ## Re: Sketch the line

Originally Posted by andyboy179
Oh I see! Thanks!
Do you know how I would be able to find the co-ordinates of A and B?
what do you mean? what are points A and B?

8. ## Re: Sketch the line

I'm not too sure, I just have to fine the co ordinates of A and B. I wasn't sure that's why I asked.

9. ## Re: Sketch the line

Originally Posted by andyboy179
I'm not too sure, I just have to fine the co ordinates of A and B. I wasn't sure that's why I asked.
what are A and B ? cite the problem in its entirety.

10. ## Re: Sketch the line

After sketching the line where the 2x+3y-12=0 cuts the x axis at A and the y axis at B, find the coordinates of A and B.

11. ## Re: Sketch the line

Originally Posted by andyboy179
After sketching the line where the 2x+3y-12=0 cuts the x axis at A and the y axis at B, find the coordinates of A and B.
A is the x-intercept ... where y = 0

B is the y-intercept ... where x = 0 (you already know this set of coordinates)

12. ## Re: Sketch the line

So A's coordinates are (0,4) and B's coordinates are (7,0)?

13. ## Re: Sketch the line

Originally Posted by andyboy179
So A's coordinates are (0,4) and B's coordinates are (7,0)?
no

2x+3y = 12

A is the x-intercept (y = 0) ... 2x + 3(0) = 12 ... solve for x

B is the y-intercept ... come on Andy, look at post #5. pay attention, son.

14. ## Re: Sketch the line

So A's coordinates are (0,6) and B's coordinates are (0,4)?

15. ## Re: Sketch the line

Originally Posted by andyboy179
So A's coordinates are (0,6) and B's coordinates are (0,4)?
Again, no- but your error may just be a typo. When x= 0, you calculated that y= 12/3= 4 so (0, 4) is a point. When y= 0, x= 12/2= 6 but x is always the first member of the pair- the other point is (6, 0), not (0, 6).

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