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.
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?
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
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).