SOLUTION: It says to write an equation in standard form for each line. I'm having trouble figuring out this one problem: the line crosses the x-axis at x=-3 and the y-axis at y=6. I was ju

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: It says to write an equation in standard form for each line. I'm having trouble figuring out this one problem: the line crosses the x-axis at x=-3 and the y-axis at y=6. I was ju      Log On


   

Question 69080This question is from textbook Algebra 1
: It says to write an equation in standard form for each line. I'm having trouble figuring out this one problem: the line crosses the x-axis at x=-3 and the y-axis at y=6. I was just wondering if anyone could tell me how to write an equation for this? This question is from textbook Algebra 1

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
One approach is to write the equation first in the "slope-intercpt" form, then convert this to the "standard form".
First, find the slope of the line from the given information. You have two points on the: (-3, 0) and (0, 6). These are the given x- and y-intercepts, respectively.
From these two points you can find the slope using: m+=+%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29
m+=+%286-0%29%2F%280-%28-3%29%29
m+=+6%2F3
m+=+2
So now you can write:
y+=+2x%2Bb but you have been given b, the y-intercept. It's 6.
y+=+2x%2B6 This is the equationin the slope-intercept form. You now need to convert it to the standard form: ax%2Bby+=+c
y+=+2x%2B6 Subtract y from both sides.
0+=+2x-y%2B6 Subtract 6 from both sides.
-6+=+2x-y or 2x-y+=+-6