SOLUTION: Graphing a line given its equation in standard form. Graph the line 4x-3y-12=0

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Question 283588: Graphing a line given its equation in standard form. Graph the line
4x-3y-12=0
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
4x-3y-12=0 Start with the given equation.


4x-3y=12 Add 12 to both sides.


-3y=12-4x Subtract 4x from both sides.


-3y=-4x%2B12 Rearrange the terms.


y=%28-4x%2B12%29%2F%28-3%29 Divide both sides by -3 to isolate y.


y=%28%28-4%29%2F%28-3%29%29x%2B%2812%29%2F%28-3%29 Break up the fraction.


y=%284%2F3%29x-4 Reduce.



Looking at y=%284%2F3%29x-4 we can see that the equation is in slope-intercept form y=mx%2Bb where the slope is m=4%2F3 and the y-intercept is b=-4


Since b=-4 this tells us that the y-intercept is .Remember the y-intercept is the point where the graph intersects with the y-axis

So we have one point




Now since the slope is comprised of the "rise" over the "run" this means
slope=rise%2Frun

Also, because the slope is 4%2F3, this means:

rise%2Frun=4%2F3


which shows us that the rise is 4 and the run is 3. This means that to go from point to point, we can go up 4 and over 3



So starting at , go up 4 units


and to the right 3 units to get to the next point



Now draw a line through these points to graph y=%284%2F3%29x-4

So this is the graph of y=%284%2F3%29x-4 through the points and