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(-1,6) , 3x+y=12

Write the slope intercept form for an equation of the line that passes through the given point and is parallel to the graph of each.
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2 Answers

Ok, let's start by writing the given equation in slope-intercept form.  We have 3x+y=12.  We need to put it in the form y=mx+b.
y=12-3x    (subtract 3x from 12)
y=-3x+12  (rearrange to match slope-intercept form)
 
We're looking for a parallel line, which means that the slope of the line will be the same as the given line -3.  The difference between the given line and the new line will be the intercept (b or 12).  So what we need to find is an intercept that will allow a line with a slope of -3 to pass through the point (-1,6).
 
To do that, we'll plug -1 for x and 6 for y into the equation y=-3x+12 and replace 12 with b.
This gives us:
6=-3(-1) + b.
 
Now we solve for b:
6=3+b   (multiply -3 and -1)
3=b       (subtract 3 from 6)
 
Finally, we put everything together:
y=-3x+3
 
This is the equation of the line that is parallel to the given line (3x+y=12) and passes through the point (-1,6).
Hi Gabby;
3x+y=12
The equation is in standard form...
Ax+By=c, A>0
Slope is -A/B.
Slope is -3/1, -3
The slope of the line parallel to this is the same slope of -3.
Point-slope formula is...
y-y1=m(x-x1)
Let's plug-in the point provided, (-1,6)...
y-6=(-3)(x--1)
Subtracting a negative number is the same as adding a positive number...
y-6=(-3)(x+1)
y-6=-3x-3
Let's add 6 to both sides...
y-6+6=-3x-3+6
y=-3x+3
This is slope-intercept format...
y=mx+b
m is the slope.
b is the y-intercept, the value of y when x=0.
y=-3x+3
Standard format...
3x+y=3