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

Question

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.

Answer 1

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).

Answer 2

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