Find the equation of the line that contains the point (3,2) and is parallel to the line 3x + y = -3.?

Question

Help with this someone!

Answer 1

y = - 3 x - 3
m = - 3
y - 2 = (- 3) (x - 3)
y = (- 3) x + 9 + 2
y = (- 3) x + 11

Answer 2

The line you are looking for has the general equation
y=mx+b. This is called the slope-intercept form for the equation of any straight line. When a straight line is in this form, m, the coefficient of x, is the slope of the line and b is the line's y-intercept

Our task is to find values for m and b for the line you seek.
We know the line is parallel to 3x+y=-3, so our line will have the same slope. Before we can determine what that slope is, we must put 3x+y=-3 into the
slope-intercept form. THIS REQUIRES +1y TO BE
ALL BY ITSELF ON ONE SIDE OF THE EQUATION.
3x+y=-3
y=-3x-3
There it is- and now we know its slope. It is -3, the coefficient of the x term.
Our desired line now looks like this: y=-3x+b
Now, to find b
We are told (3,2) lies on our line. Therefore, this point satisfies our equation. I can thus replace x with 3 and y with 2 to get 2=-3(3)+b,
2= -9+b,
2+9=b, b=11
We're done. The sought-after equation is
y=-3x+11

Answer 3

the general line parallel to your is

y = -3x +c

than I put x = 3 and y = 2

2 = -3 * 3 +c

c = 11

Hence the line is

y = -3x + 11
y +3x = 11