SOLUTION: write the equation of the line that is parallel to the line y=-3+12 and passes through the point (-1,6)

Question 562342: write the equation of the line that is parallel to the line y=-3+12 and passes through the point (-1,6)
Found 2 solutions by nerdybill, mananth:
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
I think you meant:
y=-3x+12
.
If so, the slope of the line above is -3
If the new line is parallel, it must have the same slope.
.
Plug the slope (-3) and the given point (-1,6) into the "point-slope" form:
y - y1 = m(x - x1)
y - 6 = -3(x - (-1))
y - 6 = -3(x + 1)
y - 6 = -3x - 3
y = -3x + 3 (this is what they're looking for)

Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
1 y = -3 x + 12
Divide by 1
y = -3 x + 12
Compare this equation with y=mx+b
slope m = -3 0.8
The slope of a line parallel to the above line will be the same
The slope of the required line will be -3
m= -3 ,point (-1,6 )
Find b by plugging the values of m & the point in
y=mx+b
6 = 3 + b
b= 3
m= -3
Plug value of the slope and b in y = mx +b
The required equation is y= -3x + 3

m.ananth@hotmail.ca

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