Find the equation of the line parallel to the line 3x + y = 5 and containing the point (4, – 2).
This is what I got. Is this correct?
y = –3x + 10
4 answers
Yes!
Is my evvaluation correct?
y=mx+b
3x + y = 5
y = -3x+5
y = -2
m = -3
x = 4
b = unknown
Solve For b:
y = -3x + b
-2 = -3(4) + b
-2 = -12 + b
-12 + 10 = -2
10=b
To Check:
y = -3x + b
-2 = -3(4) + b
-2 = -12 + b
-12+10 = b
10 = b
-2 = -3(4) + b
-2 = -12 + 10
-12+10 = -2
b = 10
y=mx+b
3x + y = 5
y = -3x+5
y = -2
m = -3
x = 4
b = unknown
Solve For b:
y = -3x + b
-2 = -3(4) + b
-2 = -12 + b
-12 + 10 = -2
10=b
To Check:
y = -3x + b
-2 = -3(4) + b
-2 = -12 + b
-12+10 = b
10 = b
-2 = -3(4) + b
-2 = -12 + 10
-12+10 = -2
b = 10
A real easy way to do this question is to realize that since they have the same slope, the equations must differ only in the constant.
So let the new equation be
3x + y = c
plug in the point (4,-2)
12 - 2 = c = 10
so your equation is 3x + y = 10
[ BTW, had it asked for a perpendicular line then my opening new equation would have been
x + 3y = c
(notice the slopes would be opposite reciprocals) ]
So let the new equation be
3x + y = c
plug in the point (4,-2)
12 - 2 = c = 10
so your equation is 3x + y = 10
[ BTW, had it asked for a perpendicular line then my opening new equation would have been
x + 3y = c
(notice the slopes would be opposite reciprocals) ]
So my work is correct? Thanks for the help!