Question

Can someone please help, I have been stuck on this for a while.

Write the equation of the line that is perpendicular to the line 3x + y = 7 and passes through the point (6, −1).

y = one thirdx − 3

y = one thirdx + 17

y = −3x − 3

y = −3x + 17

Answer 1

Remember the slope of the perpendicular lines: \[(\frac{ c }{ d })(\frac{ -d}{ c }) = -1\]

Answer 2

Ok well the formula you want is: \[y-y _{1} = m(x-x_{1})\]
So you have: \[3x+y=7\]\[y= -3x + 7\]
So slope would be the negative reciprocal of -3, which is: \[\frac{ 1 }{ 3 }\]
Your equation should look like: \[y-(-1)=\frac{ 1 }{ 3 }(x-6)\]
Should be easy from here.

Answer 3

Ok, I am going to do the work, sorry it took so long for me to see this.

Answer 4

Is it y= -3x + 17?