Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (anonymous):

What is the slope of the line that is perpendicular to the line whose equation is 2x + y = 4.

OpenStudy (anonymous):

Do you have choices?

OpenStudy (anonymous):

y = x - 4 y = x + 4 y = -x + 4

OpenStudy (anonymous):

those are the answers

OpenStudy (anonymous):

First we need to put the first equation into y=mx+b

OpenStudy (anonymous):

To do this simply subtract 2x from one side and to the other. It should now look like \[y=-2x+4\]

OpenStudy (anonymous):

You only have three choices?

OpenStudy (anonymous):

Okay well Im a little confused because if a line was perpendicular, it would be negative and switched. For Example: \[\frac{ 3 }{ 2 }\] would be switched to \[-\frac{ 2 }{ 3 }\]

OpenStudy (anonymous):

Which means the line perpendicular would be \[-\frac{ 1 }{ 2 }\]

OpenStudy (anonymous):

Perpendicular lines have negative reciprocal slopes

OpenStudy (anonymous):

Thanks stephen... Kinda already said that.

OpenStudy (anonymous):

im so sorry the answer choices are -2 -1/2 1/2 2

OpenStudy (anonymous):

Now those seem likes slopes

OpenStudy (anonymous):

Ok so it is there. The line perpendicular would be anything with the slope of -1/2

OpenStudy (anonymous):

What is the equation of the line described below written in slope-intercept form? the line passing through point (2, 2) and perpendicular to the line whose equation is y = x

OpenStudy (anonymous):

Close this question and make a new one

OpenStudy (anonymous):

I am not entirely sure that is correct, if it is y = -2x +4, the negative reciprocal slope would be 1/2

OpenStudy (anonymous):

Oh shoot sorry yeah stephen is right.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!