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

What is the equation of the line that passes through the point (6, 1) and is perpendicular to the line y = –3x + 2

OpenStudy (mertsj):

What is the slope of y = -3x = 2?

OpenStudy (anonymous):

y = –3x + 3 y = –3x – 9 y = x – 1 y = x + 19

OpenStudy (mertsj):

y = -3x+2, what is it's slope?

OpenStudy (mertsj):

No, sweetheart, that is not right.

OpenStudy (anonymous):

Those are the choices for answers and he/she is asking you to just pick one and give the answer.

OpenStudy (mertsj):

That's why I kept asking you what the slope is. Because the slope of the perpendicular is the negative reciprocal of the slope of the given line.

OpenStudy (mertsj):

Well none of those are correct.

OpenStudy (mertsj):

Because none have slope 1/3

OpenStudy (anonymous):

Agree. But, the "asker" is absent. So, I don't know what to say.

OpenStudy (mertsj):

Well, maybe you should just solve it for her and then we can leave.

Directrix (directrix):

y= -3x + 2 has slope -3. The slope of a line perpendicular to y=-3x+2 will have slope 1/3 because the slope product of two perpendicular lines = -1 ( with the exception of a vertical line and a horizonal line). Use the point-slope formula, y -y sub 1 = m (x - x sub 1) where (x sub1, y sub) is the point with coordinates (6, 1). y - 1 = (1/3)( x - 6) y -1 = (1/3)( x ) - 2 y -1 = (1/3) (x) - 2 y - 1 = (1/3) x - 2 y = (1/3) x - 1 ----> Choice not given

Directrix (directrix):

@ Mertsj Speaking for myself, the use of condescending language such as "sweetheart" is inappropriate here. Reference "No, sweetheart, that is not right." above. http://dailyplateofcrazy.com/2010/03/26/terms-of-endearment-let-me-call-you-sweetheart/

OpenStudy (anonymous):

its is asking for PARALLEL THAT IS WHY .

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!