Ask your own question, for FREE!
Mathematics 22 Online
OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

y = -x + 8 So we have m = -1 and b can be anything y = -x + b II y = -x + 8 y = -1 x = 4 -1 = -4 +b b= 3

OpenStudy (welshfella):

x+y=8 y = -x + 8 this has a slope of -1

OpenStudy (anonymous):

no?

OpenStudy (anonymous):

I think is y = -x + 3

Directrix (directrix):

y - y1 = m* (x - x1) where (x1,y1) = (4, -1) and m = -1

Directrix (directrix):

Parallel lines have the same slope.

OpenStudy (welshfella):

yes This form is called the Point-slope form of a straight ;ime

Directrix (directrix):

y - (-1) = 1* (x - 4)

OpenStudy (anonymous):

u could also use\[y=mx+b\]where\[y=vertical~axis\]\[x=horizontal~axis\]\[m=slope\]\[b=constant/y-intercept\]

Directrix (directrix):

Why not finish this first: y - (-1) = 1* (x - 4)

OpenStudy (anonymous):

y=x-5

OpenStudy (anonymous):

go ahead @Directrix :)

Directrix (directrix):

Yeah, well, I have a bad slope. y - y1 = m* (x - x1) where (x1,y1) = (4, -1) and m = -1 y - (-1) = -1* (x - 4)

OpenStudy (anonymous):

y = -x + 5 (4;-1) -1 = -4 + 5 3 = 5?????? I can not undertand you

OpenStudy (anonymous):

\[(4,-1)=(x,y)\]\[m=-1\]substitute this value into this equation to find b\[y=mx+c\]

OpenStudy (anonymous):

\[y=mx+b*\]

OpenStudy (anonymous):

we have y = kx + c

OpenStudy (anonymous):

\[-1=-1(4)+b\]\[-1=-4+b\]Add 4 on both sides

OpenStudy (anonymous):

\[-1+4=-4+4+b\]\[b=3\]

OpenStudy (anonymous):

YES

OpenStudy (anonymous):

y = -x + 8 So we have m = -1 and b can be anything y = -x + b II y = -x + 8 y = -1 x = 4 -1 = -4 +b b= 3 in the top of this question

OpenStudy (anonymous):

\[Now,we~can~form~a~new~equation~using\]\[m=-1~and~b=3\]substitute this value to form the equation\[y=mx+b\]

Directrix (directrix):

y = -x + 3

OpenStudy (anonymous):

yep :)

OpenStudy (anonymous):

T_T

OpenStudy (anonymous):

thanks guys so much!!!

Directrix (directrix):

It's fun when several people collaborate in a thread.

OpenStudy (anonymous):

It was strange

OpenStudy (anonymous):

Agree @Directrix yw! @cutegirl

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!