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Mathematics 20 Online
OpenStudy (anonymous):

Oh my gosh my study time is almost up 2 more questions to go! Muaha, Help me please :3

OpenStudy (anonymous):

hartnn (hartnn):

can u find slope of y=1/2 x + 1 ?

OpenStudy (anonymous):

1/2

hartnn (hartnn):

yes, and slope of parallel lines are ..... ?

OpenStudy (anonymous):

the same ? 0.0

hartnn (hartnn):

yes, so u have slope of your line as m=1/2

OpenStudy (anonymous):

yus

hartnn (hartnn):

now it passes through 12,15 u know the slope point form of equation of line ?

OpenStudy (anonymous):

i do but im sleepy so its not coming to me at the moment

hartnn (hartnn):

(y-y1) = m (x-x1) here, x1= 12, y1 = 15

OpenStudy (anonymous):

(y-15)=m(x-12)

hartnn (hartnn):

u already had m. put that also then isolate y

OpenStudy (anonymous):

y=1/2x+9?

hartnn (hartnn):

yes :)

OpenStudy (anonymous):

so which is perpendicular?

hartnn (hartnn):

now slope of perpendicular line is..?

OpenStudy (anonymous):

>.> jihx

OpenStudy (anonymous):

idk >.<

hartnn (hartnn):

negative reciprocal. so take reciprocal of 1/2 , what u get then multiply it by -1

hartnn (hartnn):

reciprocal of 1/2 is ?

OpenStudy (anonymous):

2/1

hartnn (hartnn):

and multiplying 2 with -1 will give u ?

OpenStudy (anonymous):

you get -2

hartnn (hartnn):

yes

hartnn (hartnn):

so slope = -2

hartnn (hartnn):

which option ?>

OpenStudy (anonymous):

B right?

hartnn (hartnn):

yup :)

OpenStudy (anonymous):

Omahgerd. I has one more :3

hartnn (hartnn):

okay :)

OpenStudy (anonymous):

OpenStudy (anonymous):

i only has 5 minutes before someone else needs computer :0

hartnn (hartnn):

this can't be solved in 5 min.

hartnn (hartnn):

Distance between points (x1,y1) and (x2,y2) is \(\huge d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)

OpenStudy (anonymous):

sorry i got disconnected

hartnn (hartnn):

me too :\

OpenStudy (anonymous):

Well. looks like ill have to get off in a minute. dont even think ill have time to attempt problem :(

OpenStudy (anonymous):

I just wanted to say thank you for all of your help. You've been on like every single problem of mine xD

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