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Mathematics 25 Online
OpenStudy (brucebaner):

help me plz

OpenStudy (brucebaner):

how do you find the Use point-slope form to write the equation of the line in standard form with this points:(-5,2) (1/2,4)

OpenStudy (brucebaner):

@SolomonZelman

OpenStudy (brucebaner):

@AriPotta

OpenStudy (brucebaner):

@chosenmatt

OpenStudy (brucebaner):

can you help me @AriPotta

OpenStudy (aripotta):

these are our points: (-5,2) and (1/2,4). use the slope formula to find the slope first\[\frac{ y _{2}-y _{1} }{ x _{2}-x _{1} }\]

OpenStudy (brucebaner):

sorry wrong answer

OpenStudy (brucebaner):

i don't now

OpenStudy (aripotta):

well i got \[\frac{ 2 }{ 5.5 }\]

OpenStudy (brucebaner):

how did you find 2/5.5

OpenStudy (aripotta):

\[\frac{ 4-2 }{ 0.5-(-5) }=~\frac{ 2 }{ 0.5+5 }=~\frac{ 2 }{ 5.5 }\]

OpenStudy (brucebaner):

but what happened to the 1/2

OpenStudy (aripotta):

1/2 is the same as 0.5

OpenStudy (aripotta):

so now that we know the slope, we're gonna use that and one of the points in point-slope form, which is y - y1 = m(x - x1), where m is the slope, y1 is the y part of the point, and x1 is the x part of the point

OpenStudy (brucebaner):

okay

OpenStudy (aripotta):

so what do you get when you do that

OpenStudy (brucebaner):

i dont now plz help me

OpenStudy (aripotta):

let's use (-5,2) as our point. this is point-slope form: y - y1 = m(x -1) what should we plug in for y1?

OpenStudy (brucebaner):

5

OpenStudy (aripotta):

coordinates are in the form of (x,y) so our y1 would be 2. so far we have y - 2 = m(x - x1) what should we plug in for m?

OpenStudy (brucebaner):

2/5.5

OpenStudy (aripotta):

yes. so y - 2 = 2/5.5(x - x1) what's the x1?

OpenStudy (brucebaner):

-5

OpenStudy (aripotta):

yep. so our equation in point-slope form is y - 2 = 2/5.5(x + 5) i gotta go real quick. i'll be back in like 5 min

OpenStudy (aripotta):

when i get back, we'll start putting the equation in standard form

OpenStudy (brucebaner):

okay

OpenStudy (aripotta):

alrighty. so standard form is Ax + By = C where A can't be a fraction or a negative

OpenStudy (aripotta):

what's our first step?

OpenStudy (brucebaner):

what do you mean

OpenStudy (aripotta):

ok so basically we want to get x and y on the left side of the equation

OpenStudy (aripotta):

first, we have to distribute the 2/5.5

OpenStudy (aripotta):

do you know how to distribute?

OpenStudy (brucebaner):

wait

OpenStudy (brucebaner):

2/5.5x+1.81

OpenStudy (aripotta):

yea so now we have y - 2 = 2/5.5x + 1.81. what should we do now?

OpenStudy (brucebaner):

isolate variables to one side?

OpenStudy (aripotta):

yes

OpenStudy (brucebaner):

how do i do that?

OpenStudy (aripotta):

subtract 2/5.5 from both sides

OpenStudy (aripotta):

2/5.5x*

OpenStudy (brucebaner):

wouldn't that lead me to a negative number?

OpenStudy (aripotta):

yes

OpenStudy (brucebaner):

-2.36

OpenStudy (aripotta):

y - 2 = 2/5.5x + 1.81 once you subtract 2/5.5x from both sides, you get -2/5.5x + y - 2 = 1.81 now we wanna get rid of the -2, so add 2 to both sides

OpenStudy (brucebaner):

ok

OpenStudy (brucebaner):

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