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Mathematics 21 Online
OpenStudy (anonymous):

Helpp pleaseeeeee! http://gyazo.com/6beade802ee583d2a2c208d8dc0c184a

OpenStudy (anonymous):

that line passes through (-2,0) and (0,-1) can u find the slope of that by using the slope formula and the points i gave?

OpenStudy (anonymous):

y = 2x + 4 y = - 2x + 2 y = - 1 over 2 x + 2 y = - 1 over 2 x + 4 @PFEH.1999

OpenStudy (anonymous):

i see.but i'm helping u to understand and solve it yourself...

OpenStudy (anonymous):

What formla do I use?

OpenStudy (anonymous):

so I use the slope formula? @PFEH.1999

OpenStudy (anonymous):

the slope formula is : \[m = \frac{ y2 - y1 }{ x2 - x1 }\] here,you can let y2 = -1 and y1=0,x2=0 and x1=-2 so slope = -1/2 and imagine the slope of a line in which is parallel to another line is m2 and the slope of that line is m1 ... m2 x m1 = -1 this is always true so the slope of the second line (the parallel line) is 2 because 2 x (-1/2) = -1 .. now u have slope and point...you can use slope - point formula (remember that the given point is (2,3).

OpenStudy (anonymous):

y=-2x+2 @PFEH.1999

OpenStudy (anonymous):

let me check...

OpenStudy (anonymous):

it has to have a negative because of where the line lies? @PFEH.1999

OpenStudy (zpupster):

parallel lines always have same slope.

OpenStudy (anonymous):

how careless i am ;) yes i made mistake...the same slope : -1/2 and the point (2,3) now use slope-point formula: y = m(x-x1) + y1

OpenStudy (anonymous):

y = -1/2(x) + 1 + 3 ...

OpenStudy (anonymous):

So it's y = - 1/ 2 x + 2

OpenStudy (anonymous):

no

OpenStudy (anonymous):

-1/2(x - 2) + 3 ... made sense?

OpenStudy (anonymous):

so it's y=-1/2x+$

OpenStudy (anonymous):

can u do it now?

OpenStudy (anonymous):

what? u mean 4 ?

OpenStudy (anonymous):

$ ???

OpenStudy (anonymous):

+4** sorry

OpenStudy (anonymous):

correct ;)

OpenStudy (anonymous):

i hope u have understood it well ;)

OpenStudy (anonymous):

Thanks for helping me <3 ;) @PFEH.1999

OpenStudy (anonymous):

you're welcome ;)

OpenStudy (anonymous):

http://gyazo.com/e7c86f3822f933ca0148870eae9d0588 @PFEH.1999

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