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Mathematics 59 Online
OpenStudy (anonymous):

I'm really lost on this one. Can someone show me where to start..? Type the equation of the given line in standard form. The line through (2, -1) and parallel to x - 2y = 6. Thanks so much!

OpenStudy (anonymous):

use point slope form of a line (y+1) = 1/2(x-2)

OpenStudy (anonymous):

parallel means same vector defins the line. So use <1,-2>. (x-2)-2(y+1)=0 x-2-2y-2=0 x-2y=4

OpenStudy (anonymous):

@A.Avinash_Goutham Okay, so here it is.. (-1+1)=(1/2)(2-2) 0=(1/2)0 0=0 What did I do wrong?

OpenStudy (anonymous):

x - 2y = 6. this is already standart form. So just use it.....

OpenStudy (anonymous):

why you go to slope form first?

OpenStudy (anonymous):

@myko which equation is that? Like the beginning equation to where I substitute.

OpenStudy (anonymous):

u kno that's the eqn...like u need not plg in those values.....jus simplify it to get the eqn nd u get 0=0 coz the point satisfies the line.........

OpenStudy (anonymous):

yes, just put 0 at the right side

OpenStudy (anonymous):

@myko but which equation did you originally use?

OpenStudy (anonymous):

x - 2y = 6

OpenStudy (anonymous):

@myko Ohh! I see now. Thanks :)

OpenStudy (anonymous):

yw

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