The line that contains the point Q( 1, -2) and is parallel to the line whose equation is
y - 4 = 2/3 (x - 3)
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OpenStudy (anonymous):
@helpme1.2
OpenStudy (anonymous):
\[\huge y-y_1 = m(x-x_1)\] m is the slope, since we are looking for parallel, they would have the same slope so
just plug the points to the equation above
OpenStudy (anonymous):
y + 2 = 2/3(x - 1) ?
OpenStudy (anonymous):
yup
OpenStudy (anonymous):
so now what do I do?
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OpenStudy (anonymous):
-2/3 ?
OpenStudy (anonymous):
subtract 2/3 *
OpenStudy (anonymous):
no that's about it, the question didn't specify anything
OpenStudy (anonymous):
what?
OpenStudy (anonymous):
So this is the answer? y + 2 = 2/3(x - 1)
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OpenStudy (anonymous):
yeah
OpenStudy (anonymous):
but
OpenStudy (anonymous):
@nikato
OpenStudy (nikato):
Do the same thing as before. Make it look like Ax+By=C
OpenStudy (anonymous):
so multiply 2/3 by x - 1 ?
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OpenStudy (anonymous):
you'r suppose to put it in a standard form??
OpenStudy (nikato):
Distribut
OpenStudy (anonymous):
wait so what do I do?
OpenStudy (ranga):
y + 2 = 2/3(x - 1)
multiply throughout by 3
3y + 6 = 2(x - 1)
3y + 6 = 2x - 2
2x - 3y = 6 + 2
2x - 3y = 8 is the equation in standard form.
OpenStudy (anonymous):
oh I see, thank you!
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OpenStudy (anonymous):
And Thank you @nikato and @helpme1.2 Happy New years to you both!