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Mathematics 15 Online
OpenStudy (anonymous):

Choose the equation of the line passing through the point (-4, -2) and parallel to y =1/4 x + 2

OpenStudy (jdoe0001):

parallel lines have the same slope so, what would be the slope from -> y =1/4 x + 2

OpenStudy (jdoe0001):

if you check your y = mx+b

OpenStudy (anonymous):

i ddont knooow

OpenStudy (anonymous):

@jdoe0001

OpenStudy (anonymous):

i still dont understand.. @jdoe0001

OpenStudy (jdoe0001):

can you see the slope in $$ y =\color{purple}{\cfrac{1}{4}}x + \color{orange}{2} \ \ \large ? $$

OpenStudy (anonymous):

1/4

OpenStudy (anonymous):

@jdoe0001

OpenStudy (jdoe0001):

so, now you have slope of the other line, they're both parallel to each other, thus the slope for this one is the slope for the other one now, with the slope and the point given, use that to find the equation as in http://www.teachhandson.com/calculationb.gif

OpenStudy (anonymous):

im still confused @jdoe0001

OpenStudy (jdoe0001):

well, do you happen to have the point-slope intercept formula? it'd be in the same chapter the slope-intercept is

OpenStudy (anonymous):

y=mx+b

OpenStudy (anonymous):

@jdoe0001

OpenStudy (jdoe0001):

well, that'd be the slope-intercept, the point-slope form is http://www.scs.sk.ca/cyber/elem/learningcommunity/math/math30aa/curr_content/matha30rev1/lesson3-4/PONITSLOPEFORMULA.JPG

OpenStudy (anonymous):

i bad at this @jdoe0001

OpenStudy (jdoe0001):

so your point-slope form is y2 - y1 = m(x2-x1) more or less $$ y =\color{purple}{\cfrac{1}{4}}x + \color{orange}{2}\\ \text{so the slope of the other line is "m" }\color{purple}{\cfrac{1}{4}}\\ \text{using the point-slope intercept form}\\ y_2-y_1=m(x_2-x_1)\\ \text{use }\color{purple}{\cfrac{1}{4}} \text{ for the slope, and the given point} (\color{blue}{-4},\color{green}{ -2})\\ y-(\color{green}{ -2})=\color{purple}{\cfrac{1}{4}}(x-(\color{blue}{-4})) $$

OpenStudy (jdoe0001):

solve that for "y", and that's your line

OpenStudy (anonymous):

so whats the answer ? @jdoe0001

OpenStudy (jdoe0001):

the solution by isolating "y" from that, that'd be the line

OpenStudy (anonymous):

huh @jdoe0001

OpenStudy (jdoe0001):

just solve that equation for "y", and you'd get the equation for the line

OpenStudy (anonymous):

i dont know how @jdoe0001

OpenStudy (jdoe0001):

then this exercises doesn't apply to you, and you'd need to cover the material first

OpenStudy (anonymous):

im in virtual @jdoe0001

OpenStudy (jdoe0001):

well, I'm pretty sure the material to be covered is the same, and the format is the same, the medium is just the only thing that's more flexible, but everything else is the same, you'd still be expected to cover the material/chapter of the sections

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