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Mathematics 7 Online
OpenStudy (anonymous):

Indicate in standard form the equation of the line passing through the given point and having the given slope. B(6, 2), m = -1/2 X + 2 y = _ _ Think you can help, @yummydum ?

OpenStudy (anonymous):

@lgbasallote can you take this one...i gotta go :/

OpenStudy (anonymous):

oh @saifoo.khan :D

OpenStudy (saifoo.khan):

Use the formula: y - y1 = m(x-x1)

OpenStudy (anonymous):

there u are :P

OpenStudy (lgbasallote):

just use the point slope form \[y - y_1 = m(x-x_1)\] where m is the slope x1 is the x coordinate y1 is the y coordinate

OpenStudy (saifoo.khan):

y - (2) = -1/2 (x - 6)

OpenStudy (lgbasallote):

\[\implies y -(2) = -\frac 12(x-6)\] simplify to get the final answer

OpenStudy (lgbasallote):

note: standard form is \(ax^2 + bx + c = 0\)

OpenStudy (saifoo.khan):

o_O

OpenStudy (anonymous):

You guys are on this like flies! Hang on, I'll try solving it...

OpenStudy (saifoo.khan):

Standard form: ax + by = c

OpenStudy (anonymous):

Actually wait, I don't think you guys understand the question. I'm supposed to be filling in boxes to complete an equation, hence the "_" in the question.

OpenStudy (anonymous):

So @lgbasallote I think I might need more help on this.

OpenStudy (lgbasallote):

i'll let my assistant @saifoo.khan help you for now :)

OpenStudy (saifoo.khan):

@lgbasallote ; standard form = ax2+bx+c=0 ?

OpenStudy (saifoo.khan):

i gtg now. I'm heading to bed. ^_^ @lgbasallote

OpenStudy (lgbasallote):

lol bed at 6 am :p

OpenStudy (saifoo.khan):

5! ;)

OpenStudy (saifoo.khan):

Gnite.

OpenStudy (lgbasallote):

anyway @Sephichief plug in the values for \[y - y_1 = m(x-x_1) \] first

OpenStudy (anonymous):

I did, and I got y - 2 = m(x-6)

OpenStudy (lgbasallote):

what's m?

OpenStudy (anonymous):

Oh, whoops! That would be 1/2, so that makes it y - 2 = 1/2(x-6)

OpenStudy (lgbasallote):

nope it's NEGATIVE 1/2

OpenStudy (lgbasallote):

\[y-2 = -\frac 12(x-6)\]

OpenStudy (anonymous):

Oh, forgot about the negative sign

OpenStudy (lgbasallote):

now what is the denominator in this equation?

OpenStudy (anonymous):

That would be 2 I think.

OpenStudy (lgbasallote):

right. so multiply all terms by 2

OpenStudy (anonymous):

y - 4 = -1/2(x-12)

OpenStudy (anonymous):

Then what?

OpenStudy (lgbasallote):

now you recheck what you did because you made a mistake :p haha

OpenStudy (lgbasallote):

here's a hint for you \[y-1 = \frac 13 (x-1)\] if i multiply all terms by 3 i will get \[3y - 3 = 1(x-1)\] \[\implies 3y - 3 = x - 1\]

OpenStudy (anonymous):

Silly geometry and your rules...alright, would that be: 2y - 4 = -12

OpenStudy (lgbasallote):

still no

OpenStudy (lgbasallote):

it will be 2y - 4 = -(x-6)

OpenStudy (lgbasallote):

1/2 just cancels out

OpenStudy (lgbasallote):

so the parenthesis remains unchanged

OpenStudy (anonymous):

Oh, alright.

OpenStudy (anonymous):

Got it! The answer was x + 2y = 10. Thank you for your help!

OpenStudy (lgbasallote):

welcome ^_^

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