anonymous
  • anonymous
Write an equation for the line described. Write it in standard and point intercept form(if possible) through (1,3), m=-2 through (-5,4), m=-3/2 i need a little explanation also, if you can give it... studying for a test.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
ok so you get those in standard form by first using this formula: \[(y - y_{1}) = m (x - x_{1})\]
anonymous
  • anonymous
so, for the first equation, this is (y - 3) = -2(x - 1)
anonymous
  • anonymous
y - 3 = -2x + 2

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anonymous
  • anonymous
to get it into standard form (Ax + By = 0) move 2x to the other side, giving you 2x + y - 3 = 2
anonymous
  • anonymous
depending on qhat the quiz asks you, the final answer will either be 2x + y = 5 or 2x + y - 5 = 0.
anonymous
  • anonymous
*what
anonymous
  • anonymous
Did you mean slope intercept when you said point intercept?
anonymous
  • anonymous
yes.. sorry
anonymous
  • anonymous
So the second one. Standard form : y-4=-3/2(x-(-5))
anonymous
  • anonymous
I get confused with the fractions?
anonymous
  • anonymous
And all those negatives are throwing me off..
anonymous
  • anonymous
Lots of people have trouble with it. We know that a negative negative is a positive. That gives you y - 4 = -3/2(x + 5)
anonymous
  • anonymous
Then, you multiply x and 5 by -3/2
anonymous
  • anonymous
a positive and a negative multiplied together is a negative. That gives you -3/2x - 15/2
anonymous
  • anonymous
-15/2 is reached when you multiply -3/2 and 5/1
anonymous
  • anonymous
do you understand that?
anonymous
  • anonymous
Yes, so far so good.
anonymous
  • anonymous
Then I need to put it in standard form..
anonymous
  • anonymous
Well, that wasn't a complete point slope form, so let me explain. Above you gave me y - 4 = -3/2(x - (-5)) Your complete slope form would be y = -3/2(x - 1), getting the 4 on the correct side. So, we'd have -3/2x + 3/2
anonymous
  • anonymous
do you understand how I got that?
anonymous
  • anonymous
Why dont you want it in the same form as the previous one? ax=by=0?
anonymous
  • anonymous
sorry ax+by=0
anonymous
  • anonymous
When I convert 4 into a fraction is that 8/2?
anonymous
  • anonymous
When I subtract that 8/2 from the right side I come up with -3/2x-7/2?
anonymous
  • anonymous
hold on a second
anonymous
  • anonymous
it's easier to set up standard form if you first have it in slope intercept form.
anonymous
  • anonymous
no, you want to get the constant on the right side of the equation BEFORE you multiply the whole thing by -3/2.
anonymous
  • anonymous
y = -3/2 + 3/2
anonymous
  • anonymous
constant meaning?
anonymous
  • anonymous
Constant is a number with no variables; 3a would be a coefficient, 3 is a constant.
anonymous
  • anonymous
where and how are you getting that +3/2
anonymous
  • anonymous
ok, let me explain
anonymous
  • anonymous
y-4=-3/2(x=5) can you take me from there?
anonymous
  • anonymous
Okay. (you were right, I was getting confused.)
anonymous
  • anonymous
(x+5)
anonymous
  • anonymous
first you add 4 to each side, giving you y = -3/2(x + 9)
anonymous
  • anonymous
multiply both terms by -3/2, giving you -3/2x -27/2
anonymous
  • anonymous
move -3/2x to the left side of the equation, giving you 3/2x + y = 27/2
anonymous
  • anonymous
sorry, -27/2
anonymous
  • anonymous
depending on the curriculum, your final answer would be 3/2x + y = -27/2 or 3/2x + y + 27/2 = 0
anonymous
  • anonymous
let me rewrite it out myself and see if that's the right answer
anonymous
  • anonymous
My book gives me an answer of 3x+2y=-7
anonymous
  • anonymous
Do I just multiply by 2 to clear the fraction?
anonymous
  • anonymous
no, when you're talking about lines you don't take out the fraction
anonymous
  • anonymous
but let me work it out and see if I can get it that way multiplying by 2
anonymous
  • anonymous
ok, it seems to me that your book does want you to multiply by 2, but my curriculum didn't teach me that
anonymous
  • anonymous
Ok.. thank you so much for being patient with me...this helped!
anonymous
  • anonymous
you're welcome :D

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