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ok so you get those in standard form by first using this formula: \[(y - y_{1}) = m (x - x_{1})\]

so, for the first equation, this is (y - 3) = -2(x - 1)

y - 3 = -2x + 2

to get it into standard form (Ax + By = 0) move 2x to the other side, giving you 2x + y - 3 = 2

depending on qhat the quiz asks you, the final answer will either be 2x + y = 5 or 2x + y - 5 = 0.

*what

Did you mean slope intercept when you said point intercept?

yes.. sorry

So the second one.
Standard form : y-4=-3/2(x-(-5))

I get confused with the fractions?

And all those negatives are throwing me off..

Then, you multiply x and 5 by -3/2

a positive and a negative multiplied together is a negative. That gives you -3/2x - 15/2

-15/2 is reached when you multiply -3/2 and 5/1

do you understand that?

Yes, so far so good.

Then I need to put it in standard form..

do you understand how I got that?

Why dont you want it in the same form as the previous one? ax=by=0?

sorry ax+by=0

When I convert 4 into a fraction is that 8/2?

When I subtract that 8/2 from the right side I come up with -3/2x-7/2?

hold on a second

it's easier to set up standard form if you first have it in slope intercept form.

y = -3/2 + 3/2

constant meaning?

Constant is a number with no variables; 3a would be a coefficient, 3 is a constant.

where and how are you getting that +3/2

ok, let me explain

y-4=-3/2(x=5) can you take me from there?

Okay. (you were right, I was getting confused.)

(x+5)

first you add 4 to each side, giving you y = -3/2(x + 9)

multiply both terms by -3/2, giving you -3/2x -27/2

move -3/2x to the left side of the equation, giving you 3/2x + y = 27/2

sorry, -27/2

depending on the curriculum, your final answer would be 3/2x + y = -27/2 or 3/2x + y + 27/2 = 0

let me rewrite it out myself and see if that's the right answer

My book gives me an answer of 3x+2y=-7

Do I just multiply by 2 to clear the fraction?

no, when you're talking about lines you don't take out the fraction

but let me work it out and see if I can get it that way multiplying by 2

Ok.. thank you so much for being patient with me...this helped!

you're welcome :D