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I have Points $M_1(-1,1)$ and $M_2(2,3)$. From applied formula: $$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1).$$

Solving that gives: $$y-1=\frac{3-1}{2-(-1)}(x-(-1)).$$

which when solved becomes: $2x - 3y + 5 = 0$.

I cannot understand how I get that last result. Can you please explain

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    Learn basic [MathJax](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) at least for posting here. And change the tag. – 2017-01-28
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    What should be the tag? Sorry for not well formatted question – 2017-01-28
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    I think 'algebra-precalculus' should suffice. – 2017-01-28
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    @ΘΣΦGenSan and from here? how I get last result? – 2017-01-28
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    @StubbornAtom, there's no need to be quite so rude in your original comment. This is OP's first question here. Although the post didn't use MathJax, it was clearly legible and well specified. Saying "And change the tag" does not help the OP understand what a better tag would be. Why not take sixty seconds to fix the MathJax and add a tag, as ΘΣΦ GenSan did, and then leave a helpful comment indicating what you changed? – 2017-01-28
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    @wchargin Because I wanted him to have a look at the reference page and do all of that himself. If he didn't then certainly I would have done that for him. It doesn't do much harm to a new user to try and figure things out for himself. In any case I never intended to be rude. – 2017-01-28

3 Answers 3

5

You can proceed as -

$$y-1=\frac{3-1}{2-(-1)}(x-(-1))$$

$$y-1=\frac{2}{3}(x+1).$$

$$3(y-1)=2(x+1)$$

$$3y-3=2x+2$$

$$2x-3y+5=0$$

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    Why you send 3 from first side? – 2017-01-28
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    Cross multiply. – 2017-01-28
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    If we have fraction on one side in multiplication with other terms. We can multiply its denominator with terms on other side. – 2017-01-28
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    Clear to you now? – 2017-01-28
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    Yes, 100% thank you all – 2017-01-28
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    Happy to help you. – 2017-01-28
4

you can compute the slope at first $$m=\frac{3-1}{2+1}$$ and we get by the Ansatz $$y=mx+n$$ with the $$m=\frac{2}{3}$$ the equation $$y=\frac{2}{3}x+n$$ plugging $$x=-1,y=1$$ in the given equation we have $$1=-\frac{2}{3}+n$$ this gives: $$n=\frac{5}{3}$$ and our equation will be $$y=\frac{2}{3}x+\frac{5}{3}$$ multiplying by $5$ we get $$3y=2x+5$$

2

Just simplify to get $$y-1=\frac{2}{3}(x+1)$$ which means that $$3(y-1)=2(x+1).$$ And so, we get $$3y-3=2x+2.$$ Transpose all the terms at the right side, we get $$0=2x+2-3y+3.$$ Finally, $$2x-3y+5=0.$$

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    Can you explain how you get to step 2 please – 2017-01-28