1. Common denominator

Hi, my algebra is quite rusty and I can't figure this one out.

$c_3 = \frac{y_3-y_1}{(x_3-x_1)(x_3-x_2)} - \frac{y_2-y_2}{(x_3-x_2)(x_2-x_1)}
$

My book says that "by taking a common denominator we get",

$c_3 = \frac{\frac{y_3-y_2}{x_3-x_2}-\frac{y_2-y_2}{x_2-x_2}}{x_3-x_1}$

If I was to find a common denominator for this, it would look something like this,

$c_3 = c_3 = \frac{y_3-y_1(x_2-x_1)}{(x_3-x_1)(x_3-x_2)(x_2-x_1} - \frac{y_2-y_2(x_3-x_1)}{(x_3-x_2)(x_2-x_1)(x_3-x_1)}$

I've played with it for a while, but can't seem to arrive at the right expression..

Any tips besides finding another hobby?

$c_3 = \dfrac{\frac{y_3-y_2}{x_3-x_2}-\frac{y_2-y_2}{x_2-x_2}}{x_3-x_1}$

Now in the part where is says $x_2 - x_2$ then it will equal 0. And you know that anything divided by 0 is undefined.

$c_3 = \dfrac{y_3-y_1}{(x_3-x_1)(x_3-x_2)} - \dfrac{y_2-y_2}{(x_3-x_2)(x_2-x_1)}$

Now we multiply everything by $(x_3-x_2)$ giving us:

$c_3(x_3-x_2) = \dfrac{(y_3-y_1)(x_3-x_2)}{(x_3-x_1)(x_3-x_2)} - \dfrac{(y_2-y_2)(x_3-x_2)}{(x_3-x_2)(x_2-x_1)}$

Cancel out like terms leaving:

$c_3(x_3-x_2) = \dfrac{y_3-y_1}{x_3-x_1} - \dfrac{y_2-y_2}{x_2-x_1}$

And then divide everything by $(x_3-x_2)$ to get the answer:

$c_3 = \dfrac{\frac{y_3-y_1}{x_3-x_1} - \frac{y_2-y_2}{x_2-x_1}}{x_3-x_2}$

Now we see the part where it says $\frac{y_2-y_2}{x_2-x_1}$ we know that it will be 0 because $y_2-y_2$ = 0. (I should have cancelled that out from the start...)

This leaves:

$c_3 = \dfrac{\frac{y_3-y_1}{x_3-x_1}}{x_3-x_2}$

$c_3 = \dfrac{y_3-y_1}{(x_3-x_1)(x_3-x_2)}$

Can you carry on from there? Or am I wrong?

3. I'm really sorry, I made a typo. It is supposed to be: $y_2-y_2$ would not make much sense I guess.
Thanks though!
$c_3 = \frac{y_3-y_1}{(x_3-x_1)(x_3-x_2)} - \frac{y_2-y_1}{(x_3-x_2)(x_2-x_1)}
$

4. The first nominator should be $y_3-y_2$. This book has a lot of errors. Should have seen it at once, but so many terms are giving me a headache.

5. So from what I gather, this is the formula:

$c_3 = \dfrac{y_3-y_1}{(x_3-x_1)(x_3-x_2)} - \dfrac{y_3-y_2}{(x_3-x_2)(x_2-x_1)}$

If that's the case then it would simplify to:

$c_3 = \dfrac{\frac{y_3-y_1}{x_3-x_1} - \frac{y_3-y_2}{x_2-x_1}}{x_3-x_2}$