+108 We have a line that passes through points (2,3) and (5,-1).
Using the formula's:
\(y-y_{a}=m(x-x_{a})\)
\(m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
We get 2 equations.
Equation (a)
\(y+1=-\frac{4}{3}(x-5)\)
Equation (b)
\(y-3=-\frac{4}{3}(x-2)\)
Both of these equations are equal to:
Equation (c)
\(3y=-4x+17\)
Can you please explain the process of changing equations a and b to equation c?
Thanks !
+118608 Hi Kreyn,
I like the way you worded your question.
Let's see.
\(y+1=\frac{-4}{3}(x-5)\\ \mbox{Multiply both sides by 3}\\ 3(y+1)=-4(x-5)\\ 3y+3=-4x+20\\ 3y=-4x+20-3\\ 3y=-4x+17\)
Maybe you would like to try the second one yourself?
Ask if you do not understand any of my answer. :)
+108 Oh okay, so multiply equation b by 3 and add 9 to each side!
\(y-3=-\frac{4}{3}(x-2)\)
\(3(y-3)=3(-\frac{4}{3}(x-2))\)
\(3y-9=-4x+8\)
\(3y=-4x+17\)
Is there is a rule I can use for different cases. For example If I did not know equation (c), what would I start with to try to equalize the 2 equations? I see that 3 is the y coordinate of the first point. Would you try to multiply each equation by one of the point coordinates and then see which 2 best fits? Does each equation have to be multiplied by the same variable?
Thanks heaps!
+118608 Hi Kryn,
Sorry about the answering delay :/
You will get the hang of these types of equation rearranging quite easily I think.
It just takes a bit of practice.
The idea is always to keep the 'equation balanced'
So you decide what will help you get to the end that you want and then you have to ALWAYS do the same thing to both sides.
So say you want to get from
\(y+1=-\frac{4}{3}(x-5)\)
to
\(y-3=-\frac{4}{3}(x-2)\)
Lets take a look 
\(y+1=-\frac{4}{3}(x-5)\)
First I want y-3 on the left so if I subtract 4 on the left that will work. BUT I must do the same on the right side.
\(y+1-4=-\frac{4}{3}(x-5)-4\\ y-3=-\frac{4}{3}(x-5)-4\\ y-3=-\frac{4}{3}(x-5)-\frac{4*3}{3}\\ y-3=-\frac{4}{3}(x-5)-\frac{4}{3}*3\\ \mbox{Factorize out the }\frac{-4}{3}\\ y-3=-\frac{4}{3}(x-5+3)\\ y-3=-\frac{4}{3}(x-2)\\ \)
I could also have expanded the bracket as the 3rd step and then collected like terms and factorize the -4/3 again afterwards. The answer would have been the same.
I hope that helps but by all means aske more questions if you want more help or clarification. 
----------------------------------------------
When you sove equations you are 'undoing' calculations so you do it in the opposite order to BODMAS. :)
