1. ## Standard Form Problem

I'm having some trouble with this problem in my test review. Here it is:

Find the equation of each line in standard form (where A > or = 0, and A, B, and C are integers):
Slope is (c/d), contains (a, b)

This was in a group of problems, and I didn't have any trouble with the others (they all had numbers instead of variables). To start this one, I put it into point-slope form like I had with the others.
So I had: y - b = (c/d) * (x - a)
y - b = (cx/d) - (ac/d)
y = (cx/d) - (ac/d) + b

I'm at a loss as to how I can put this (if it's even correct so far ) into the standard form. I checked the key and the final answer should be cx - dy = ac - bc. Any help on how to get there would be much appreciated! Thanks. Let me know if anything I wrote was unclear

2. Originally Posted by justjess
I'm having some trouble with this problem in my test review. Here it is:

Find the equation of each line in standard form (where A > or = 0, and A, B, and C are integers):
Slope is (c/d), contains (a, b)

This was in a group of problems, and I didn't have any trouble with the others (they all had numbers instead of variables). To start this one, I put it into point-slope form like I had with the others.
So I had: y - b = (c/d) * (x - a)
y - b = (cx/d) - (ac/d)
y = (cx/d) - (ac/d) + b
....Next step: Multiply through by d to clear fractions
I'm at a loss as to how I can put this (if it's even correct so far ) into the standard form. I checked the key and the final answer should be cx - dy = ac - bc. Any help on how to get there would be much appreciated! Thanks. Let me know if anything I wrote was unclear
Using upper-case letters for the standard form, the line is
$A\ x +B\ y = C$

The given line in point-slope form is as you said:
$y - b = \dfrac cd \times (x - a)$
Multiply through by $d$ to clear fractions:
$d(y - b) = c(x-a)$
$dy - db = cx - ca$
Transpose x and y terms to left, constants to right
$-cx + dy = -ca +db$
Change signs so coefficient of x is positive
$cx - dy = ca - db$

Equate coefficients of the standard form:
$A = c$,.....$B = -d$,.....$C = ca - db$

3. Originally Posted by justjess
y = (cx/d) - (ac/d) + b
Ok dokay Jess; what do you get if you multiply above equation by d ?

4. Originally Posted by Denis
Ok dokay Jess; what do you get if you multiply above equation by d ?
I did:
y = (cx/d) - (ac/d) + b
yd = cx - ac + bd
yd - bd = cx - ac
-cx + yd = -ac + bd
cx - yd = ac - bd
cx - dy = ac - bd
(I just swapped the yd for dy here to make it look better)

I just checked mine with yours, DrPhil, and it looks the same, but somehow the teacher got cx - dy = ac - bc. Well, the key has been wrong before! What's more important is I know how to do it now. Thanks all!

5. Originally Posted by justjess
cx - dy = ac - bd (I just swapped the yd for dy here to make it look better)

...but somehow the teacher got cx - dy = ac - bc.