Question

For finding the equation of a straight line do you use y1-y2/x1-x2 or y2-y1/x2-x1?

Answer 1

the second one

Answer 2

http://www.purplemath.com/modules/slope.htm according to this site it's the other one?

Answer 3

Both are correct.

Answer 4

can you explain breifly?

Answer 5

the negatives will cancel out no matter which way you do it

Answer 6

thank you

Answer 7

Let me give u an example. consider a straight line y= x

2 points on the line will be (1,1) and (2,2)

y1-y2/x1-x2 = 1-2/1-2 = -1/-1 = 1

y2-y1/ x2-x1 = 2-1/2-1= 1/1= 1

Answer 8

this explains it pretyy well
http://www.khanacademy.org/video/slope-of-a-line?topic=core-algebra

Answer 9

yeah they work out the same... thanks. Are you familiar with boolean algebra?

Answer 10

post your specific question and I'm sure someone will answer.

Answer 11

i did 20 min ago

Answer 12

b(1+a'd) + a(b'c +c'd+d') = b+a(b'c+c'd+d') =?

to simplify
b'c+c'd+d'

use DeMorgan's rules
not(a+b)= a'b'
not(ab)=a'+b'

c'd+d' = not( not(c'd)d)= not((c+d')d)= not(cd + d'd)= not(cd)= c'+d'

so b'c + c'd+d' -> b'c + c'+ d'
similarly b'c + c = not(not(b'c)c')=not((b+c')c)=not(bc+c'c)= not(bc)=b'+c'
which makes
b'c + c'+ d' -> b'+c'+d'
and the original expression is
b+a(b'c+c'd+d') = b+a(b'+c'+d')= b+ab' + a(c'+d')
simplify b+ab':
b+ab'= not(b'(not(ab'))=not(b'(a'+b))= not(a'b'+b'b)=not(a'b')=a+b

we now have b+ab' + a(c'+d') -> a+b+a(c'+d')= b+a(1+(c'+d'))=b+a

Answer 13

wow thanks a lot

Answer 14

let me fix this line (in case you go through this carefully)
similarly b'c + c = not(not(b'c)c')=not((b+c')c)=not(bc+c'c)= not(bc)=b'+c'
should be
b'c + c' = not(not(b'c)c)=not((b+c')c)=not(bc+c'c)= not(bc)=b'+c'