Question

I haven't done Algebra in over 5 years, so I need help with this problem: What’s the equation of a line that passes through points (0, -1) and (2, 3)?

Answer 1

First to find the slope of the line you use the equation:

\[\frac{ y _{2}-y _{1} }{ x _{2}-x _{1} }\]

pick one of the pairs to plug into \[y _{2}\] and \[x _{2}\] just make sure to keep the order after you start plugging numbers in. Does this make sense so far?

Answer 2

you want to obtain y=mx+b and that formula is the formula for slope (aka "m") @masukasu

Answer 3

How do I know what numbers to assign to y and x?

Answer 4

so this is what it will look like once you plug the points in:

\[\frac{ (3) - (-1) }{ (2)-(0) }\]

Answer 5

You have the points (0, -1) and (2, 3); you know that 0 and 2 are x-values and -1 and 3 are y-values, correct?

Answer 6

I used the second set of points to use first only so the equation won't look messy. Putting the -1 in the \[y _{1}\] spot allows it to become positive because two negatives make a positive.

Answer 7

So, (0, -1) and (2, 3) would look like this with their appropriate values: (x, -y) and (x, y) right?

Answer 8

But you will get the same answer no matter which set of points you plug first, but always be sure to keep the order and correct to your above question

Answer 9

so go ahead and solve for the slope (m) and tell me what you get

Answer 10

But, the example of the order of my values(in my last comment) are correct, right?

Answer 11

yes you are correct = )

Answer 12

Ah! Thank you. I'll try to solve the problem. One minute (:

Answer 13

now that is only part of it, there is another equation to arrive to the final equation

Answer 14

once you find the slope I can tell you the next part = )

Answer 15

And I would multiply \[x{2}\] and \[y{2}\] like any other number with power, right? i.e. 2 x 2 and 3 x 3

Answer 16

Oh I'm sorry, the subscript 1 and 2 are only there to decipher between the different values those aren't exponents

Answer 17

Ohhh ok, thank you. I couldn't remember the right terminology either, sorry. But, I think it's: \[\frac{ 4 }{ -1 }\]

Answer 18

here let me do this
\[\frac{ y-y }{ x-x }\]
do you see how this can look like the answer would be 0 right? that's why they put the 1 and 2 on the x and y to determine between the two sets of points

Answer 19

your numerator is correct double check your denominator (2-0) = ?

Answer 20

Pff... Sorry. I mixed my numbers up >.<

Answer 21

no apologies = ) I was just on here earlier getting help and I did the same thing lol ; D

Answer 22

Essentially it comes out to \[\frac{ 2 }{ 1 }\]

Answer 23

lol I guess we all make mistakes haha! It's how we learn.

Answer 24

correct = ) so that is your "m" or slope

Answer 25

yay! Alright, so that should be written out like so: \[y =2x +b\]

Answer 26

And b is the 'base', right?

Answer 27

the next equation you will need is:

\[y-y _{1} = m (x-x _{1})\]

Now don't get confused that the \[y _{1}\] and \[x _{1}\] are the same as the ones above you can choose either set of points to plug in. The regular y and x are the ones that will remain in the equation after plugging in the set of points which will give you the y=mx+b. Does this make sense so far?

Answer 28

I can't remember if it is called the base but that will be your y-intercept once you graph it

Answer 29

Ok so i chose the points (2,3) to use bc I like positive numbers lol Plug in the set of points and what you got for m into the equation \[y-(3)=(2)(x-(2))\] Tell me what you get after solving that

Answer 30

@masukasu

Answer 31

Sorry. Someone had to use my laptop.

Answer 32

that's ok = )

Answer 33

Ok, I'm going to solve that. One minute (:

Answer 34

Wait... Do I NEED the parenthesis around the numbers? Or does it not really change anything?

Answer 35

It's just easier to see if you have any sign changes. Now say you used the other set of points instead of the positive numbers you would end up with this:
\[y-(-1) = (2)(x-0) \]
It's safe to have that parentheses around the -1 bc the sign of it changes to y+1 which will screw things up when you have to solve for y

Answer 36

but it will only screw things up when you have to solve for y if you forget to change the sign of the -1.

Answer 37

This equation is a little confusing. I'm used to it looking like \[2\left( x -2 \right)=y -\left( 3 \right)\] but even then, it's a little difficult.

Answer 38

ok so the only difference is the sides of the equation are on opposite sides = ) do you know how to factor? I'll help you step by step = )

Answer 39

I would bring the 3 over and it would become a +3, then it would look like this:\[2\left( x-2 \right)+3=y\] I think

Answer 40

Factoring... Uh lol Is that factoring? I can't remember. Am I able to reverse the equation like I did though?

Answer 41

that is a correct move but that's not factoring. Take a look at the 2(x-2):
when you see something like this you use factoring where you take the front number and multiply it by both numbers inside the parentheses.

So you would have 2*x - 4 then tack on what's left in the equation:

2x-4+3=y

Then simplify

Answer 42

OH! I remember now!

Answer 43

Well... that part at least