Question

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Answer 1

It's lke the change in \(y\) over the change in \(x\).

Answer 2

1) Pick any two points. Let's say you picked \((x_1,y_1)\) and \((x_2,y_2)\).
2) The slope is given by:\[\dfrac{y_2- y_1}{x_2 - x_1}\]

Answer 3

So can you pick two points from the line you were given?

Answer 4

Well a slope is just the rise over run of a function

or in other words,

\[\huge\ m=\frac{ \Delta y }{ \Delta x }=\frac{ y_2-y_1 }{ x_2-x_1 }\]

There are a few other slope formula which you may need to know ,

\[\huge\ m=\frac{ -coefficient \ of \ x }{ coefficient \ of \ y }\]

Answer 5

@hba: That font is too small. =P

Answer 6

@ParthKohli Next time i will make sure it is smaller than that :P

Answer 7

Too much sarcasm to handle =/

Answer 8

Well, the word “slope” per se is self-explanatory.

Answer 9

Aarrrhhhmagerd this must be Mt. Everest

Answer 10

And BTW, the asker didn't ask for the concept of a slope. =P

Answer 11

^ Point.

Answer 12

I didn't ask for you to start you own conversation on my question either :P

Answer 13

Google it Bruv >.<

Answer 14

@ParthKohli

Answer 15

What?!

Answer 16

Sal khan has already given the concept.

Answer 17

Because they don't know about sal khan.

go to khanacademy.com @nincompoop

Answer 18

It's being asked because the students are too lazy to complete their homework.

Answer 19

^ Exactly.

Answer 20

No?

Answer 21

I was looking for someone to explain it, and help me out.

Answer 22

LMGTFY doesn't get old... does it? :D

Answer 23

I've already solved the problem by time hba started commenting

Answer 24

Just for the sake of complicating it:
Calculate the vector between two arbitrary points \(a,b\). Suppose the vector is \(\langle h_1, h_2\rangle\). The slope is \(\dfrac{h_2}{h_1}\) =P

Answer 25

@cherrycool What do you get for the answer?

Answer 26

And never mind that response.

Answer 27

You can still respond if a question is closed. That's why they are allowing us to respond =)

“If it's a feature, it can be used.”

Answer 28

Well, if you'd like to continuing posting then go ahead :)