How would you find the equation of a line in standard form and slope-intercept form, that passes through (-1,5) and (1,9)?
M = y2-y1/x2-x1 plug in values
That'll give you your slope :p
So the slope is 2?
Well, 9-5 = 4 1-(-1) = 2 so 4/2 = 2 yes
So naw that you have your slope, put it into y-y1=M(x-x1)
The equation would be y=2x+7 then right?
So its: y-1=2(x-(-1) y-1=2x +2 +1 +1 y=2x + 3
How did you get 7 lol xD
Oh I put in a wrong value, for the y1 I put in 5 by accident
Do you know what standard form is?
Isn't that ax^2+bx+c?
Thats quadratic xD Ax + By = C
So how would you convert y = 2x +3 into Ax + By = C
On a side note, do you know how to do quadratic? :O
No not really
Not really as in for Ax + by = c or the quad thingy i asked lol <- descriptive
The quadratic one
Oh lol kk, well that was irrelevant to this so, to convert y = 2x + 3 you would subtract 2x from both sides and get -2x + y = 3 .... in most cases you would multiply by -1 to get the x coefficient positive
And that would be the standard form of the equation?
Idk some teachers allow the x coefficient to be negative, some require you to multiply is by -1 to make it positive
So if I were to turn it positive it would be 2x-y=-3
If you'd like you just have to multiply every term by -1 -2x(-1) +y(-1) = 3(-1)
Usually its A>0
Okay, thank you so much!
no prob yw :}