How do you find an equation of the line that passes through the point and has the indicated slope m? (2,4); m = -1 With this one, isn't the formula y-y1=m(x-x1)
thats as good as any other form of the equation of a line
Okay, but when I figure it, I can't seem to get it to come out to what you got
Is (2, 4) = 2 = x1 and 4 = y1?
-1 would = m correct?
y-4 = -1(x-2) ; do you wanna shift it to the slope intercept now?
Okay, I switched the y-4 and had -4-y, and just input 0 That is where my mistake was then...So you can have many different equations then for these types of problems?
I just have to show an equation at this point...
yes; you can have as many equations as there are ways to move everything around. All the forms of the equations do is move stuff without changing any values
Wow, and I noticed that y-4=-1(x-2) cannot be simplified; therefore will be an equation correct?
its like having a box of crayons; it doesnt matter how you arrange them in the box; you still have the same box of crayons
Nice... Your always such a good help.
How is the family?
yes, the point slope form that is there is a valid form of a line equation :)
You helped me before, personally. JaniSmit...remember
if its been longer than 3 days, its merged into the past memories lol
family is ok; daughter decided to sprain her ankle tho
Nice...poor little thing...hope she recovers quickly
:) yeah, shes bound to break other parts tho
I told you my husband is military and I'm helping him with his math classes...this is the last one..thank God
So when I have to find an equation of the line that passes through the given points such as (2,1) and (2,5) I would use the same formula right?
slope formula; which is just telling you to subtract one point from another
Or, m=y2-y/x2-x1
Gotcha...awesome...i caught that one ..woohoo..
yeah, that one... (x , y ) -(x1 , y1) --------- (x-x1, y-y1)
Okay, so with (2,1) and (2,5) I would do... m= 5 -1 / 2 - 2 = 4 / 0 = 4?????
(2,5) -(2,1) ------ 0,4; slope = y/x = 4/0 The ONLYtime you get this is when you have a vertical line.
when you have a vertical line then the equation is simply: x = whatever your x value was in your points
So it equals 0 then? As if a fraction 4/0=0
not zero; undefined
Undefined when x1=x2............
0 slope is like driving on a flat road; its just flat, no elevation changes
4/0 is like jumping off a cliff; you only go in one direction
That slope could be going in any direction correct? Up and down or sideways right..
Funny, I got you exactly..i understand
I have another tough one that i'm lost with ..you game?
sure :)
This one says to find an equation of the line that has slope m and y-intercept be? for the equation m = -2; b= -1 I am unsure which formula to use really..
Maybe the y-y1=m(x-x1)
the one thats called slope intercept form :) since it gives a slope and an intercept
y = mx +b
y = -2x -1
Darnit, you beat me..I found some formulas just now....it said exactly what you said..your awesome
yeah :)
Wow, so these are literally just chosing equations that do not need to be solved..
well, the solution IS the line that is created by them
which is why they are called 'line' equations :)
one solution is called a point; all solutions are called a line
So, If I am trying to find 2 different slopes that are parallel to one another, do I just create 2 different equations?
I do not recall this kind of formula
yep; but they should have the same slopes
ax+by = c ax + by = d are parallel lines
when c=d they are the same line
Basically, if i had to find an equation for (-1,3) that is parallel to the line passing through points (-2, -3) and (2,5), would that formula fit that description?
perpendicular lines are: ax + by = c bx - ay = d
you are given a point; and some slope information; use slope point form after using the slope formula to determine the slope
Okay, I see one formula in my book that says "General Form" Ax + By + C = 0
yep; that is a general form
Wow, this one is actually confusing me..darnit..i'm really getting it too
(-2, -3) -( 2 , 5 ) --------- -4, -8; slope = y/x = -8/-4 = 2
lol ..... i messed that one up
Wow, i'm lost..ugh
y-3=2(x+1)
Okay, so point-slope formula first
slope formula first; cant plug in a slope unless you get a slope first
Okay, just to verify....y=mx+b (point slope comes first)
Then, the Actual slope-intercept form? y=mx+b?
Okay, so y-3=2(x+1) is the actual point slope formula....and y=mx+b is the slope-intercept form
yep; slope formula, then point slope, then slope intercept
y - 2x +2 +3 y = 2x +5
I guess that is where i'm getting confused..I am using the (-2, -3) and (2, 5) for this formula right?
The second part assuming...i'm actually lost here
Hey, I know you are busy and you have been an enormous help. Thank you so very much. I appreciate all your hard work. Makes so much more sense now.
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