find the equation of the line that contains the point (8,1) and has slope -5....thanks for the help!
y = -5(x-8)+1
y = -5x +41
do i add -5x to both sides to solve for y?
its already solved for y. Do you mean solve for x in terms of y?
I guess I wanted to know if anything further needed to be done or if this was the complete answer. Sorry...I am horrible at math! :)
there are many ways to express the equation of a line; this is the most common form tho.
yes the usual is y=mx+c in wich this is already displayed
5x+y = 41 y = -5x +41 y-1 = -5(x-8) 5x +y -41 = 0
m being the gradient and c the intercept, usually the way thats most useful
gotcha...thank you for the help and for the extra examples. I really appreciate it!
no problem :D
one more question....would the equation be the same if I have two points rather than a slope? Examples points (0,4) and (1.-3)
sorry that should be (1,-3)
The slope is found using diff in y / diff in x
\[\frac{4-(-3)}{1}\] = -7 so the equation becomes y=-7x+4
so you subtracted the two y points from each other and the two x points from each other, than divide that to find the slope? so in the equation y represents the slope?
sorry missed the -1 in the denominator
The y represents the slope and the intercept y=Mx+C where M is the slope and C is where it crosses the y axis
So the equation represent the line
so in its fullness we get \[y= \frac{(4-(-3))}{(0-1)}x + 4\]
I see it now! :D
kewl :)
thank u for ur patience with me!
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