#1 (permalink)

Show work plx

Write and equation of the line that contains the points (-6,9) and (-11,-4)

#2 (permalink)

shit, wish i was in highschool. hang on i'll get some people to help you with this

#3 (permalink)

i'm confusing myself. anyone else gettin 2.6 for the gradient? :S

the way im taught how to do these is: to find the gradient y2-y1/x2-x1 = -4 - 9 / -11 + 6
= -13/-5
= 2.6

then use the equation y-b=m(x-a) and put in the values (-6,9) = (a,b) and (-11, -4) = (x, y)

so it'll be -4 - 9 = 2.6(-11 -(-6))


but i think i've done it wrong =/ sorry. cause we normally use the equation to find out the 2 points you don't know :S.

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#4 (permalink)

slope=change in y value/change in x value=rise over run or slope

so

(-6,9) and (-11,-4)

9-(-)4=13 (change in y value)

-11-(-)6=5 (change in x value)

so 13/5=y/x=rise over run

so the slope is 13/5 so rise 13 run 5, slope =13/5

so use one of your coordinates and plug in with y=mx+b

-6,9
x, y

y=13/5(-6)+9=-6.6, which you have to convert the decimal to an improper fraction, then you have your equation solved...

I hope i helped, its been about 4 years since my last math class in highschool. Im in calc now and i cant remember the last time i did algebra

#5 (permalink)

dont you only use y=mx+b for a straight line?

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#6 (permalink)

god no, y=mx+b is slope blah blah

and the only line that isn't "Straight" is a parabola, and i bet his aint dealing with parabola's in alg I

#7 (permalink)

http://www.math.com/school/subject2/.../S2U4L2DP.html

here is a useful link

#8 (permalink)

meh, you's do it differently and have different names for stuff so im just getting more confused lol.
i think i get it. you's use the word "slope" for the gradient? never mind im just gonna do my calculus homework ¬_¬.

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#9 (permalink)

You made a mistake when you subbed it into y=mx+b. You used the y value for b; the value of b should be the y-intercept value.

Here's the correct equation with explanations:

(-6,9) and (-11,-4)

First you find the slope between the two points:

m=delta y/delta x
m = (-4-9)/(-11-(-6))
m = -13/-5
m = 13/5

Then, since you have the slope and you also have two usable points that the line goes through, you sub it into this formula:

y=m(x-p)+q
Where m is the slope, y and x are left as they are, p is the x value of one point (TIMES -1, so it's sign will be opposite of what the x value is for the graph), and q is the y value of the same point.

y=m(x-p)+q
y=13/5(x+6)+9
y=13/5x+78/5+45/5
y=13/5x+123/5
y=2.6x+24.6

So, the equation of the line that contains the points (-6,9) and (-11,-4) is y=2.6x+24.6.

#10 (permalink)

Great post Knox.

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