Someone please help, even if just for one of them! 1.Write the equation of the line with a slope of -5 and a y-intercept of (0,3). 2.Write the equation of the line with a slope of -1/3 and passing through the point (6, -4). 3.Write the equation of the line passing through the points (0, -4) and (-2, 2). 4.Write the equation of the line passing through the points (-6,1) and (-4,2). 5.Write the equation of the line with an undefined slope, passing through the point (2, 5).
how far did you get?
I don't know how to start
let's start with #1
1.Write the equation of the line with a slope of -5 and a y-intercept of (0,3).
The given slope is -5, so m = -5 (m is the slope) the y-intercept is (0,3) or just 3, so b = 3
y = mx + b ... start with slope intercept form y = -5x + b ... replace m with -5 (since m = -5) y = -5x + 3 ... replace b with 3 (since b = 3)
so the equation with a slope of -5 and a y-intercept of (0,3) is y = -5x+3
make sense?
Okay thank you! so do you use the same formula for all of them?
yes, but the ones that don't have a given slope, you need to find it first
and if you don't know the intercept, you need to find that as well
once you know the slope and y-intercept, you know the equation based on what is done above
okay so what if they give me a line on a graph and tell me to write the equation?
identify two points on that line find the slope using those two points then use one of those points (along with the slope) to find the y-intercept
to make life easier, you can visually look for the y-intercept so you don't have to do much work
does that make sense?
kind of
what are two points on that line?
-1 and 3
write them as ordered pairs for instance, the point (3,1) is on the line (see attached image)
what other point is on the line?
(-3,-3)
good
now use these 2 points to find the slope
you do so by using the slope formula \[\large m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\]
okay!
tell me what you get
,m = -3-1/-3-3
keep going
-2
no
-3 - 1 is equal to ???
-4 /-6
reduce that
-2/-3
keep going
two negatives divide to get a _____
positive
so the slope is 2/3
so given any two points, you can find the slope of the line through those two points (using the slope formula) so you'll be using this formula for #3 and #4
we now have a slope of 2/3, so m = 2/3
how do we find the y-intercept? ie how do we find the value of b?
i dont know
remember we have the general equation y = mx+b
we know the slope is m = 2/3 we also know that the point (-3,-3) is on the line so this means that x = -3 and y = -3
plug all this in, then solve for b \[\large y = mx+b\] \[\large y = \frac{2}{3}x+b\] \[\large -3 = \frac{2}{3}(-3)+b\] \[\large -3 = \frac{2}{3}*\frac{-3}{1}+b\] \[\large -3 = \frac{2*(-3)}{3*1}+b\] \[\large -3 = \frac{-6}{3}+b\] \[\large -3 = -2+b\] \[\large b = ???\]
-5?
no
you have to add 2 to both sides to isolate b
oh crao i added them, it would be -1
good, look at the graph and notice where it crosses the y axis
oh wow thats alot easier hah
yes BUT you won't always have the graph so it helps to know how to do this without a graph
besides, the line may intersect the y-axis at some point that's not a whole number
anyways, we know the slope is 2/3 the y-intercept is -1 so this means m = 2/3 b = -1
Now plug these pieces of info in \[\large y = mx+b\] \[\large y = \frac{2}{3}x+b\] \[\large y = \frac{2}{3}x+(-1)\] \[\large y = \frac{2}{3}x-1\] So the equation of the line is \[\large y = \frac{2}{3}x-1\]
thank you !
you're welcome
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