Hello where not going to do some more slope and then maybe some Y intercept problems as well, let’s get started. So let me make up a problem. Let’s say we have the points 2 comma 5. The other point let’s make that negative 3 comma negative 3. Well first let’s just graph those 2 points. I’m going to graph them in yellow, so 2 comma 5 lets see that’s 1 2, 1 2 3 4 5 so 2 comma 5 is going to be right over there. It’s 2 comma 5 okay, and then let me graph negative 3 comma negative 3, so it’s 1 2 3, 1 2 that’s a negative 3 comma negative 3 is right over there and then now let me draw a line that will connect them.

When you take and draw two pieces. I think that’s good enough. Okay so let’s see if we can at least first figure out the slope of the line then if we have a time we’ll try to figure out the Y intercept and then we’ll know the whole equation for the line. Let me pick a slightly thinner color and we’ll get started.

So the slope if you see, if you saw the last module that just introduce us how we calculate the slope, that just rise over run or change in Y. I hope I’m still using a line tool. Change in Y over change in X this is a Y. Change in Y over change in X, so let’s just do that real fast so let’s think this is our starting point so change in Y could be 5, remember Y is the second coordinate. 5 minus negative 3 and that’s this one over, now you do the change in X, 2 minus—this is also a negative 3. Well 5 minus negative 3 that’s 5 plus plus 3. So that equals 8 and then 2 minus negative 3 once again it’s 2 plus plus 3, so that equals 5.

So we figured out the slope of this equation it’s 8 over 5. Let’s see if that make sense, let’s figure out what’s the rise and the run is. If we were start to at this point right here, let’s see how much we have to rise to get to the same Y coordinate at the other point. So let’s see, we’re here and the other point is up here. So let’s see how much we have to let’s figure out what this distance is. So now it’s a good time to use a fact—oh men I have a shaky hand. Okay let’s figure out what that distance is. That distance is Delta Y which is change in Y.

So it’s 1 2 3 4 5 6 7 8 that equals 8, and that make sense because if you think about what did we just do? We just took Y equals 5 which is up here minus Y equals negative 3 and so we obviously we just calculated that distance just by looking at the 2 quadrants 5 minus negative 3. When you do this calculation, it actually gives you this distance right here so that’s how we figure out how much we have to rise. So now let’s do the run. Well the run to go from this point to the other point we went this far. We went that far and let’s count how far that is. Well it’s 1 2 3 4 5 units, so we can say Delta X is equal to 5. That’s exactly we did Delta Y over Delta X was equal to 8 over 5 or rise over run is equal to 8 or 5. And it would’ve been the same thing we calculated run here of we calculated rise here. It’s the same thing. I hope that’s making sense to you.

I hope that also make sense that if you if the rise for given run becomes more than the slope and the line become stepper or become a bigger number. So let’s see what we have so far for the equation of this line. So far we know the equation to this line is equal to Y is equal to the slope 8 over 5X plus B. So we’re almost done we just got to figure out—I’m sorry I’m using this improperly, we just have to figure out this B right here.

When that B just so you remember that’s the Y intercept and that’s were intersect the Y axis. And since this graph is pretty neat we can actually inspect it then see that well it looks like where intersecting a Y axis at 2 so my guess is we’re going to come up to B equals 2 but lets solve it just in case we didn’t have this neatly drawn in graph here.

So how we can solve for B? Well we can substitute values that we know that work for X and Y. Well either of this points are on that line, so we can substitute them in for X and Y. So let’s use the first one since where—okay so the Y, we get 5 will equal 8 over 5 times X. Well X there is 2 times 2 plus B. Well now we just get 5 is equal to, that is 16 over 5 plus B and then we get B equals of 5 is 25 over 5 right. 5 is 25 over 5 minus 16 over 5 equals 9. Alright see so, I was actually wrong when I looked at, when I looked at this graph I said oh that looks like almost 2 so yeah it’s probably going to be it’s probably going to be 2 but when we actually did it using algebra than we did it analytically we actually solve that B is equal to 9/5. So all, it’s almost to 9/5 is 1 4/5 so 1.8.

So that’s almost 2 but it actually turns out that is not, it’s at 1.8, I can write it down as a decimal 1.8. So the final equation for the line I’m going to try and squeeze it in at the bottom of this page, its going to be Y is equal to—oh we know the slope, 8 over 5X. Now we just add the Y intercept plus 9 over 5 there we solved it.

Let’s do another one and so alright that’s 9 over 5. I don’t want to be to repetitive. Let’s solve—let’s do another problem. We intend to do another problem and let me put that graph back there again. Graph there you go. Alright I’m going to think of 2 random numbers again. I’ll try do this fast because YouTube put a 10 minute limit on me. So let’s say I had the points 2 comma negative 3 I had the point negative 4 comma 5. So 2 comma negative 3, lets plot that sucker real fast.

So X is 2 so it’s here and the negative 3, 1 2 3 so 2 comma negative 3 is there and negative 4 comma 5, so it’s 1 2 3 4, 1 2 3 4 5, I have to count like this because this graph is unlabeled but if we actually were draw in the coordinates you would see this 5 and this is negative 4 and so on and this is 2 and this is negative 3. And now let’s just draw a line. I’ll start right there with my shaky hand. Okay there you go. Good line and another good line. Alright so first we need to figure out the slope, well we could just do that doing the algebra. So slope is just Delta—oh I’m still using the line tool again—Delta Y or Delta X change in Y over change in X.

Let’s take this Y as the first point now so let’s say 5 minus this Y negative 3 over—now since we use the 5 first we’ll choose the negative 4 first as well—negative 4 minus 2 well 5 minus negative 3 that equals 8 and 4, negative 4 minus 2 well that equals negative 6 and negative 8 over 6 well that equals they both are divisible by 2, so that equals minus 4 over 3 and let’s see, does that make sense as a slope? Well if we were go down 4 from this point, so the rise is negative 4 1 2 3 4, so if we go down—oh oops I’m using white so that’s why you can’t see it—we go down by 4 here and then we go to the right 3 positive 3, we still end up on the line, so it works.

It looks good to me, let’s see if we can figure, if I can solve the Y intercept in 30 seconds otherwise I’ll start on the next module. So we get Y is equal to minus 4 over 3X plus B and actually what will do is leave off here and I’m going to solve for B and you could try that on your on and in the next installment in this presentation.