Understanding Cartesian Coordinates and Linear Equations

In this video, we'll cover the prerequisite concepts from basically Cartesian coordinates. We'll look at points, lines, and slope of lines in two space. So the objectives here, we want to be able to find delta y and delta x, find the distance d between two points on a straight line, find the slope of the line segment. Then given vertices of a parallelogram, try to find the fourth vertex that is sort of find the parallelogram. Then given some properties of the line, find its equation, find where one line intersects with another, and then find the distance from a point to a given line. So let's kind of refresh ourselves with what coordinates, incremental change, slope, and distance are about in the Cartesian plane. So the point in the xy plane or the Cartesian plane is defined by some sort of ordered pair. We typically denote that with p or x. Y are coordinates right X is the distance from the origin Y is along the x-axis Y is the distance From the origin along the y-axis and then we plot those points. So here's a point in another point Delta X is the incremental change in X as you go from P 1 to P 2 So Delta X can be negative as in the case here as we go from X 1 to X 2 Delta X is going to be negative And delta y is the change from y1 to y2, so this offset here. d is the distance between the two points. That's the length of this line. We're going to use the Pythagorean theorem and find the distance by taking the square root of delta x squared and delta y squared. Notice that even if delta x is negative, Delta x squared will be positive, so the sum of the squares of delta x and delta y will always be positive. And the square root of that is the distance. This is the triangle theorem or Pythagorean theorem. We can also talk about the slope of the line. We typically denote slope with the letter m. And slope is the change in y over the change in x. Rise over run, if you want to think about that. And slope is nice because if I'm going to start it. P1 here, right? And I can go, you know, so this little delta Y, delta X here. I can use M to just walk myself along that line, to find any of the other points along that line. Okay, so that's kind of what slope gives you. We typically denote slope, or we will find slope as delta y over delta x, with the understanding, or with the restriction that delta x cannot be zero. When delta x is zero, you have vertical lines which have undefined slope. So let's work a couple of simple examples here. Given a point, let's find delta y, delta x, the distance between the two points, and the slope of the line connecting the two points. So we are going to go from P1 to P2 here. So let's just find delta X is going to be X sub 2. We're going to let the X sub 1, Y sub 1, X sub 2, Y sub 2 there. Let's get a little space here. X sub 2, Y sub 2. Okay, so delta x is going to be, change in x is going to be from 1 to 2, we do x sub 2 minus x sub 1 minus 1 minus 2 is minus 3. Delta y is going to be y2 minus y1, that's 4 minus 2. I'm sorry, that's 4 minus a minus 3. All right, so that's 4 plus 3, and that's 7. So we found delta y and delta x. The distance between the two points is the square root. Delta x squared plus delta y squared, that's the square root of minus 3 squared. squared plus 7 squared. Delta x was minus 3. That's the square root of 9 plus 21. So that's the square root of 7 squared. I'm sorry, it's 49. So that's the square root of 58. All right, let's see. We found delta x, we found delta y. We want to find slope. Slope is delta y divided by delta x, and that would be 7 over minus 3 or minus 7 thirds. All right, let's do now, let's see here, we'll do the same problem, except now we're going to go from P2 to P1, all right, so same two points, just in this case. We're going from one ordered pair to the other. OK. So now delta x is going to be, we're going to go from p2 to p1. I'm going to mark those again just so we're clear. This is now x sub 2, y sub 2. And this is x sub 1. y sub 1. So delta x, going from p2 to p1, is going to be x1 minus x2. That's 2 minus minus 1, which is 3. Notice a sign switched on us here. Delta y will be y1 minus y2. Y1 is minus 3 minus 4 equals minus 7. So distance is going to be the square root of 3 squared plus minus 7 squared. And that's what? 9 plus 49 square root of 58. And what do we have last? M. Slope is going to be delta y over delta x. Delta y, we said, was minus 7. x was 3. So that is minus 7 thirds. So notice the slope and distance between the points don't change. However, delta x and delta y change in sign. All right, so now let's talk a little bit about straight lines. If we have a straight line, we have two points on a plane. Those define a straight line. Any two points, you connect them, you get a straight line. That line runs. You want to think about that in both directions to infinity there. So, phi, we can talk about the angle of inclination of phi, which is basically this angle between the positive x-axis and the line. Okay, and any horizontal line is going to cross the x-axis. The slope of that line is tangent, which we'll talk about trig functions later, but I'm sure most of you are familiar with that. So slope is going to be tangent of phi with phi between 0 and 180. For horizontal lines, phi is 0, which means the slope is 0. For vertical lines, phi is 90 degrees, which means m is undefined. And as we know... The tangent of 90 degrees is also undefined. All right, so on to parallel and perpendicular lines. So we're just given some notation here. So angle of inclination there. Parallel lines. So any two non-vertical lines are parallel if their angle of inclination is the same. All right? And If their slopes are the same, then their angle of inclination is the same. So if you know the slope of two lines, if they're the same, the lines are parallel. For vertical lines, you have to have both M1 and M2 be undefined, okay, for them to be parallel. For perpendicular lines, you really want to have that the angle of inclination phi differs by 90 degrees between the two. So let's get some red here. There's line L1 and line L2. This is 90 degrees here. Okay? So they're perpendicular. And one of the easy tests for perpendicular. is that the slopes are the negative reciprocals. So in this case, if m1, let me get some blue here, m1 is the slope of l1. Since L2 is perpendicular to L1, that's what the little symbol means there, is perpendicular to, right? That's what this little symbol means. Oh, that was not good. Okay. Then m2 is minus 1 over m1. So lines that are perpendicular to each other, their slopes are the negative reciprocals of either other, with the exception of when m1 is a vertical line and has no slope. In this case, when m1 is 0, m2 is undefined. When m1 is undefined as a vertical line, as I said, then m2 is 0. This you can't find by finding negative reciprocals. It's just something we know, right? If one line has a slope of 0, then the other line, then the perpendicular line has an undefined slope. And if a line has an undefined slope, then the perpendicular line to that has a slope of 0. Here we're just showing the equation. Here's the equation for a horizontal line. It's red. Equation for a vertical line, y equals b and x equals a. We can see there are slopes there. So now let's do a little work here where we're going to take the parallelogram and we're going to find the fourth vertex of a parallelogram. All right, so remember parallelograms are opposite sides are parallel and of equal length and the special case of that would be a rhombus where all the sides are the same length or where the diagonals form a perpendicular line, form perpendicular lines. So pretty much if you want to do one of these, we're going to be given coordinates. We're going to have to draw this out. So I'm going to come over here, make myself a little coordinate system. So we've got coordinates in the first quadrant, the second quadrant, and the fourth quadrant. So I'm going to need all my quadrants. I just looked at the sign. B is in the first quadrant because both x and y are positive. C, x is negative, y is positive, that's quadrant two. And D, that's five, x is positive, y is negative, that's quadrant four. So now let's start plotting this and we'll just plot B. I'm just going to kind of, let's see, most of these guys are within, what's my longest one out there going to be? Yeah. So 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. 1, 2, 3, 4, 5, 6, 7. 1, 2, 3, 4, 5, 6, 7. So we'll try to be mostly the scale. So b is at 3, 4. I'm going to go ahead and write the coordinates on there. C is at minus 5, 2, 1, 2, 3, 4, 5. Clean that up a little bit for you there. D is at 5 minus 3. Okay, so we want to find the coordinates for A. All right, so let's find the slope here of the line BC. So I'm going to put that in blue. And then I'm going to come under here and go over. And so here's delta, oops, let's code that better. Delta y here, delta x. So that means there's some point that show color, well we'll make a be some color out here. So a is out here somewhere. And to find a, we're basically going to do this perpendicular line here, which is going to have the same delta x and the same delta y. So I'm going to come over here and I'm going to say that I'm going to say that Delta x here is going to be, we're going from c to b, and we want to go from d to a. So that's going to be 3 minus minus 5, and that's minus 8, sorry, that's positive 8. Delta y is going to be... 4 minus 2, so that's 2. So now if we let x of a be the coordinate of a and y sub a be the coordinate of a be x of a, y sub a, then x of a is equal to. D's x sub D plus 8 and x sub D is 5 plus 8 equals 13 and y sub a is equal to y sub D plus Delta Y so let's do And that's going to be minus 3 plus what was delta y was 2. And that's minus 1. So a is 13 minus 1. And if you really want to see the parallelogram, we can connect the other two lines there. So we found point A that completes the parallelogram. All right, so now we're going to move on to. Forming straight lines, all right, and the equations for straight lines. So we've got lots of different forms for the equations of straight lines. So if we've got two points, we know how to find delta y and delta x. I've already done that. So one form for the equation of a straight line is the point. slope form. This will work for non-vertical lines. Here we're just going to do m is equal to, so I'm going to write out to the side. You can do m is equal to y minus y1 over x minus x1. And that's going to be the point slope form, except I'm going to multiply through here, and we get y minus y1 equals m times x minus x1. That's the point slope form right there. Okay, I like to remember it. like this just because I know that's what M is. If this was y2 minus y1 and x2 minus x1, I'd find slope. All right, so then we have the slope intercept form. Here we're just replacing b with y1 minus mx sub 1. We might be given the intercept. in which case we're given the intercept in the slope. So this is the slope intercept form where we're given y equals mx plus b. And if I wanna plot something with the slope, given the slope intercept form, I plot the point b and then I just make another point using the slope. Let's do that in red. Right, so then... This is going to be 0 plus delta x and b plus delta y. And we're going to get delta x and delta y from the slope. From M right M is equal to Delta Y over Delta X Alright That's point slope form or slope intercept form. So I'm gonna write the name underneath here for you Slope Intercept this is the one most students this is called point slope Point slope, slope intercept, what else have we got here? We've got the two-point formula. Here we're going to basically do, we're just going to put in the computation for slope, m. So y minus y1. Over x minus x 1 here's the equation piece and y 2 minus y 1 over x 2 minus x 1 gives me slope as long as X 2 and x 1 aren't the same point. That's really nice if you're given two points That's pretty much a good form to write the equation of the line Nice piece about this one is you can kind of rearrange some things here and we can get This form just sort of the two intercept form So a and b are the x and the y intercepts. So we rearrange the slope form. So if you think, let's see. If y equals mx plus b and we divide through by b here, then I get y over b minus... M over BX equals 1. So X over A plus Y over B equals 1 with A equal to minus B over M. So if you're given two points, two intercept pieces, then X over A plus Y over B equals 1 is the equation for the line. A general form for the equation of the line is finally ax plus by is equal to c, where a and b are not both 0. If a is equal to 0, we have a horizontal line. If b is equal to 0, we have a vertical line. Alright, so this is the general form. ax plus by equals c. This is the general form, and I will note that often we impose the restriction that a must be positive, and you multiply through by minus 1 if a is negative, just to make the first leading coefficient of x be positive in the general form. All right. Alright so now let's do some stuff with lines here. So we're going to work on an objective. We want to do given some prop. We want to give it a specific line. We want to find its equation. We want to find where some lines intersect and then we want to find the distance from the point to a line. So these kind of all work. work together so let's I'm gonna kind of trace something out here on the side and we'll see if we can't work most of this and keep all of this on the same page here for us all right so we want to find the line that passes through the point one five and it's perpendicular to the given line 3x plus 4y plus 2 equals 0 All right, so let's go to how are we going to graph 3x plus 2y equals 0. Well, I say we move that 2 over and we move that 3x over and we write this thing as y equals minus 3 fourths x. Minus one half. Okay, that gives me my slope information. Let's go ahead and plot this thing. Alright, so when x is equal to 0, y is equal to minus 1 half, and our slope is minus 3 fourths, it's almost minus 1, so it's just going to be something about like that. We're not going to really worry about... And I'll just give it in standard form, 3x plus 4y. So there's our line. And we know one intercept. And we know m of line 1 is minus 3 fourths. All right, so let's find the equation. for line L. So L is perpendicular to this line and passes through the point 1, 5. So let's put the point 1, 5 on here. So 1 over, let's just call this P1, 5 here. Right, and I'm just gonna go over and put my perpendicular mark there. I don't know if I have that intersection in the right place or not. I'm close. So we want to find the equation for line L. This is line L. So the slope of line L, since this is perpendicular to line 1, is minus 1 over the slope of line 1. And that will be 4 thirds. We know a slope and we know a point. Therefore, the nice form for writing the equation of this line would be to start with a point-slope form. So 4 thirds equals y minus y1, which is 5, over x minus x1, which is 1. Let's multiply through by the denominator and that gives me y minus 5 is equal to 4 thirds times x minus 1. We could stop there, but I think I'm going to go ahead and take this on to slope intercept form. So that'll be multiplied through by the 4 thirds and add 5 to both sides. So I'm going to put a therefore sign out front here. That's what the double arrow or the equal sign with the greater than sign in front of it looks like. Is it therefore it implies. So y equals, let's see, and we're going to have what, four thirds x. And then we're going to have what, we're adding five to both sides. So that's going to be 15 minus four thirds. That's plus 11 thirds. I'm going to go ahead and box my answer there. Alright, so there's the equation for line L. Now we want to find the intersection point between line L and line L1. That'll be this intersection point here. Well, to find that, we want to find the value of x that makes both those equations true. So what I'm going to do, since I have y equals, I'm going to find the value of x. So I'm going to find 4 thirds x plus 11 thirds and I also have y equals minus 3 fourths x minus 1 half. I'm going to set those two things equal to each other. So it'll be minus 3 fourths x minus 1 half equals 4 thirds x plus 11 thirds. And now we just want to solve for x. So we're going to add. 3 fourths to both sides and we're going to subtract 11 thirds from both sides. So that's minus 1 half minus 11 thirds is equal to 4 thirds plus 3 fourths x. Combine all of those, I'm going to get what common denominator is going to be 6. So that's going to be... right, I'm going to multiply 2 by 3, that's going to be minus 3 minus 22 over 6, and that's going to be equal to, a common denominator here is going to be 12, that's going to be 16 plus 9. All right, so therefore I'm going to get what minus 25 over 6 equals 16 and 9 is... ah! Sorry about that. 16 over plus 9 is 25 so that's equal to 25 12x. Therefore, I'm going to get a little more space here. Reciprocate, that's going to give me x is equal to 12 over 25 times minus 25 over 6, and that's equal to 2. Lost my minus sign. Equal to minus 2. And if x is equal to minus 2, I can plug in almost anywhere. I'm going to plug into the original equation. So y equals minus 3 fourths minus 2 minus 1 half. And that's 6 fourths minus 2 fourths. And that's going to be 1. So the point of intersection, P, is x is equal to minus 2, y is equal to 1. I'm going to come back up here and label that in red. This is P. What did I find? Minus 2, 1. It's believable that that's the point. All right. Be a little off there. Finally, now, we want to find the distance. From the point to the line. Well, the distance from the point to the line is the shortest distance from the point to the line. So that's the distance from the line, from the point P15 to the point on line L1 that is perpendicular, such that the line that connects those two is perpendicular. That's the point we just found. So get myself a little space here. Whoops. Let's get back to blue here. Distance is the square root of, let's just write it as delta x squared plus delta y squared. In this case if I look up here that's going to be what? Delta x 1 minus minus 2 so that's 3 squared. And 5 minus 1 is 4. So that's the square root of 9 plus 16. And that's equal to the square root of 25, which is equal to 5. So the distance is 5 units from the point P1 to the line L1 that we started with. All right? Okay, so finally, let's look at a problem where we can use the linear relationship between things to solve a problem. All right, so the amount of heat in joules required to convert one gram of water into vapor is linearly related to temperature. That means that the relationship is a line, which means that we can use slope kinds of thinking to solve this problem. At 10 degrees centigrade, this conversion requires 2,480 joules. And each increase in temperature of 15 degrees Celsius lowers the amount of heat needed by 40 joules. So we want to sort of find... The amount of heat needed to convert one gram of water into vapor when the temperature in the atmosphere is 35 degrees Celsius. Okay, so the first thing to do is let's write a formula or an expression that gives H in terms of T. So if the relationship between the amount of heat in joules to require to convert one gram of water into vapor is linear related to temperature, then that means that delta H over the change in temperature, that's dy and dx, right? equals, so that'll give me slope information. Now I just need h minus some initial point over temperature minus some initial temperature. We're given some initial information there, so that means I'm going to have... Each increase in temperature of 15 degrees Celsius, that would be my delta T, lowers the amount of heat by 40 joules. And that's going to give me a formula for heat. And now I know my initial starting point is... At 10 degrees Celsius, this conversion requires 2,480 joules, and the initial temperature there is 10 degrees Celsius. All right, so I need to do some simplification there. This is going to give me what? Minus 8 thirds. Times t minus 10 equals h minus 2480. So divide through by 5, this is 8, and this is 3. So minus 8 thirds and then I just multiply by the numerator. I can now solve this for H because we want to express H completely in terms of T. H equals minus 8 thirds T minus 10 plus 2480 joules. Okay? If you think about it, we knew the joules at this initial point when t is equal to 10. So when t is equal to 10, which I read, when t is equal to 10, Then H is 2480 joules because this will be zero. So that makes sense. And then as we increase the temperature, the slope keeps walking down the total number of joules. We're going to be subtracting off the number of joules. So now we want to find... H at 35 degrees and this is equal to minus 8 thirds T minus let's see 35 minus 10 degrees Celsius plus 2480 joules that's going to be 2480 35 minus 10 is 25 25 times 4 is 100 times 8 is 200 so that's minus 200 thirds and that's going to be what three times of 2480 is 7440 minus 200 over 3 so that gives me H is equal to 72 thousand forty over three Joules. Clean that up a little bit. 7240. Okay. Alright, so that concludes this video.

Then given vertices of a parallelogram, try to find the fourth vertex that is sort of find the parallelogram. Then given some properties of the line, find its equation, find where one line intersects with another, and then find the distance from a point to a given line. So let's kind of refresh ourselves with what coordinates, incremental change, slope, and distance are about in the Cartesian plane. So the point in the xy plane or the Cartesian plane is defined by some sort of ordered pair. We typically denote that with p or x.

Y are coordinates right X is the distance from the origin Y is along the x-axis Y is the distance From the origin along the y-axis and then we plot those points. So here's a point in another point Delta X is the incremental change in X as you go from P 1 to P 2 So Delta X can be negative as in the case here as we go from X 1 to X 2 Delta X is going to be negative And delta y is the change from y1 to y2, so this offset here. d is the distance between the two points.

That's the length of this line. We're going to use the Pythagorean theorem and find the distance by taking the square root of delta x squared and delta y squared. Notice that even if delta x is negative, Delta x squared will be positive, so the sum of the squares of delta x and delta y will always be positive. And the square root of that is the distance.

This is the triangle theorem or Pythagorean theorem. We can also talk about the slope of the line. We typically denote slope with the letter m. And slope is the change in y over the change in x. Rise over run, if you want to think about that.

And slope is nice because if I'm going to start it. P1 here, right? And I can go, you know, so this little delta Y, delta X here. I can use M to just walk myself along that line, to find any of the other points along that line.

Okay, so that's kind of what slope gives you. We typically denote slope, or we will find slope as delta y over delta x, with the understanding, or with the restriction that delta x cannot be zero. When delta x is zero, you have vertical lines which have undefined slope.

So let's work a couple of simple examples here. Given a point, let's find delta y, delta x, the distance between the two points, and the slope of the line connecting the two points. So we are going to go from P1 to P2 here. So let's just find delta X is going to be X sub 2. We're going to let the X sub 1, Y sub 1, X sub 2, Y sub 2 there. Let's get a little space here.

X sub 2, Y sub 2. Okay, so delta x is going to be, change in x is going to be from 1 to 2, we do x sub 2 minus x sub 1 minus 1 minus 2 is minus 3. Delta y is going to be y2 minus y1, that's 4 minus 2. I'm sorry, that's 4 minus a minus 3. All right, so that's 4 plus 3, and that's 7. So we found delta y and delta x. The distance between the two points is the square root. Delta x squared plus delta y squared, that's the square root of minus 3 squared.

squared plus 7 squared. Delta x was minus 3. That's the square root of 9 plus 21. So that's the square root of 7 squared. I'm sorry, it's 49. So that's the square root of 58. All right, let's see.

We found delta x, we found delta y. We want to find slope. Slope is delta y divided by delta x, and that would be 7 over minus 3 or minus 7 thirds. All right, let's do now, let's see here, we'll do the same problem, except now we're going to go from P2 to P1, all right, so same two points, just in this case. We're going from one ordered pair to the other.

OK. So now delta x is going to be, we're going to go from p2 to p1. I'm going to mark those again just so we're clear.

This is now x sub 2, y sub 2. And this is x sub 1. y sub 1. So delta x, going from p2 to p1, is going to be x1 minus x2. That's 2 minus minus 1, which is 3. Notice a sign switched on us here. Delta y will be y1 minus y2.

Y1 is minus 3 minus 4 equals minus 7. So distance is going to be the square root of 3 squared plus minus 7 squared. And that's what? 9 plus 49 square root of 58. And what do we have last?

M. Slope is going to be delta y over delta x. Delta y, we said, was minus 7. x was 3. So that is minus 7 thirds.

So notice the slope and distance between the points don't change. However, delta x and delta y change in sign. All right, so now let's talk a little bit about straight lines. If we have a straight line, we have two points on a plane. Those define a straight line.

Any two points, you connect them, you get a straight line. That line runs. You want to think about that in both directions to infinity there.

So, phi, we can talk about the angle of inclination of phi, which is basically this angle between the positive x-axis and the line. Okay, and any horizontal line is going to cross the x-axis. The slope of that line is tangent, which we'll talk about trig functions later, but I'm sure most of you are familiar with that.

So slope is going to be tangent of phi with phi between 0 and 180. For horizontal lines, phi is 0, which means the slope is 0. For vertical lines, phi is 90 degrees, which means m is undefined. And as we know... The tangent of 90 degrees is also undefined. All right, so on to parallel and perpendicular lines. So we're just given some notation here.

So angle of inclination there. Parallel lines. So any two non-vertical lines are parallel if their angle of inclination is the same.

All right? And If their slopes are the same, then their angle of inclination is the same. So if you know the slope of two lines, if they're the same, the lines are parallel.

For vertical lines, you have to have both M1 and M2 be undefined, okay, for them to be parallel. For perpendicular lines, you really want to have that the angle of inclination phi differs by 90 degrees between the two. So let's get some red here.

There's line L1 and line L2. This is 90 degrees here. Okay? So they're perpendicular. And one of the easy tests for perpendicular.

is that the slopes are the negative reciprocals. So in this case, if m1, let me get some blue here, m1 is the slope of l1. Since L2 is perpendicular to L1, that's what the little symbol means there, is perpendicular to, right? That's what this little symbol means. Oh, that was not good.

Okay. Then m2 is minus 1 over m1. So lines that are perpendicular to each other, their slopes are the negative reciprocals of either other, with the exception of when m1 is a vertical line and has no slope. In this case, when m1 is 0, m2 is undefined.

When m1 is undefined as a vertical line, as I said, then m2 is 0. This you can't find by finding negative reciprocals. It's just something we know, right? If one line has a slope of 0, then the other line, then the perpendicular line has an undefined slope. And if a line has an undefined slope, then the perpendicular line to that has a slope of 0. Here we're just showing the equation. Here's the equation for a horizontal line.

It's red. Equation for a vertical line, y equals b and x equals a. We can see there are slopes there. So now let's do a little work here where we're going to take the parallelogram and we're going to find the fourth vertex of a parallelogram.

All right, so remember parallelograms are opposite sides are parallel and of equal length and the special case of that would be a rhombus where all the sides are the same length or where the diagonals form a perpendicular line, form perpendicular lines. So pretty much if you want to do one of these, we're going to be given coordinates. We're going to have to draw this out.

So I'm going to come over here, make myself a little coordinate system. So we've got coordinates in the first quadrant, the second quadrant, and the fourth quadrant. So I'm going to need all my quadrants.

I just looked at the sign. B is in the first quadrant because both x and y are positive. C, x is negative, y is positive, that's quadrant two. And D, that's five, x is positive, y is negative, that's quadrant four.

So now let's start plotting this and we'll just plot B. I'm just going to kind of, let's see, most of these guys are within, what's my longest one out there going to be? Yeah. So 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. 1, 2, 3, 4, 5, 6, 7. 1, 2, 3, 4, 5, 6, 7. So we'll try to be mostly the scale.

So b is at 3, 4. I'm going to go ahead and write the coordinates on there. C is at minus 5, 2, 1, 2, 3, 4, 5. Clean that up a little bit for you there. D is at 5 minus 3. Okay, so we want to find the coordinates for A. All right, so let's find the slope here of the line BC.

So I'm going to put that in blue. And then I'm going to come under here and go over. And so here's delta, oops, let's code that better.

Delta y here, delta x. So that means there's some point that show color, well we'll make a be some color out here. So a is out here somewhere.

And to find a, we're basically going to do this perpendicular line here, which is going to have the same delta x and the same delta y. So I'm going to come over here and I'm going to say that I'm going to say that Delta x here is going to be, we're going from c to b, and we want to go from d to a. So that's going to be 3 minus minus 5, and that's minus 8, sorry, that's positive 8. Delta y is going to be... 4 minus 2, so that's 2. So now if we let x of a be the coordinate of a and y sub a be the coordinate of a be x of a, y sub a, then x of a is equal to. D's x sub D plus 8 and x sub D is 5 plus 8 equals 13 and y sub a is equal to y sub D plus Delta Y so let's do And that's going to be minus 3 plus what was delta y was 2. And that's minus 1. So a is 13 minus 1. And if you really want to see the parallelogram, we can connect the other two lines there.

So we found point A that completes the parallelogram. All right, so now we're going to move on to. Forming straight lines, all right, and the equations for straight lines. So we've got lots of different forms for the equations of straight lines. So if we've got two points, we know how to find delta y and delta x.

I've already done that. So one form for the equation of a straight line is the point. slope form. This will work for non-vertical lines. Here we're just going to do m is equal to, so I'm going to write out to the side.

You can do m is equal to y minus y1 over x minus x1. And that's going to be the point slope form, except I'm going to multiply through here, and we get y minus y1 equals m times x minus x1. That's the point slope form right there.

Okay, I like to remember it. like this just because I know that's what M is. If this was y2 minus y1 and x2 minus x1, I'd find slope.

All right, so then we have the slope intercept form. Here we're just replacing b with y1 minus mx sub 1. We might be given the intercept. in which case we're given the intercept in the slope. So this is the slope intercept form where we're given y equals mx plus b. And if I wanna plot something with the slope, given the slope intercept form, I plot the point b and then I just make another point using the slope.

Let's do that in red. Right, so then... This is going to be 0 plus delta x and b plus delta y. And we're going to get delta x and delta y from the slope. From M right M is equal to Delta Y over Delta X Alright That's point slope form or slope intercept form.

So I'm gonna write the name underneath here for you Slope Intercept this is the one most students this is called point slope Point slope, slope intercept, what else have we got here? We've got the two-point formula. Here we're going to basically do, we're just going to put in the computation for slope, m.

So y minus y1. Over x minus x 1 here's the equation piece and y 2 minus y 1 over x 2 minus x 1 gives me slope as long as X 2 and x 1 aren't the same point. That's really nice if you're given two points That's pretty much a good form to write the equation of the line Nice piece about this one is you can kind of rearrange some things here and we can get This form just sort of the two intercept form So a and b are the x and the y intercepts.

So we rearrange the slope form. So if you think, let's see. If y equals mx plus b and we divide through by b here, then I get y over b minus... M over BX equals 1. So X over A plus Y over B equals 1 with A equal to minus B over M.

So if you're given two points, two intercept pieces, then X over A plus Y over B equals 1 is the equation for the line. A general form for the equation of the line is finally ax plus by is equal to c, where a and b are not both 0. If a is equal to 0, we have a horizontal line. If b is equal to 0, we have a vertical line.

Alright, so this is the general form. ax plus by equals c. This is the general form, and I will note that often we impose the restriction that a must be positive, and you multiply through by minus 1 if a is negative, just to make the first leading coefficient of x be positive in the general form.

All right. Alright so now let's do some stuff with lines here. So we're going to work on an objective. We want to do given some prop. We want to give it a specific line.

We want to find its equation. We want to find where some lines intersect and then we want to find the distance from the point to a line. So these kind of all work. work together so let's I'm gonna kind of trace something out here on the side and we'll see if we can't work most of this and keep all of this on the same page here for us all right so we want to find the line that passes through the point one five and it's perpendicular to the given line 3x plus 4y plus 2 equals 0 All right, so let's go to how are we going to graph 3x plus 2y equals 0. Well, I say we move that 2 over and we move that 3x over and we write this thing as y equals minus 3 fourths x.

Minus one half. Okay, that gives me my slope information. Let's go ahead and plot this thing. Alright, so when x is equal to 0, y is equal to minus 1 half, and our slope is minus 3 fourths, it's almost minus 1, so it's just going to be something about like that. We're not going to really worry about...

And I'll just give it in standard form, 3x plus 4y. So there's our line. And we know one intercept.

And we know m of line 1 is minus 3 fourths. All right, so let's find the equation. for line L.

So L is perpendicular to this line and passes through the point 1, 5. So let's put the point 1, 5 on here. So 1 over, let's just call this P1, 5 here. Right, and I'm just gonna go over and put my perpendicular mark there.

I don't know if I have that intersection in the right place or not. I'm close. So we want to find the equation for line L. This is line L.

So the slope of line L, since this is perpendicular to line 1, is minus 1 over the slope of line 1. And that will be 4 thirds. We know a slope and we know a point. Therefore, the nice form for writing the equation of this line would be to start with a point-slope form. So 4 thirds equals y minus y1, which is 5, over x minus x1, which is 1. Let's multiply through by the denominator and that gives me y minus 5 is equal to 4 thirds times x minus 1. We could stop there, but I think I'm going to go ahead and take this on to slope intercept form.

So that'll be multiplied through by the 4 thirds and add 5 to both sides. So I'm going to put a therefore sign out front here. That's what the double arrow or the equal sign with the greater than sign in front of it looks like.

Is it therefore it implies. So y equals, let's see, and we're going to have what, four thirds x. And then we're going to have what, we're adding five to both sides.

So that's going to be 15 minus four thirds. That's plus 11 thirds. I'm going to go ahead and box my answer there. Alright, so there's the equation for line L. Now we want to find the intersection point between line L and line L1.

That'll be this intersection point here. Well, to find that, we want to find the value of x that makes both those equations true. So what I'm going to do, since I have y equals, I'm going to find the value of x.

So I'm going to find 4 thirds x plus 11 thirds and I also have y equals minus 3 fourths x minus 1 half. I'm going to set those two things equal to each other. So it'll be minus 3 fourths x minus 1 half equals 4 thirds x plus 11 thirds. And now we just want to solve for x. So we're going to add.

3 fourths to both sides and we're going to subtract 11 thirds from both sides. So that's minus 1 half minus 11 thirds is equal to 4 thirds plus 3 fourths x. Combine all of those, I'm going to get what common denominator is going to be 6. So that's going to be... right, I'm going to multiply 2 by 3, that's going to be minus 3 minus 22 over 6, and that's going to be equal to, a common denominator here is going to be 12, that's going to be 16 plus 9. All right, so therefore I'm going to get what minus 25 over 6 equals 16 and 9 is...

ah! Sorry about that. 16 over plus 9 is 25 so that's equal to 25 12x. Therefore, I'm going to get a little more space here.

Reciprocate, that's going to give me x is equal to 12 over 25 times minus 25 over 6, and that's equal to 2. Lost my minus sign. Equal to minus 2. And if x is equal to minus 2, I can plug in almost anywhere. I'm going to plug into the original equation.

So y equals minus 3 fourths minus 2 minus 1 half. And that's 6 fourths minus 2 fourths. And that's going to be 1. So the point of intersection, P, is x is equal to minus 2, y is equal to 1. I'm going to come back up here and label that in red.

This is P. What did I find? Minus 2, 1. It's believable that that's the point. All right. Be a little off there.

Finally, now, we want to find the distance. From the point to the line. Well, the distance from the point to the line is the shortest distance from the point to the line. So that's the distance from the line, from the point P15 to the point on line L1 that is perpendicular, such that the line that connects those two is perpendicular. That's the point we just found.

So get myself a little space here. Whoops. Let's get back to blue here.

Distance is the square root of, let's just write it as delta x squared plus delta y squared. In this case if I look up here that's going to be what? Delta x 1 minus minus 2 so that's 3 squared.

And 5 minus 1 is 4. So that's the square root of 9 plus 16. And that's equal to the square root of 25, which is equal to 5. So the distance is 5 units from the point P1 to the line L1 that we started with. All right? Okay, so finally, let's look at a problem where we can use the linear relationship between things to solve a problem.

All right, so the amount of heat in joules required to convert one gram of water into vapor is linearly related to temperature. That means that the relationship is a line, which means that we can use slope kinds of thinking to solve this problem. At 10 degrees centigrade, this conversion requires 2,480 joules. And each increase in temperature of 15 degrees Celsius lowers the amount of heat needed by 40 joules.

So we want to sort of find... The amount of heat needed to convert one gram of water into vapor when the temperature in the atmosphere is 35 degrees Celsius. Okay, so the first thing to do is let's write a formula or an expression that gives H in terms of T. So if the relationship between the amount of heat in joules to require to convert one gram of water into vapor is linear related to temperature, then that means that delta H over the change in temperature, that's dy and dx, right? equals, so that'll give me slope information.

Now I just need h minus some initial point over temperature minus some initial temperature. We're given some initial information there, so that means I'm going to have... Each increase in temperature of 15 degrees Celsius, that would be my delta T, lowers the amount of heat by 40 joules.

And that's going to give me a formula for heat. And now I know my initial starting point is... At 10 degrees Celsius, this conversion requires 2,480 joules, and the initial temperature there is 10 degrees Celsius. All right, so I need to do some simplification there.

This is going to give me what? Minus 8 thirds. Times t minus 10 equals h minus 2480. So divide through by 5, this is 8, and this is 3. So minus 8 thirds and then I just multiply by the numerator. I can now solve this for H because we want to express H completely in terms of T.

H equals minus 8 thirds T minus 10 plus 2480 joules. Okay? If you think about it, we knew the joules at this initial point when t is equal to 10. So when t is equal to 10, which I read, when t is equal to 10, Then H is 2480 joules because this will be zero. So that makes sense.

And then as we increase the temperature, the slope keeps walking down the total number of joules. We're going to be subtracting off the number of joules. So now we want to find...

H at 35 degrees and this is equal to minus 8 thirds T minus let's see 35 minus 10 degrees Celsius plus 2480 joules that's going to be 2480 35 minus 10 is 25 25 times 4 is 100 times 8 is 200 so that's minus 200 thirds and that's going to be what three times of 2480 is 7440 minus 200 over 3 so that gives me H is equal to 72 thousand forty over three Joules. Clean that up a little bit. 7240. Okay.

Alright, so that concludes this video.

Transcript for:Understanding Cartesian Coordinates and Linear Equations

Transcript for:
Understanding Cartesian Coordinates and Linear Equations