Midpoint, Magnitude and Gradient Coordinate Geometry

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[00:00:12]So, here today we're going to be looking at the midpoint, midpoint, the distance of a straight line, distance of a straight line, and the gradient.

[00:00:28]So, coordinate geometry we're just going to be looking at lines.

[00:00:33]How the line um how the line will be able to produce a straight line, the distance, and all that. But, for this video we're just going to look at the midpoint, distance, and the gradient.

[00:00:49]Then, in another video we're going to continue from there.

[00:00:53]So, what are some of the things that you need to know?

[00:00:56]So, let's talk about formulas.

[00:00:58]So, we have the midpoint. How do you know how do you find the midpoint? So, the midpoint, if let's say if I was to ask you the number between 1 and 3 is 2.

[00:01:11]How come? If you add 1 + 3 is 4.

[00:01:16]If you divide to it will give you the number in between. So, this is consistent. If you have 4, 5, and 6, the number between these is if you add 4 and 6, you divide by 2 it will be 4 + 6 is 10 / 2 = 5. So, this is the formula for the midpoint. So, the midpoint midpoint is equal to we're going to have x1 + x2 / 2, {comma} y1 + y2 / 2. Now, we're just introducing coordinate geometry in which we're going to be using the X and Y coordinates. So, uh that is how we find the midpoint. Let's look at the question.

[00:02:07]So, find the midpoint of these.

[00:02:10]So, how are you going to do it? So, if you look at the midpoint, you're going to have the midpoint should be equal to you're going to have your X1 plus your X2 over the Y1. Sorry, not Y1. Over the two {comma} your Y1 plus your Y2 over the two like that. So, if you have a line like that, you're looking for the midpoint. You're looking at the center. What are these coordinates?

[00:02:42]What is your X and Y for the midpoint?

[00:02:47]So, that is what we are looking at. So, you add the X1 and the X2. So, if you have the X2 Y2 here, if you have your your Y1, your your X1 {comma} Y1, you're going to add them. X and X, you divide by two, it will give you the midpoint like that.

[00:03:10]So, it's a very simple concept. So, we have the midpoint.

[00:03:17]So, the midpoint will be equal to we have what is our X1.

[00:03:22]So, our X1 in this case is we have coordinates such as let me write them here. It is three {comma} positive four.

[00:03:32]Then, we also have a seven -7 {comma} two like that.

[00:03:38]So, we are going to have our three X1 will be three. So, we are going to say three plus our X2 is -7. So, we have -7 over a two like that, comma, the Y1. Remember the formula is Y1 + Y2.

[00:03:57]So, Y1 is four. So, we are going to say four plus Y2, which is a two.

[00:04:06]Then, divided by two. So, that is the midpoint. So, the midpoint will be three. This by this, it will be -7 over two, comma, 4 + 2 will be equal to six.

[00:04:20]We divide it by two like that.

[00:04:24]Then, finally, we are going to get a value that is equivalent to 3 - 4 is -7, -4 divided by a two, comma, 3 divided by 6 divided by two is a three. So, finally, the midpoint for this one will be equal to -2, comma, positive three. If you divide, -4 / 2 is -2. So, -2, comma, three. So, that is one type. The second type where they can ask you the midpoint is this type.

[00:05:05]Let's say we have been told, if this is the midpoint of AB, where A is this, then we are required to find the coordinates of B. So, to do that, we know that the midpoint of AB.

[00:05:29]So, we have a straight line.

[00:05:31]Let me just draw it there. We have A.

[00:05:35]We have B.

[00:05:37]Then, they have provided us with the midpoint coordinates, which are -2, 1 like that.

[00:05:44]They want us to find that we also have the coordinates of A which are um -3,2.

[00:05:52]We need to find this.

[00:05:55]So, what are we going to do? So, we are just going to use the midpoint formula since they have provided us with that.

[00:06:01]So, we have the midpoint is equal to X1 + our X2 over 2, we have our Y1 + our Y2 over our 2 like that.

[00:06:14]So, our midpoint midpoint of AB should be equal to we have what is our X1? So, our X1 is um found to be X1 is equal to We have our X1 which is for A which is our 3.

[00:06:38]Is it positive or negative? Anyways, it's okay like this. So, we have 3 whatever you are doing here.

[00:06:46]So, positive 3.

[00:06:49]So, we are going to say it will be 3 + X2 we don't know it is at B so we don't know.

[00:06:57]Then over the 2, we have our 2 like that.

[00:07:00]For Y1, it is the A coordinate of Y which is our 2. So, we can say 2 + our Y2 over we have our 2 like that.

[00:07:14]So, with this done, the midpoint of AB we have been provided that it's -2,1. So, we are going to write like that. So, we have -2,1 should be equal to we have our we have written our 3 + X2 over 2, we have uh 2 + Y2 over 2.

[00:07:39]So, now if you or if you look at this, this is the X coordinate, so that is to be equal to another X coordinate. So, we are going to say -2 should be equal to because whatever you will get whatever you will get from here is the one that is producing this -2, so you equate them. So, two should be equal to 3 + X2 over two like that.

[00:08:07]From here, you are going to cross multiply. 2 * -2, it will be -4 should be equal to 3 + X2, which will be if you take this this side, it will be -4 - 3 should be equal to our X2 like that.

[00:08:25]Which will give us a -7 should be equal to our X2 like that.

[00:08:32]So, this is how you find the midpoint for that problem or the value of your B for X. Then, you are going to do the same. You are going to get this positive one, equate it to that.

[00:08:46]So, whatever you are you are you are going to get here is the one that is producing this one.

[00:08:53]So, you are going to say one one should be equal to you are going to say 2 + Y2 over the two. Which will be which will be you cross multiply.

[00:09:08]We are going to have 1 by 2, it will be 2 should be equal to 2 + the Y2.

[00:09:14]Then, when you do that, we are going to have this two goes this side, we are going to have 2 - 2, which will be equal to Y2.

[00:09:23]This time, Y2 is equal to zero like that. So, this will be our Y2. So, we can be able to say that our we have our X2 and Y2, so that means to say B has coordinates at -7, which is your X coordinate, {comma} 0 like that.

[00:09:43]So, that is how we find uh the answer under the midpoint. If we are required to find the distance now, distance between two points.

[00:10:01]So, the distance between two points, remember this is a line and it's not starting at zero. So, that means to say whatever X value is here, let's say for you to find the distance from 5 to 10.

[00:10:15]It is from 5 to 10. How do you find the distance from 5 to 10? If it was just a straight line like this, there it is not a hypotenuse, it would be very easy in the sense that if this is 5, this is 10.

[00:10:29]So, for you to move from here up to here, it means to say you just have to cover a distance of 5 m. How do you find that? By subtracting 10 minus 5. So, which will give you 10 minus 5, it will give you 5. So, it means to say for you to find the distance between two points that are not starting at zero, subtract like that. So, that means to say that's how we find the distance of a straight line. Now, the pressure with uh coordinate geometry is this line already is is uh an hypotenuse. You can be able to split it in the X components and the Y components like that. So, that means to say you need to subtract the X and the Y, then square them. So, therefore, the distance is given distance is equal to the square root of uh X two, because there are two of them. We have one X value at this point. So, we have uh X2 minus uh X1. Remember you have to subtract for you to find the distance between two points. Squared minus or plus. Now you can add the x and the y, but you have to subtract them like that. Y 2 - Y 1, then you square them.

[00:11:48]So that is how you find the distance.

[00:11:51]Let's say we have been told to say find the distance Find the distance between A and B.

[00:12:02]A has got 2 {comma} -3 and B has got -3 {comma} a positive 4. So how do you find this? We're going to say distance is equal to the square root of x 2 - x 1 squared plus our y 2 - y 1 squared like that. So that we have the square root of our x 2. X 2 is equal to -3. So this is our x 2 y 2 um y 1 x 1. So we're going to replace. So we have x 2 is -3 minus our x 1 is 2. So we square it.

[00:12:54]Then we have plus our y 2 is 4. So we're going to say 4 minus our y 1 is 3. So we're going to say -3 like that. Then we square them. We're going to have the square root of our negative 3 minus 2 is -5 squared. Then we have plus our 4 This by this we're going to have plus our 3 like that. Then we are going to have the square like that. So this is going to be the square root of -5 squared if it's in brackets it will be positive 25 then we have plus 4 plus the 3 that will be a 7 so we have 7 we square it this is going to give us the square root of 25 plus 7 squared is 49 so 49 plus 25 will give us a value of the square root of 25 plus 49 it will be equal to 74 so this is how we find the the length like that.

[00:14:05]Uh let's move on to the gradient.

[00:14:08]So how do you find the gradient? So let's say find the gradient.

[00:14:15]So the gradient already is if you have a line it is the change in Y how far it is moving in the Y direction.

[00:14:25]So this is the Y this is the X so in in other words the gradient is the simpler way of uh of finding the gradient is this the change in Y over the change in X so how it is moving up in terms of Y how it is moving in this direction in terms of X so the gradient gradient should be equal to what is our Y coordinate? What is our X coordinate?

[00:14:53]So since it's a change you can only find the change by subtracting so you're going to say Y2 minus Y1 over X2 minus X1 so that is the gradient. So let's find the gradient between let's say we have been told between -2,7 and our 4,5 5 like that.

[00:15:21]So to find the gradient which is our M, should be equal to we are going to say Y2 - Y1 over X2 - X1. So, our This is our X1, Y1, X2, Y2. So, we have our Y2 is equal to 5 minus our Y1 is equal to 7 over We have our X2 is 4 minus X1 is equal to open brackets negative two like that.

[00:15:59]So, this is going to give us 5 minus 7. We are going to have a negative two like that over We have four. Negative by negative is positive.

[00:16:11]So, positive two like that.

[00:16:13]So, we have negative two over six.

[00:16:16]Dividing these, we are going to have negative one over and a three like that.

[00:16:21]So, this will be our gradient in that form. So, we have the gradient like that.

[00:16:29]So, there are two ways of finding the gradient. If the angle is known, if the angle is known, like for example, if you are looking at uh if you are looking at the equation Y is equal to X, which is an equation that looks like this.

[00:16:53]So, this equation cuts this line.

[00:16:57]Just passes like that. So, this is the equation of Y should be equal to X. So, this line cuts uh the 45°.

[00:17:13]It also here 45° like that.

[00:17:17]So, if want to find the gradient of this line, you're just going to say the gradient if the angle is known, it is equal to the tan of theta. Why tan? Because already if you recall, tan theta is equal to opposite over adjacent. But, what is the opposite in terms of in terms of this? The opposite side is equal to the X coordinate.

[00:17:47]Then, the one that is uh not the X, the opposite is the Y in this case since Y is going up. So, the opposite is Y, the down is X. But, this is the gradient Y2 - Y1 over X2 - X1.

[00:18:06]This means to say you're looking at tan already because tan is equal to opposite, which is tan of theta is equal to the change in Y over the change in X. So, that's how come we have uh the gradient to be equal to tan of theta. Now, for this for this line, it is 45. So, we can find the gradient to be 45, which will be the value, which will be tan of 45° So, the tan of 45° This is a special angle. So, we have the tan of 45° is just equal to one. So, that's the gradient. You can verify it by taking some coordinates like for example here.

[00:18:51]This one is 0,0.

[00:18:54]We can have another coordinate. Y is equal to X. We'll have if you draw a table of values, it will have if X is equal to one, Y will be equal to one. If X is equal to two, Y will be equal to two. Let's uh take Let's take this. Let's say we have we take any of the two we have 1 1 and a 2 2. You can find that the gradient which will be equal to y2 - y1 over the x2 - x1. We are going to have what is our y2? Our y2 will be 2 the y1 which will be 1 over x2 which is this 2 the x1 which is 1. So for any two coordinates you take like that it will be 1 over 1 which will be your gradient.

[00:19:49]So your gradient will be equal to 1. So that is how you deal with that.

[00:19:55]Thank you very much for watching and see you in the next tutorial. Don't forget to subscribe to avoid missing out these powerful contents.