Coordinate Geometry | Past Papers | IGCSE/O-Level Mathematics | Class 24

Generated automatically from the lecture captions, so spoken maths may be transcribed imperfectly.

Okay. So we are going to continue with the topic of coordinate geometry. The things we did in the last class. I want to do some past questions on that and then we will move on to the new thing for today's class. So here is one of the passovers and look the things that we did exactly the same thing. So a line L passes through the point the two point the two points are given to you they're asking you to for to find the equation of the line. So we know the equation of the line is y = to m x + c. So you need to find the value of m. So m will be x1 x2 - x1 and y2 - y1. So, x1 y1 x2 y2. So, y2 - y1 would be 9 - - 3 upon 6 - 0 which

would get you uh 12 upon 6 which will be 2. So now you have the gradient. So you can replace the value of m. So y is equ= to 2x + c. And now to find the value of c you can use any one of the points. So I'm using this one. So y is - 3 and x is 0. So you're getting your value of c as minus c sorry - 3. So this means that your complete equation is 2x - 3 and they are giving you three marks for this thing. Three marks. So you can see how straightforward the questions are. Exactly the same thing that we did. And then for the second part find the equation of the line that is perpendicular to the line L. So now

there is another line which is perpendicular to the line L and passes through the point to 0A 2. So line L gradient we already know it is two. And there's another line for which I'm I'm calling it the second line 2 M2. This would be because it is perpendicular. So this would be negative reciprocal. The thing we did in the last class negative reciprocal gradient would be negative reciprocal. So 2 would become -1 upon 2 and so this will become y is equ= to - 1 upon 2x because your m for the second line is - 1 upon 2 plus c and for c you have the point they have given you the point. So y is 2 and x is 0.

So your c will come out as 2. So this means that your equation of line has become -/ x + 2. This is it. So easy. Five marks total. Okay. You can see it's in front of you. Easy. Easy. Very easy. I think if there's any questions, anyone ask ask >> sir -2 positive. It would be which one? Negative 1 upon two. >> G sir. No sir. Answer two. >> Uh sorry. Sorry. My bad. Plus two. Mistake. >> Good. Good. Okay. This is it. Now let's find let's see if we have something interesting from P2. This is from P4. Sorry. uh in P4 the versions are you can say bit longer. Oh this looks like a good one. Let's do this complete question and then we'll

move on to the new thing and this would include everything that we have done today. Oh, and the last class this is of March 23. So, a very recent past 2023 only I think two years ago. So, they have given you look the point A. A is 0A 4. Let me label it down. And B is the point 8 comma 0. And they're telling you line L1 is parallel to the x-axis. It is horizontal. You can see it is parallel to the x-axis. And L2 L2 is the line passing through A and B. We can see it as well. Write down the equation of L1. So for L1, it is a horizontal line. And I told you what what happens for the horizontal line. The y the equation is

simply y is equals to four. The y intercept. Simple because you can see at any point on this line the value the coordinate of y will always be four. At any point on this line the ycoordinate would always be four. So y intercept you can also see that the y intercept is four and the gradient is zero because it is horizontal line. So gradient is zero. Horizontal line is a greater than zero and y intercept you can see it is four. So 0 x + 4. So this becomes four. Simple. Then for line two they're asking you to get the uh sorry this equation of line two. So we would need the gradient and for the gradient you would need x1 y1

x2 y2. You can choose any of them. So your gradient for the line 2 will come out as y2 - y1. So 4 - 0 upon x2 - x1 0 - 8 which will be 4 upon - 8 which will be - 1 upon 2 after simplification. So the your equation would become y = - 1 upon 2x y= to mx plus. So your m is this m value is 1 upon 2 and you can see that the gradient sorry the y intercept for the other nine it is also four. You can see four. So your c value is four. So your equation becomes y = to - 1 upon 3 sorry 1 upon 2x + 4. This is it >> sir. Simple. >> Yes. >> Line one gradient similar. >> It is a horizontal line. I told you a

horizontal line is a gradient zero. >> Okay sir. >> Because horizontal. Let me write now because it is horizontal. Now they have given you another point C 2a 3. The line C the line three passes through C and is perpendicular to L1. So we know the gradient of L sorry L2 the gradient of L2 was neative 1 upon2. We found it in the previous part. Gradient of L2 was 1 upon2. And then they're telling us that gradient that there is line three which is perpendicular perpendicular perpendicular to L2. And I told you what's the rule that when it is perpendicular the gradients are negative reciprocal. So do the negative

reciprocal. So negative negative becomes positive and >> yes positive two. So now we have the gradient of line two. Line three. Line three. Gradient of line three. And we have the point through which the line three is passing. So you would use y is equals to 2x + c. So y is 3, x is 2 + c. So 3 = 4 + c and 4, c is equ= to 3 - 4 which will be minus 1. So the your equation becomes y = to mx + c where your m is 2 and your c is minus 1. the exact same thing that they are asking us to show. Look, they're asking us to show that the equation of line 3 is y= 2x - 1. So easy three marks. Exactly the same thing that

we did I gave you the homework in the last class. We have they have just changed the values for the gradient the point through which the line is passing through. Okay. Then we do that not they're telling us that the line three. So line three we know the equation of line 3 is 2x - 1. It is crossing the x-axis. Crossing the x ais means the x intercept. And at the x intercept we know that the value of y is zero. So it is crossing the x-axis at point d. So we know the point d the value of y would be zero. We want to find the value of x at that point. So but your y is 0. 2x - 1. So 1 is equ= to 2x and so your value is

coming out as 1 upon 2. This means the the point d is half comma 0. You can see it is l3 read the question again. L3 the line three which is this line. It is crossing the x-axis at d. So wherever it is crossing the x-axis which means that your y would be zero. When a line or any curve crosses the x-axis the y is zero. So you put in the value of y as zero and then you found the value of x and you got that when your y is zero your x is coming out as half. So now you have the point d and then they're asking you to find the length of the d. So you have the C, you have the D and you know the distance formula for C

for between any two points. You have done this which is a square root X2 - X1 whole square plus Y2 - Y1 square. So you can call any one of them x1 y1 x2 y2 and then just input in the things over here. So x2 is half - x1 is 2 square plus y2 is 0 - y1 is 3 whole square and then this is p4 your calculator is allowed so use your calculator you have used your brain now use your calculator the square root half - 2² + 0 - 3 square which is coming out 3.35 four which would be 3.35 correct to three significant figures this is your answer so five marks I don't know why easy adding it wasn't very difficult

what do you say any questions anyone explain should I explain it again >> yes >> line three for which the equation we this >> part >> part B this part >> and this they're asking you to find the gradient the equation of L2 so you have the two points of L2 L2 is passing through 0a 4 and 8 0 so you found the gradient of L2 with these two points this point and this point you found the gradient comes out at minus 1 upon 2 so you put in the value of the gradient is 1 upon 2 and you can see the y intercept Y intercept is the point where your graph or the curve is crossing the y- axis. It is crossing it at four. So no

need to for the calculation. You can see it over here. It is crossing the y-axis at four. C is where the the graph or the graph crosses the y- axis. And you can see it is crossing at four. So no need for the calculation. So you replace the value of c is four and the gradient you found is negative half. And then you replace them both. You got the answer. >> Okay. >> Okay. Fine. Understood? Right. >> Yes sir. >> Now all other questions would be very very very very 90% of the questions would be similar to the things that we have done of the past in today's class. So we will do more past in the next

class as well. But right now I want to finish this topic which includes finding the equation of perpendicular bis sector between two points. So there is very there are simple steps that you need to follow. You'll get the answer. Ah here it is. Look. Oh no. Wait wait wait wait. We won't I won't take this one. Yes this one is. So look five marks. And I think these are the easiest five marks you can you just have a simple procedure to follow. Understand the situation what they're asking you and then you'll get the answer. This is the question that we would do later on. I'm keeping it over here. But

right now understand what are they asking you. So they will give you two points. So let's suppose you have this A point and you have this B point. These are given the only thing that they would give you are these two points. So whatever is in the red the red this is the given they have given you these two red part red A and B coordinates and you have to find the perpendicular basis. You know perpendicular passctor is definitely a line which is bicting it and bicting means it is dividing the distance between these two points equally. So it would be exactly in the middle passing through the middle

and perpendicularly perpendicular means that it would be crossing like this at 90° from the middle. This is your perpendicular bis sector of AB. So now you know what you are trying to find. They will give you the two points and you have to find a line which is passing through the exact middle of the two points perpendicularly. So you will find the thing which is in the blue and the only thing that they will give you are the two points. Now how do you how do you find this? So you would need a point on this line. You would need the gradient. So you would need the gradient of perpendicular bis sector

and you would need the uh the point through which it is passing through which it is passing and we can see it is passing through the midpoint. point AB. It is passing through the midpoint of AB which is this point. And how will you get the gradient of the perpendicular bisector? You know the formula for the midpoint. It's an easy one. You add the X, you divide by two. You add the Y, you divide by two. You will get the midpoint coordinate. But how would you find the gradient of the perpendicular b as sector? You will get that by doing the negative reciprocal of the gradient of the two points that they

have given you. So you find the gradient of AB and we can see that the perpendicular bis sector is perpendicular to the AB line and so if you do the negative reciprocal of the gradient of AB you will get the gradient of the perpendicular bis sector. So you understand the situation this is what you are going to find now everything is in front of you. Now you can start from anywhere that you want. You either find the gradient first or you find the midpoint first. It's up to you. But you need a point. You need the gradient. And if you get those, you can put in the formula and then you'll get the equation. So here is an example.

So they have given you these two points and they're asking you to find the perpendicular capacitor. So let's suppose I'm changing the names over here. This is f let's suppose this is f they have given you 3a 2 and this is g which is - 5a 4. So I think the easiest part first of all is to get the midpoint midpoint of FG which will be add X iside by 2. So 3 - 5 upon 2 and 2 + 4 upon 2 which will be -2 upon 2 is - 1 and 6 upon 2 is 3. So now we already have the midpoint of FG which is - 1a 3. Visualize it. Okay. Whatever the information that they have given you, whatever the things that you find out, visualize it. Have that in

your mind. The diagram what is happening actually. And then you would need the gradient. And for the gradient of the perpendicular bisector, first I will find the gradient of FG which will be Y2 - Y1 X2 - X1. So, x1 y1 x2 y2 So, y2 - y1 upon x2 - x1 which will be 2 upon -2 - 8 which will be - 1 upon 4. And now for the gradient of the perpendicular bis sector it will be negative reciprocal. So negative negative becomes plus and one upon four flip flip it. It will become four. So now you have the gradient. Now you have the point through which the perpendicular bis sector is passing. So what do you do now? y= to mx

+ c. y = to m is your 4x 4 and you have 4x + c. And now you have the point. So y is 3, x is - 1 + c. So 3 - 4 + c. So 3 + 4 is equ= to c and so your c is 7. So finally your equation is y = to 4x + 7. This is your answer. That's it. And they are giving you five marks for this. Five marks. Think about it. Less than five steps. So it's a very much what do you can say score gaining part that's it this half page of working of understanding can get you five marks out of the total CI exam. So work on this part very very carefully very do the practice properly and you will get so five marks easy five marks

if if there's any questions ask ask right now this is your time and connect everything whatever that you're trying to find you found the midpoint of FG you found the gradient of FG build the steps in your mind first I'm doing this okay then I will get this and then I will add that and then I will get this. Not just blindly follow the steps. Okay, understand what they are asking you to find and then understand what would you need to get the answer and get those things separately and then combine them everything you will get the final answer. It's giving you six marks for the same thing. adding.

So you can see the first uh they have given you a point a point b they're ask they're telling you m is the midpoint find the m they're asking you to get the midpoint so simple you would add the x 1 + 3 upon 2 you would add the y is 5 + 9 upon 2. So 4 upon 2 is 2 and uh 14 upon 2 is 7. Then find the equation of the line that is perpendicular to AB and passes through the point M. They are actually talking about the perpendicular bis sector. So they're asking to find a line which is perpendicular to AB and passes through a midpoint of AB. M is the midpoint of AB. So it is actually the perpendicular bisector. So for this you

would need the gradient of AB X1 Y1 X2 Y2 use your uh brain y2 - y1 9 - 5 upon 3 - 1 4 upon 2 to 2 and then the gradient of the perpendicular would be negative reciprocal -1 upon And then y = to -/2x + c and it is passing through m which is 2a 7. So y is 7 x is 2 minus sorry 2 and 2 cancel out. So 7 + 1 which is 8. So y is equ= to - 1 upon 2x + 8. This is your answer. So six marks. Easy six marks. You have any questions? Ask part A is exactly the same we did in the last class. They have given you one line. They're asking to get the equation of the other line which is perpendicular to the line that they have given you and

pass it through 1, 3. So you do the negative reciprocal of the gradient. The gradient is half negative reciprocal it. So you will get -2 and y is equ= to -2x + c and then you would put in the x as 1, y as 3, find a c and that is your equation. So easy easy easy easy easy I think easy. Okay you people can do the practice properly. If you have any questions ask I don't see any difficult questions. It's finished. Okay, this looks good. A little bit different type of question from corner geometry. Perpendicular B sector is done. Just people have to do the practice. In this version they have given you that

the gradient a line has gradient five. So the gradient is five. M and n are two points on this line. M is the point and N is the point. They're asking you to find an expression of X in terms of K. So you have to make X the subject. First of all, they have given the gradient and they given you two points. So you can use the gradient formula X1 Y1 X2 Y2. So gradient is 5. 5 equ= to y2 minus y1 which is x upon uh x2 which is k sorry sorry sorry sorry y1 sorry y2 is 23 y1 is 8 8 and your x2 is k x1 is 8 x so this would go with your multiply so 5 k - >> sorry k - x is equals to 15 and this five we can go down and divide

15 / 3 >> kide >> yeah so your k is your x is coming out is >> kus that's it >> algebra nothing sus value. >> I move the x - x on the other side and three over here. >> Okay, sir. >> - x becomes plus x and 3 becomes minus 3. Easy, easy, easy.