CHAMPION MATHS is live

Añadido: 2026-06-02

[00:02:37]All right. All right. All right. Uh my people, let's go straight to KZN. Let's go straight to KZN.

[00:02:47]Uh this question paper was written in in KZN actually by learners KZN common test June 2024 grade 11 let's try to um answer all these questions actually uh the focus here it will be gray A no no no no no the focus here it will be um analytical geometry right after doing this analytical geometry I will move to to free state analytical geometry limpopo analytical geometry as well as analytical geometry so this is what I will be doing today guys so please stay tuned um let Let me see. Yeah. Okay.

[00:03:51]Let's go straight to analytical geometry. Guys, people, please, please, please, please, please uh make sure that you you write the correct uh maths.

[00:04:04]Write the correct maths and you get a distinction, you move to the next level just like that. So I'll be starting with question number one and that's saying in the diagram below uh the coordinate of D E F and G are the vertices of a quadrilateral. Right? F is a point on the X-axis.

[00:04:31]The diagonals of the quadrilateral biseect each other at K. The equation of diagonal eg is 3x - y + 6 = 0.

[00:04:46]So we are given this um equation right we're given this equation. So please do not forget and b and beta is the angle of ination of eg.

[00:05:01]So let's try to determine guys. I hope everything is clear on on your sides.

[00:05:08]I hope everything is clear and then please comment if ever you can see what I'm talking about because I don't want to I don't want to confuse people and they might think that no man this guy now is is giving us our time can't be able to see the question paper but I believe the question paper it's it's clear now so let me answer all these questions so in this diagram we are given the coordinate of D and the coordinate of G only right and please people do not be afraid of anything just apply what we taught you right so um right now they're saying calculate the size of angle B beta uh the size of angle beta, right?

[00:06:10]Uh let me take out this uh okay they want us to calculate the size of angle beta in this case they want this size this size here they're looking for this size right so what to determine this size how do we determine this size we know that um the equation of EG is given right so we are going to use the equation of EG so in this case we're going to use uh 3x 3x - y + 6 = 0. Okay, we are using this equation is the one that we are given for what? For eg. So in order for us to determine the angle beta, you need to know the gradient of this line. Right?

[00:07:01]So now the gradient we are to solve this to let we are to make y the subject. So y it is equ= to -3x -3x - 6 and then you multiply both sides by negative so it remain with y it is equ= to 3x + what + 6. Therefore this is the equation now this is the equation of eg right by making y the subject of the formula and then from this part now um you can do what now you can write the gradient of the line eg. So the gradient of the line eg the gradient of the line eg it is equals to it is equals to three. Now you can determine the angle of beta. Right? The angle of beta in this case it's tan beta which is to three. Okay. Now you can use your calculator to determine angle beta. Right. To determine angle beta. Now you are going to use your calculator. You say shift turn.

[00:08:18]Shift turn. Let's shift turn.

[00:08:23]turn three and then we're getting 71 comma 57. So this 71 comma 5 what? 7 what degrees. So now we know this angle right we have already determine this angle of beta. Yeah. So let's move to the next question. The next question they are saying calculate the coordinate of f if the equation of df is x + 3 y which is equals to 8. We are to determine the coordinate of of f the coordinate of f here. So to determine the coordinate of f we using the equation that we are given now which is x + 3 y = 8. Right? It's the equation of what? is the equation of df. So you will realize that f is the what? f is the x intercept. So from this equation that we given which is x x + 3 y it= to 8. We are going to let y = 0. Okay. So our y here it will be equals to z. So this will be three open bracket open bracket uh 0 which is equals to 8 right and then from here you solve for x. So x it is equals to 8.

[00:09:49]So the coordinate of f here it's the coordinate of f is what? It's 8 8 and zero. Okay let's move to uh 1.3. Calculate the coordinates the coordinates of E. To determine the coordinates of E more this X and Y basically X and Y, right? To determine the coordinates of E. Uh, it's easier. It's easier. We can use uh the the midpoint, right? We know that the midpoint of DF and the midpoint of EG are equal. So in this case you're going to use M. The midpoint the midpoint of what? Of DF.

[00:10:46]The midpoint of DF we know that it is close to the midpoint of E what? E G or GF. G I mean let me use G.

[00:10:58]Okay G E. So in this case the midpoint of DF this will be what? This will be uh -10 + 8 / 2 and 6 + 0 / 2. Okay. So these are your x1 and x2 y1 and y2 right equals to the midpoint of g. The midpoint of g is what? The midpoint of g is 1 + x / 2 is to one is to 9 + y / two in that order. Right? So now what do we do?

[00:11:52]We solve for x and y simultaneously. So or let me not say simultaneous but we need to solve x and y. Sorry for that.

[00:11:59]So it's - 10 + 8 - 10 + 8 it's -2 which is -1. So yeah it will be -1 and 6 + 0 is 6 / 2 is 3. Okay. Equals c = what? 1 + x / 2 and 9 + y / 2. So right now you can just equate the x's and the y's. So now you're going to say 1 + 1 + x / 2 it is = to -1. And then you solve for x. and 9 + y / 2 it is = to 3.

[00:12:49]So you solve for for x and y. So this will be 1 + x = -2. You cross multiply again and then x will be = -2 -1 which is = to -3.

[00:13:06]So x is -3. And then what about y? Our y it will be what? It will be 9 + y it = to 6 2 * 3 it's 6. Okay. And then from this part now you can just solve for y.

[00:13:22]y = 6 - 9 which is = to -3. Therefore, the coordinates the coordinates of E is -3 and -3 just like that. So, you do not have to uh stress yourself. Okay, this what you need to do. But sometimes if if if you are told that it's a quadrilateral to determine E, it's easier. You can just use change in X and change in Y. Okay, you can try to use this. Let me show you how we do it.

[00:14:04]Here you will name this your X1, your X2, X3, and X4. Okay, so this is what you need to do if you want to determine the coordinates of E. To determine the coordinates of E, you're going to say instead of instead of determining the midpoint, you can say X1 - X2, it is equals to X3 - X4. Okay?

[00:14:36]Right? And then now we're going to say x1 is - 10 - x because x2 is x here is the one that we want to calculate, right? And x3 is 1 - x4 which is 8. So in this case you're going to have -10 - x it is 1 - 8 it's - 7.

[00:15:05]Right? And then from this part you solve for x. So - 7 + 10 you are getting what?

[00:15:12]You are getting ne you are getting positive3.

[00:15:17]So you multiply by negative both sides and then you remain with x which is equals to -3. Easier. Do you see that?

[00:15:24]Very much interesting. To determine y very same thing applies. Y1 Y2 Y3 Y4 Okay. So now we are going to say Y1 - Y2 it is equals to Y3 - Y4. Our y1 it's it's 6 y 2 it's - y = = 1 = y3 y3 is 9 - y4 which is 0. Now you can determine y so - y = 9 - 6. Okay, there is there was no need for us to include this 0 because 0 means nothing. So - y = to 9 - 6 is 3 and then you multiply by negative both sides you remain with y it= to -3. Therefore the coordinates of e it's -3 and3.

[00:16:32]So that's how you can determine the coordinates of e. Okay, the vertex E because uh it's what we are we are given here and told that this is a quadrilateral.

[00:16:48]So if ever it's a quadrilateral, you can use uh this method change in x and change in y, right? Or if ever you do not want to do this, you can use uh the method that we have been taught by your teachers to say that if ever you given a quadrilateral the the diagonal the midpoint of diagonal G and the midpoint of diagonal DF are equal. So you can firstly determine the the the midpoint of the diagonal DF and then after that you equate to the to the midpoint of diagonal what g. So it's very easy.

[00:17:30]Let's move to the next question.

[00:17:32]The next question they are saying here we need to prove guys we need to prove that D E FG is a rhombus.

[00:17:44]You can prove using two things uh two methods here.

[00:17:49]The first meth uh method it will be uh since uh we were given the equation of uh this equation of df.

[00:18:02]The equation of df. Do you still remember the equation of df? And then we're also given um the gradient of GE.

[00:18:15]The gradient of GE.

[00:18:18]If we can prove that the gradient of DF multiply by the gradient of G E the answer is -1. Therefore, we have proven that this angle right here is what? It's 90°. So, basically, we need to prove that uh DF is perpendicular to what to G because these diagonals bisect at 90°.

[00:18:51]So, let's prove that they bisect at 90°.

[00:18:55]So this what you need to do guys you are going to say since you are given this equation here 3 y = 8 - x and then you divide both sides by 3 remain with y = - uh x / 3 + 8 / 3 okay so I have rearranged this it means you can write as -1 / 3x + 8 / 3. Okay.

[00:19:42]So now now we know the gradient of what? We know the gradient of df, right? The gradient of DF it's negative. The gradient of DF the gradient of DF it is equals to -1 / 3. But the gradient of G, the gradient of G, we have already the grad of G E we know that when from the equation that we're given right so the gradient of G E it is equals to three. So let's try to multiply these two. When we multiply them and then we get -1, it means that DF is perpendicular to G. So let's calculate M DF mult* by M. Let's see. MDF is -1 / 3 * 3 and the answer in this case we're getting -1. Therefore, we conclude by saying we conclude by saying our DF is perpendicular to what 2 G E done.

[00:21:06]We have proven that the quadrilateral it's a rhombus. Thank you very much. But sometimes uh you you can also say that um um d uh dg the distance of dg and then de you can calculate the distance dg and then you also calculate the distance d e if these distance are equal. Therefore, therefore D E FG is a rhombus.

[00:21:51]The very same thing applies if you can calculate the distance from G to F and F to E and then you find that they are equal. Therefore, D G F E is a rhombus. Thank you very much.

[00:22:09]Let's move to the next question.

[00:22:13]Question number two.

[00:22:15]We are moving to question number two guys.

[00:22:20]We are moving to question number two.

[00:22:22]And with this question number two in the diagram below you are given the coordinates of A, coordinates of B and C and then told that are the vertices of a triangle.

[00:22:38]BC is produced to D.

[00:22:42]AC car the Yaxis at F.

[00:22:45]E is a point on the X-axis such that CE is perpendicular to X-axis.

[00:22:55]So in this case guys the diagram or or the the the sketch it has been drawn for you which correspond with the statement. So now we are to determine number one.

[00:23:14]Calculate the length of AC.

[00:23:18]Leave your answer in simplified said form.

[00:23:24]So in this case the distance of of a ac the distance of AC it's a equal to the square root of the square root of 1 the square root of 1 + 2 squared. Okay.

[00:23:51]+ 5 + 1 squared. So I'm using positives here because uh -2 it was my x2.

[00:24:07]I started by saying this will be my x1.

[00:24:11]Even from the distance formula I said one is the x1 -2 is the x2. So when I when I um replace our x1 and x2 I just realize that the x coordinate here is negative. So I multiply with the negative from the formula. That's why I wrote plus here 1 + 2 2 5 + 1 2. Okay.

[00:24:42]And then from this part use your calculator.

[00:24:45]Let me use the calculator. The square root of the square root of um open bracket 1 + 2 close 2 + open bracket 5 + 1 close bracket what squared and then we're getting 3 square<unk> of 5. So the answer here is 3 square<unk> of what? Of 5.

[00:25:19]So that's the distance of AC. We are done with number one. And then you get your two marks for doing that. We are done with this one. Done.

[00:25:30]Okay. Let's move to 2.2.

[00:25:34]2.2 saying calculate the gradient of BC.

[00:25:38]The gradient of BC B to C. This was 2.1.

[00:25:44]our 2.2 2 BC. The gradient of BC the grad of BC it is equals to our Y 2 is 4 Y1 is 5 - 5.

[00:26:05]Our X2 is - 3.

[00:26:09]Our X1 is is one which is minus1. So it doesn't matter the the position of your your x and y. Okay, you can start by saying this was my x2 and this was my x1 doesn't matter. So now the gradient of BC if we use your calculator you know you are getting one 1 / 4 -1 / 4.

[00:26:44]So let me confirm here for no we're getting 1 / 4. So this will be the gradient. Okay. So after determining this gradient now you you are to move to number number 2.3 done with this one we are done with this one we have calculated this and then you get your two marks as well 2.3 calculate the value of y if b c and d are colinear what point b c d are collinear points and then we know that if we we have a colinear if we have colinear points it means that the gradient of BC it is called the gradient of CD equals the gradient of what of BD so in this case allow me to use BC and CD so you're going to say the gradient since we're told that they are colinear points The grad of BC. It is called the gradal of what? Of CD. So let me just use uh D C or B D is fine. I can use B D or C D.

[00:28:04]Anyhow it's fine. Right. So let me just use CD. Yeah. The gradient of CD. Right.

[00:28:11]So BC we know the the gradient of BC it's 1 / 4 = 2. Now we are to substitute. I will start by using y followed by 5. It will be y - 5 / 5 - 1.

[00:28:33]So this will be 1 / 4. It is y - 5 / 4. Right? So since the denominators are the same, we just forget about them and then we equate the numerators, right? Because if we multiply by four both sides, if we multiply by four both sides, we are getting rid of these fours, right?

[00:29:03]And then remain with 1 = y - 5.

[00:29:06]Therefore, y will be equals to what? y will be equals to um 6. y will be equals to 6. Right? So you get your three marks for doing that and one mark for for the gradient of of BC and the gradient of of CD.

[00:29:32]And then you get another mark for equating the gradients.

[00:29:38]One mark for the answer. And then we're done with this question. Again, let's move to 2.4.

[00:29:49]2.4 They're saying if h is a point such that a h is perpendicular to bc detine the equation of a hiccular line.

[00:30:06]I know that perpendicular lines they form what they form a 90° angle. I'm looking for a ruler here so that uh I can just construct the line of of ah so that you guys uh can see what I'm talking about. Unfortunately, I can't find the ruler here.

[00:30:29]Okay.

[00:30:38]>> Okay. No, it's fine. Even if there's no ruler, it's fine. Even if there there is no ruler, it's fine. I will just use a pen. Okay. What I'm trying to say is that they saying we need to determine the equation of of a h. So if a if h lies lie here on the straight line what b c d since they're saying the ah is perpendicular to to bc so it means we're going to have something like this perpendicular line right there right this line is perpendicular to bc right so therefore here it will be H. Okay, that will it will be what? It will be H.

[00:31:28]So, we know that this is 90 what? 90 uh°.

[00:31:33]It's 90° this part here. So, since we know that it's 90°, it means now I want to determine the equation of a h.

[00:31:43]So to determine the equation of a h we know that a h a h a to h the gradient of a h it's what it's let me just write the 2 point um 24 the gradient of a H since we know that the gradient of BC it's 1 / 4 it means the gradient of a H of A H will be -4 because we are told that AH is per perpendicular to BC. So if we multiply the two gradients we're getting -1.

[00:32:33]So in this case now we're saying the equation the equation will be in this form. You are going to use um let me just calculate the value of c y = -4x + c. So we're going to use the coordinates of a here because uh it's what we have through this line a h a it's -2 and 1. So y is -1 = to -4 - 2 + c and c will be equals to -1 - 8. Okay. So this will be minus um we're getting - 9. So therefore the equation in this case of 2.4 it will be y = -4x - 9. So this is the equation I'm talking about.

[00:33:38]This is the equation I'm talking about.

[00:33:40]Right? We have already determined the equation and then we are done. You get your how many marks? You get your three marks. Three marks for that. One mark for the gradient of ah, another mark for substitution and another mark for calculating one.

[00:34:00]Right? So the equation is that for 2.5. Let's move to 2.5. Calculate the coordinates of G if C B A G in that order is a parallelogram.

[00:34:18]Okay. So they're saying C is this one, B is this one, A is this one, and then G. If ever they're saying it's a parallelogram, it means we're going to have a line.

[00:34:38]We are going to have this line here.

[00:34:42]You're going to have a line like this.

[00:34:45]So this will be G.

[00:34:48]This will be G.

[00:34:52]Right? So if this is G, it means now we have what? We have our parallelo what?

[00:34:59]We have our parallelogram.

[00:35:02]This will be a parallel a parallelogram.

[00:35:05]Okay. So we need to do what? We need to calculate the coordinate of G. The coordinates of G which is X and Y. But with this G, we know that Y is zero.

[00:35:19]Okay, we know that Y is zero. So if you know that Y is zero, it means you have to determine your X.

[00:35:30]You need to determine your what your X.

[00:35:33]So in this case I have already taught you how to determine. So this what you can do.

[00:35:40]You are going to follow this order.

[00:35:42]Right?

[00:35:43]This what we can do. Let me use this paper here.

[00:35:48]So you are going to say x1 - x2 it is equals to x4 x1 in this case will be - 3 - 3 - - 2 it is equ= to 1 - x. Therefore now you can determine -3 - 2 it's -1. Okay - 1 -1 = -x.

[00:36:28]So x will be equ= to 2.

[00:36:33]So the answer here is 2.

[00:36:36]And then we move to y.

[00:36:40]But there was no need. But we can try to to show that y is 0.

[00:36:46]y1 - y2 = y3 - y4.

[00:36:52]y1 in this case is 4 - -1 = 5 5 - y. So this uh we are getting 5 it= to 5 - y. So 5 - 5 it is = to - y.

[00:37:12]Therefore y it is= to z. So that's how you can determine the coordinates of g.

[00:37:19]All right we are done with this one but done with this question and then we are moving to 2.6.

[00:37:30]2.6 six. That's where you need to um determine the we need to determine the area the area of C E O F need the area in this case this is basically 2.6 2.6 six. Okay.

[00:38:00]Um, they're saying C E O F this part.

[00:38:10]Let me use a different color.

[00:38:14]We're looking for the area of this C E O F. The area of this part.

[00:38:23]Okay. So if we are looking for the area of this part now we we can uh draw a perpendicular line from E F we draw a perpendicular line right and we know that this will be 90° okay so if we know that this will be 90° and then we know the coordinates of f the line AC the equation of the line AC.

[00:38:59]So f in this case we know that um it will be zero 0 and uh and three 0 and three and we know that uh let me just produce any light here. It can be uh let me use I. Okay, it will be I in this case.

[00:39:32]So with the I we know the coordinates here.

[00:39:37]Yeah here it will be one.

[00:39:40]Even the coordinates of E it will be one and one end.

[00:39:48]So you you you know that uh from from uh 0 to this point is three from O to E.

[00:40:07]O to E.

[00:40:09]It's one and what? It's one and and zero. Okay. Yeah, it's one and zero. So now to determine um to determine the the area determine the area you know that you are going to use uh area of C E O F it is also have base half base times the height. Right? So in this case you're going to use our our base here our base in this case.

[00:41:05]It will be EC our EC multiply by the height.

[00:41:13]The height in this case is what is O E.

[00:41:18]Okay. Or let me say O E O F because the distance from O to E is the same distance from O to F. Okay.

[00:41:33]Same applies to O to I. So you can use o e >> multiply by >> multiply by what?

[00:41:50]>> No disconnect.

[00:41:53]Disconnect. Network is a problem.

[00:42:00]Network is a problem. I'm really sorry for that. Just use your calculator. find this and then you write the answer. That will be the the answer for for the area of C E O F.

[00:42:14]Okay, this one even though we did not include this one, it's fine. Can just say half base times the height. Okay, our height is OE. O2 is the