Coordinate Geometry Essentials for Revision

Hello students, in this week's video, I'm going to provide you with a complete and effective revision guide to coordinate geometry. So please stay till the end. Right, let's look at some of the concepts that are needed in this chapter. Right, first of all, we all know that to find gradients of a line segment, basically you can just take y2 minus y1 over x2 minus x1. But there's another way of finding the gradient of the line segment, and that is if you know the angle theta that the line makes with the positive x-axis, what it means is that if you measure the angle theta from the x-axis in the anti-clockwise direction, right, the gradient of the line is just tangent of the angle. I'm going to provide you with an example how we can actually apply this concept. So let's look at this question. Right, supposing in this question here, you are given three points ABC and they're collinear. Collinear means that these three points lie on the same line, right? So you're given the coordinates B and you're told that C lies on the x-axis and you're given the tangent theta equals negative 2 and theta is the angle between the positive x-axis and the line and you're asked to find the equations of AC. Right, so to find the equations of the AC, we all know we need to have the gradient. The general form of a straight line equation is y equals to mx plus c. So basically, I need to know the gradient, and I need to know one point to substitute into this y equals to mx plus c to find my y-intercept. So where is it? How is it that? What is it? What is the gradient of this line AC? Right, yes. So this gradient will come from this information that is given here highlighted in yellow. That is tangent theta equals to negative 2. So this is where we make use of the concepts that we have just mentioned. m equals to tangent theta, right? So since m is equal to tangent theta, and tangent theta is equal to negative 2, then the gradient of this line AC is negative 2. And to find the equations of the line AC, we can just apply, substitute the gradient equals negative 2 and this point here into the equations y minus y1 equals to n bracket x minus x1. where x1, y1 is the point B. And from there, we can just find the equations of the line AC to be y equals to negative 2x plus 4. So this is just one example how we can actually use this concept m equals to tangent theta. Let's look at another example on how we can actually make use of this concept m equals to tangent theta. So the questions can be phrased in this way, such that using the same diagram, using the same concepts I thought we've done earlier, but in this case here, we're not given tangent theta equals to negative 2. It is not so direct now. We're given the angle alpha and alpha is 45 degrees. And angle alpha is between the x-axis and the line. So we're asked to find the equations of AC. So in these questions here, we can see that it is slightly more challenging, right? So we need to know the connections, right, between alpha and theta. So what exactly is theta, right? So theta has to be measured from the x-axis in the anticlockwise directions. So we can see that theta is actually 135 degrees. Just take 180 minus 45 to give you 135. Right. And tension 135, if we just use a calculator, is just negative 1. Therefore, the gradient of this line segment AC is negative 1. And using the same concepts of what we've done earlier, we just substitute the gradient to be negative 1 and point B, 1 comma 2, into y minus y1 equals to m bracket x minus x1. and we can find the equations of the line a c to be y equals to negative x plus 3. So this is a different example of how you can actually be tested on this concept m equals to tangent theta. So this question is actually an O-level question that's been tested in the year 2017. This question actually tests you on the concepts of m equals to tangent theta. So if you look at these questions here, you're given a trapezium with the vertices AB and C and point D. As a trapezium, meaning we have one pair of parallel sides. So in this case here, you're given that AB is parallel to DC and the angle DAB is 90 degrees. You're also given the angle ABO is equal to angle CBO. So on the diagram, I've let the angle ABO to be theta and therefore the angle CBO is also theta. Right, so we are asked to find the gradients of AB and CB in terms of P. Okay, what is P? Right, P is the y coordinates of point B. So finding gradient is not difficult. So let's go on to find the gradient of AB. Okay, before we go on to find the gradient of AB, let's see what happens if I extend the line BC longer and extend the line BA longer. Right, so in this case here, you see that we actually have two congruent triangles there. Right. So the next thing is that let's go on to find the gradient of BC, oh sorry, the gradient of AB first. So basically it's just y2 minus y1 that is p minus 1 over 0 minus bracket negative 2. So that gives you p minus 1 over 2 and let's go on to find the gradient of BC. Alright, so now the next thing is that we need to find a connection between the gradients of AB and the gradients of BC. So let's go back to the diagram. Alright, so I'm going to let, I've already explained earlier on, I've let the angle ABO to be theta and therefore CBO is also theta. Right, so using the concepts of m equals to tangent theta, we know that the gradient of AB would just be tangent alpha. And the gradient of BC would be this angle here. Because if you were to, let's recap, m equals to tangent theta again. So theta is the angle that makes with the x-axis measured in an anticlockwise direction. So therefore, the gradient of BC is this angle theta here, which is 180 minus alpha. And tangent alpha is equal to negative tangent 180 minus alpha. when your alpha is between 0 to 180 degrees. So therefore your tangent bracket 180 minus alpha is just basically minus tangent alpha. Right, so from there we can conclude that the gradient of AB is actually the negative of the gradient of BC. Right, so from there we can form an equation, right, where we let the gradients of AB equals to negative the gradients of BC, and we can find that P is equals to 5. So in this example here, this is the most challenging. problems involving the concepts of m equals to tangent theta. So the three examples I've just shown you basically will cover most of the things that needed to apply this concept m equals to tangent theta. So it's useful to learn these three different methods well. Right, what's the next concept that are needed? Right, so if we're given two lines, y equals to m1x plus c1, and another line, y equals to m2x plus c2, and if these two lines are parallel to each other, then the gradients must be equal to each other. Right, so I have an example there to show you how we can actually apply this concept. So supposing I have a line L, and this line L is parallel to another line, which is x plus 2y equals to 6. And given that this line passes through a certain point 3,4, we are asked to find the coordinates of the point where the line L meets the x-axis. Right, so how do we go on to apply the concepts of m1 equals to m2? Yes, so let's start off from x plus 2y equals to 6. From there, we can see that the gradient is negative half. First of all, I must remind you, whatever equation is given, you have to rewrite it in the format y equals to mx plus c. Only by rewriting it in the format y equals to mx plus c, then you can actually find out what is the gradient. And here, we found that the gradient of this line is negative half. And since... L is parallel to this line, L will also have gradient negative half. Alright, so we're going to substitute the gradient negative half and the 0.3 comma 4 into this formula, which is very, very useful to find equations of any line. So from here, we find that the equations of this line L is y equals to negative half x plus 11 over 2. And we are asked to find the coordinates of the point where L meets the x-axis. So on the x-axis, this is where your y-coordinate is 0. So we're going to let y equals to 0 to find the x-coordinate. And we found out to be x equals to 11. So therefore the point where the line L meets the x-axis will be 11,0. Right, so this is how you can actually make use of this concept. M1 equals to M2 when the two lines are parallel to each other. Right, sometimes in the exam questions, they will not... tell you that the two lines are parallel rather they will say that three points are collinear and when they say that it means that they lie on the same straight line and by saying that they lie on the same straight line it means that the gradients of each line segment will be equal to each other so let's look at an example on you how we can actually apply this concept. Right, so in these questions here, we are asked to show that these three points P, Q and R are collinear. So if these three points are collinear, means that the gradient of this line segment P, Q will be equal to the gradient of the line segment QR. If we can show that they have the same gradient, then It means that the points PQR lies on the same line. So let's start off by finding the gradients of PQ. right which turn up to be 1. Very important thing is that to bracket this one here okay anything that involves more than one term you should always bracket it and let's go on by finding the gradients of qr which also turns up to be 1. So we can conclude that since the gradient of PQ is the same as gradient of QR, and since Q is a common point, therefore the three points are collinear. Right, let's look at example 6. Sometimes the questions can be phrased in such a way that you are told that the points are collinear. and then from there you are asked to find a certain unknown value. So this unknown value is the value k. So when these three points are collinear, they are all on the same straight line, it means that the gradient of two points of the gradient of the line segment joining first and second point will be the same as the gradient of the line segment joining the second and the third point. So let's start off by naming, giving a name to all these three points. So we give cov,,,,. So let's go on to find the gradients of and the gradients of. And since the line on the same straight line means the gradient must be equal to each other, And therefore, from there, we can find the values of k to be negative 3. So this is another example on how we can actually apply the concepts of three points being collinear to one another. Right, more concepts needed. In this chapter, Coordinate Geometry, we also need to know the midpoint formula. So if you're given two points A and B, so the midpoint of A and B will be just x1 plus x2 divided by 2 and y1 plus y2 divided by 2. So this midpoint formula, it's very commonly used to find the fourth vertex of a special chord such as a parallelogram, rhombus, square and rectangle. This special collinear lateral has the same midpoint of the diagonal. So let's see how we can actually apply this concept. Right, so let's look at example 7. So we're given the parallelogram and if we come up the vertices point A, B, C and D. and you're given the point M, it's the midpoint of AC. So we ask to find these coordinates first. And then in part 2, we ask to find the coordinates of D. So this is where we apply the midpoint formula. So let's go on to find M first. which is just negative 8 which is your x1 plus x2 which is 12 divided by 2 and 0 plus 10 divided by 2 so the midpoint of this diagonal ac is 2 comma 5 and the midpoint of the diagonal bd will also be 2 comma 5 so let's go on to find the midpoint of this diagonal bd So p over 2 will be equals to 2. That makes p equals to 4. Negative 6 plus q over 2 equals to 5. That makes q equals to 16. Therefore, your point D would just be 4 comma 16. Right. So do remember that besides parallelogram, rhombus, rectangle, square also share the same property. and you can use this midpoint formula to find the fourth vertex. Right, another popular concepts that are needed in this chapter is that when you have two lines that are perpendicular to each other, then the gradients of these two lines multiplied together will give you negative one. Right, and sometimes you'll be asked to find the equations of the perpendicular bisector. So what do you understand by this word perpendicular bisector? Right, it means that from the diagram that's shown, your XY, this line XY will be the perpendicular bisector of AB. So perpendicular means XY. intercept AB at 90 degrees, that's called perpendicular. Bisector means it divides the line AB into 2 equal half. Right, so we would have to make use of concept number 6. If we're given the gradient of AB, then the gradient of the perpendicular bisector will be negative 1 divided by the gradient of AB. meaning from here, m1 is equal to negative 1 divided by m2, or m2 is equal to negative 1 divided by m1, depending on what gradient you're given. And we also need to know how to find the distance of a line segment joining two points. So I'm going to show you more examples on how we can apply these concepts in the next few slides. Right, so let's look at example 8. So example 8, we're given two points A and B, and we're asked to find the equations of the perpendicular bisector of AB. We're also given further that M is the midpoint of AB, and this perpendicular bisector meets the y-axis at point C, and we're asked to calculate the length of CM. Right, so first of all, let's have a rough idea where point A, B are. So from the diagram, we can clearly see that we are able to find the gradient of A, B, which comes up to be negative one-third. And therefore, the gradient of the perpendicular bisector will be negative 1 divided by negative 1 third. That gives you 3. And the next thing is that we need to find M, which is the midpoint of AB. And with all this information, I can just substitute into my y minus y1 equals to M bracket x minus x1 to find the equations of the perpendicular bisector, which come out to be y equals 3x minus 7. And on the next slide, I'll show you how we can find the length cm. Alright, so given all this information here, c actually lies on the y-axis. So therefore, the x coordinates of point c is 0. So we're going to substitute x equals to 0 into the equations that we found earlier on. and therefore the point C will have coordinate 0 and negative 7. Right, and from there we can apply the distance formula and substitute in the values for x2, x1 and y2, y1 and that comes out to be 15.8 units. Right, so finally we need to know how to find areas of our triangles or collider laterals using this method called the shoe lace method. And in a lot of coordinate geometries questions, you're asked to find areas of the given collider laterals. So it is very important that you learn how to apply this. Shoe Lace Method correctly. So in the next slide, I'm going to show an example of how we can apply it. All right, let's look at a slightly harder question. So in this case here, I'm given a polygon. Okay, this polygon have how many sides? One, two, three, four, five, five sides. Is it five sides of the polygon? All right, yes, it's a five-sided polygon. We are asked to find the area of this polygon and then part two is to show that the area of a c d e is 25.5 units squared all right so first thing first we're going to just sketch the points okay let's sketch the points see where are all the points so let's start off with point a point b c d and e all right next thing we draw join them up draw all these points up using line segment to get the polygon so one two so we have uh let me use a highlighter one two three four five is this is a five-sided polygon all right okay so what do we do yes let's arrange them okay arrange them so i start with point a all right use take it in the anti clockwise direction and the clockwise direction so start with point a which is negative two negative five not to forget the half must be always there some students forget about the half there Okay, start with the half and the vertical line, then you arrange all your coordinates, starting off with the x coordinate first, followed by y coordinate. So negative 2, negative 5, followed by 3, 5, 1, 8, 0, 9, negative 3, 1, go back to negative 2, negative 5. Just remember, when you come out of the house, you always like to go back to the house, right? So same thing, you start with point A, you end off with point A. You start with point E, you end off with point E. Okay. Right. So what do we do? So that will be equals to half. All right. How do I get 38? So we're going to simplify the calculations. OK, sometimes if you if you were to do what I have done earlier on, basically by writing bracket negative 2 times 10 give you negative 10 and then plus 3 times 8, which is 24 and then plus 1 times 9, which is 9. 0 times 1 is 0 and then negative 3 times negative 5 is positive 15. Okay, you see that it's a rather tedious way of doing it. So instead of doing it, we would really, really try to make use of the calculator as much as possible. Especially in exam, time is very limited. So we should make use of calculator as much as possible. So use a calculator to just key in negative 2 times 5 plus 3 times 8 plus. 1 times 9 plus 0 times 1 plus negative 3 times negative 5. And you will end up getting 38. Yes. Okay. You will get 38. All right. So then minus away the other direction. So the other direction is, again, use a calculator. Okay. Negative 5 times 3 plus 5 times 1 plus 8 times 0 plus 9 times negative 3 plus 1 times negative 2. And you end up get negative 39. Okay, and again, use a calculator. Okay, right, use a calculator as much as possible to help you in the calculation. All right, you get 38.5 units square for the polygon ABCDE. All right, so the next step is that we need to show that the area of ACDE is 25.5. So one way of course is going to find ACDE, right? But ACDE, I think we'll have more points to find. So instead, I go on to find ABC. Eh? Yeah. I go on to find ABC or AEC. What did I find? Wait, let me see. Area of triangle. Oh, yes. So that means I highlight the diagram wrongly. Okay. Oh, no, no. I didn't highlight the diagram wrongly. I went on to find the area of ABC. And then I used 38.5 to minus away the area of ABC to get the area of. A, C, D, E. Okay, the area of A, C, D, E, which is the one highlighted in blue, right? So I need to find, show that the area that's highlighted in blue is 25.5. So what I do, what I've done is I find the area of A, B, C first, okay, which come out to be 13 units square. and then i take 38.5 minus away 13 to get the area of ACDE. This one is ACDE. Alright and that is 25.5 unit square. Okay. Alright last but not the least, we need to know some properties of this quadrilateral. Alright let's start off with the rectangle. which is quite easy that the rectangle have four right angles and opposite sides are equal and the square has four right angles and four equal sides. Palaeograms do pass a parallel size the opposite side is equal. Rhombus have all equal sides and the diagonals intersect each other at 90 degrees. Alright trapezium have two sides parallel. Right, and the kite have two pairs of adjacent sides of the same length. And also the diagonals actually intersect each other at 90 degrees. Right, and just to mention again that for rectangle, for square, for parallelograms, and for rhombus, we can use the midpoint formulas to find the fourth vertex. So finally, In the last few slides, we're going to apply all the concepts we've learned into solving problem sums in this chapter, Coordinate Geometry. Right, instead of looking at these very complicated questions, so I've actually have broken up these questions into different parts so that you will not feel suffocated by it. Usually for coordinate geometries, it's around 9 to 10 marks questions. And usually in this such of questions, there will always be a phrase here saying that solution to these questions by accurate drawing will not be accepted. Meaning you have to apply the concepts we've learned. You can't actually draw out and say that, oh, Point D is so and so by drawing. You can't do that. In this question here, we are given a trapezium. Your BC is parallel to AD. If you are given point AB, if you are given a midpoint M, M is a midpoint of AB. And then the line CD is parallel to the x-axis. These are very important information you have to take note. The line CD parallel to the x-axis means the gradient of CD is 0. It's a horizontal line. Horizontal lines have 0 gradient. Right. C lies on the y-axis. And then we are asked to find the coordinates of B, C, D. So first thing first, let the point B be x, y. Let point C be 0, B. Why is it 0, b? Because c lies on the y-axis, so the x-coordinate is 0. And because cd is parallel to the x-axis, that makes cd a horizontal line. So that makes the y-coordinate of d be b. Okay, so let's see which points we can find first. Right. And quite obviously, we can find point B. Because you're given point A and you're given the midpoint, so we can just apply the midpoint formula to find point B first. So point B is negative 4 comma 1. So now we have to decide. We are left with point C and D. So we have to decide which point is easier to find. And in this case here, I think it will be easier to find point C because because we are able to find the gradient of AB which turns out to be 2. And having found gradient of AB, we know that the line AB is perpendicular to the line AD. And therefore we can use the concept of m1 times m2 equals to negative line or m1 times m2 equals to negative 1. for two perpendicular lines. So from there we can actually find the gradients of AD which is actually the same as gradient of BC because your AD is parallel to BC and using interior angles property your angle ABC is also 90 degrees. So therefore your line BC is also perpendicular to the line BA. Right, so you can just take negative 1 divided by 2. So the gradient of BC will be negative half. And we can use Y2 minus Y1 over X2 minus X1 to find the gradient of BC. and these two values must be the same because they both refers to the gradient of the same line. Right, so here you just take note. Negative half means negative 1 over 2 or 1 over negative 2. It can only belongs, the negative can only belongs to the numerator or the denominator but not both. So here what I have done is I check the negative with my numerator. So therefore I take negative 1 times negative 4. And hence my denominator will be a positive 2. So from there, I managed to find the values of b to be negative 1. So hence the coordinates of c will be 0, negative 1. and therefore the coordinates, the y coordinates of d will also be negative 1 and we can go on to find the gradients of ad using y2 minus y1 over x2 minus x1 and let it equals to negative half and from there we can find that a is 20. So therefore, the coordinates of D will be 20, negative 1. Right, so we have managed to find the coordinates of all the three points B, C, and D. Right, so the next thing is to find the area of the trapezium. So we're going to use the shoelace method. So to use the shoelace method, first of all, make sure that you... take the directions of the coordinates in the anti-clockwise direction. We start off with having a half there, arrange it in the anti-clockwise direction. We start with 0, 9 or end off with 0, 9. And I always like to do it this way, right? So how do I fill in the values in the bracket, right? It's basically just 0 times 1 plus negative 4 times negative 1 plus 0 times negative 1. plus zero times 20 plus zero times uh sorry plus 20 times nine you just use your calculator to work it out you get 184. likewise for the other bracket it's just the the other the other directions and then from there you can just use a calculator to work out the values to be 120 units square so that would be the area of the trapezium abcd right so let's look at another question again this one is is rather complicated So I've broken it up into parts by parts. So the first part is to find the coordinates of A and of C. So you're given the quadrilateral ABCD. You're given point B. Point B lies on the y-axis, 0,3. Point D lies on the x-axis, that's 7,0. And you're given the equations of the line AD to be 5y equals 2x minus 14. and we are also told that point C lies on the line y equals to x. Right, so what it means is that if I'm going to let my x coordinate be a, then my y coordinates is also a. So if my x is 1, my y is 1. If my x is 2, my y is 2. If my x is 3, my y is 3. That means any points that lie on this line y equals to x will have the same x and y coordinates value. Alright, so your CD is parallel to the Y-axis, so that makes CD a vertical line, and your angle CDO is 90 degrees. So that's a very important observation. Alright, so since CD is a vertical line, C will have the same X coordinate as point D, which is 7. And since the point C lies on y equals to x, that both x and y coordinate will be the same. So if the x coordinate of point C is 7, the y coordinate will also be 7. So therefore, your point C will have 7,7 as the coordinates. Right, so that's point C. That's quite easy to find. So point A is harder because point A lies on the perpendicular bisector of BD. So we need to find two equations and solve them simultaneously to find point A. So we need to draw the diagram, the perpendicular bisector of BD. Yes, so we have to find the equations of this perpendicular bisector and solve it simultaneously with 5y equals to 2x minus 14. Okay, so as usual, to find equations of a perpendicular bisector, we need to find the gradients of BD first, and then take negative 1 divided by the gradients of BD to find the gradients of your perpendicular bisector. Then you need to find the midpoint, and then substitute into y minus y1 equals to gradient bracket x minus x1 to get the equations of a perpendicular bisector, which I'm going to call it equations 1 and then the equations of AD is equation 2 and solve these two equations simultaneously and we can find the x coordinates of point A which is 2 then you can substitute into equations of AD or the perpendicular bisector to find the y coordinates which turns out to be negative 2. Therefore your point A is 2 comma negative 2. Right the second part is rather easy. Once we found the coordinates of point C and A, we can just use the shoelace method to find the area of your quadrilateral ABCD. So again arrange it in the anticlockwise direction. You start with 2 negative 2 and it ends with 2 negative 2. Right so basically you don't really have to write down 2 times 0 plus 7 times 7 plus 7 times 3 plus 0 times negative 2. You just key in this value into a calculator. Make sure you key in correctly. You'll come out to be 70 and then for the other bracket you just do it during the other directions. and come up to be negative 8 and you use a calculator to work out again you come up to be 39 units square right the last part of the questions require you to explain whether the quadrilateral ABCD is a kite so for that we need to know the properties of a kite so if you were to recall if ABCD is a kite right so this are the properties that the kite must have. AB is equals to AD, BC must be equals to DC and the diagonals BD and AC must intersect each other at 90 degrees. So we're going to find the gradients of BD and find the gradients of AC right and we're going to take multiply these two gradients together and you see that it doesn't gives you negative one right so it does not satisfy the conditions of a kite So therefore, ABCD is not a kite. Right, let's look at the last questions. So in these last questions here, okay, you're given a triangle PQR. You're given point P, and you're given that PQR passes through the origin, and OQ is perpendicular to PR. Okay, and you're also given that area of triangle POQ is 15 units square. Find the equations of OQ. So what you need to take note is the origin, the O meaning refers to the origin, and the origin has x coordinate 0 and also y coordinate 0. So that's something very important. And because OQ is actually perpendicular to RP, right, so we know that the gradients of This two perpendicular lines multiplied together gives you negative 1. So these are the concepts that you need to know. So let's start off by finding the gradients of OP. And from there, we will know that the gradients of OQ would be 2. And if you look at the diagrams, the line OQ actually passes through the origin. So it means that the y-intercept is 0. So therefore, the equations of OQ will just be y equals to 2x plus 0. That gives you y equals to 2x. Okay, let's look at the second part of the questions, where you are asked to find the coordinates of Q. Right. So since the equations of OQ, it's y equals to 2x, what we are saying that any points on the line OQ, if your x coordinate, let's say it's 1, then your y coordinates will be 2. If your x coordinate is 2, your y coordinates will be 2 times 2, 4. So if I let the x coordinate of Q be a, then the y coordinate will be 2 times a, gives you 2a. Let's explain this part here. and the origin is just 0 comma 0 right so i'll be using this information here the triangles of p o q is 15 unit square to find the unknown a so here i'm going to use the shoelace method make sure you take it in the anti-clockwise direction six times 2a plus a times 0 plus 0 times negative 3 so that gives you 12a do the other directions that gives you negative 3a so you will end up with half 12a plus 3a equals to 15 so that will be 15a equals to 30 and therefore a is equals to 2 so from there we can find the coordinates of q to be 2 comma 4 Let's look at part 3. In part 3, we are told that the length of OP is 3 times the length of OR. So imagine that you are at the point P, you would have to walk 3 units to the left and 3 units up to reach point O right? And then you would need to walk 1 unit to the left and 1 unit up to reach point R. Because the ratio is 1 is to 3. So these 3 units... actually represent 6 because the x coordinates of p is 6. So if 3 units represent 6, then 1 unit will represent 2 steps. So take the x coordinates of p, take it as 6 steps. So 3 units is 6 steps, so 1 unit is 2 steps, right? So therefore here, you will have to walk 2 steps. two steps to the left and it's on the negative side of the x-axis so therefore it's negative 2 that means your x coordinates of your r is negative 2 okay and you need to work one step up one step is positive right so one step it's two unit sorry it's a different story just now we're talking about x coordinate so that is six steps y coordinate is not six steps it's three steps here this is three steps for the y coordinate so this should not be two but rather it should be one because three steps represent three unit three step represent three unit so one step is one unit okay let me repeat now here this one here let's look at your x coordinate first So 6 steps represent 3 units. So 1 unit will represent 2 steps. So you're walking 2 steps to the left. So therefore, it's negative 2, right? Now let's look at your y coordinate. Y coordinate is 3 steps represent 3 units. So 1 unit will be 1 step. So you're walking 1 step. Therefore that will be 1. So your coordinates of r will be negative 2 comma 1. Okay so that's filled up what we have known so far. We have followed the coordinates of q to be 2 negative 2 comma 4. coordinates of r to be negative 2 1. So we asked to find a point S such that any point on the line PR, that means any point here, will be same distance from Q, also same distance from the point S. That means anywhere here, same distance. This is what it means. so that means your line RP is like a mirror line it's like a mirror line right so where do you think your point S is yes your point S if you were to join your point S to O and to Q QS is actually your perpendicular or rather RP is actually the perpendicular bisector of QS so that makes all the midpoint of QS So using the midpoint formula, so your point S is negative 2, negative 4. All right, so here we conclude the whole video on how you can apply all the concepts that are needed to solve any problem sums in this chapter, coordinate geometry. Right, so happy studying.

Right, first of all, we all know that to find gradients of a line segment, basically you can just take y2 minus y1 over x2 minus x1. But there's another way of finding the gradient of the line segment, and that is if you know the angle theta that the line makes with the positive x-axis, what it means is that if you measure the angle theta from the x-axis in the anti-clockwise direction, right, the gradient of the line is just tangent of the angle. I'm going to provide you with an example how we can actually apply this concept. So let's look at this question.

Right, supposing in this question here, you are given three points ABC and they're collinear. Collinear means that these three points lie on the same line, right? So you're given the coordinates B and you're told that C lies on the x-axis and you're given the tangent theta equals negative 2 and theta is the angle between the positive x-axis and the line and you're asked to find the equations of AC.

Right, so to find the equations of the AC, we all know we need to have the gradient. The general form of a straight line equation is y equals to mx plus c. So basically, I need to know the gradient, and I need to know one point to substitute into this y equals to mx plus c to find my y-intercept.

So where is it? How is it that? What is it? What is the gradient of this line AC?

Right, yes. So this gradient will come from this information that is given here highlighted in yellow. That is tangent theta equals to negative 2. So this is where we make use of the concepts that we have just mentioned. m equals to tangent theta, right? So since m is equal to tangent theta, and tangent theta is equal to negative 2, then the gradient of this line AC is negative 2. And to find the equations of the line AC, we can just apply, substitute the gradient equals negative 2 and this point here into the equations y minus y1 equals to n bracket x minus x1.

where x1, y1 is the point B. And from there, we can just find the equations of the line AC to be y equals to negative 2x plus 4. So this is just one example how we can actually use this concept m equals to tangent theta. Let's look at another example on how we can actually make use of this concept m equals to tangent theta. So the questions can be phrased in this way, such that using the same diagram, using the same concepts I thought we've done earlier, but in this case here, we're not given tangent theta equals to negative 2. It is not so direct now.

We're given the angle alpha and alpha is 45 degrees. And angle alpha is between the x-axis and the line. So we're asked to find the equations of AC.

So in these questions here, we can see that it is slightly more challenging, right? So we need to know the connections, right, between alpha and theta. So what exactly is theta, right? So theta has to be measured from the x-axis in the anticlockwise directions. So we can see that theta is actually 135 degrees.

Just take 180 minus 45 to give you 135. Right. And tension 135, if we just use a calculator, is just negative 1. Therefore, the gradient of this line segment AC is negative 1. And using the same concepts of what we've done earlier, we just substitute the gradient to be negative 1 and point B, 1 comma 2, into y minus y1 equals to m bracket x minus x1. and we can find the equations of the line a c to be y equals to negative x plus 3. So this is a different example of how you can actually be tested on this concept m equals to tangent theta. So this question is actually an O-level question that's been tested in the year 2017. This question actually tests you on the concepts of m equals to tangent theta. So if you look at these questions here, you're given a trapezium with the vertices AB and C and point D.

As a trapezium, meaning we have one pair of parallel sides. So in this case here, you're given that AB is parallel to DC and the angle DAB is 90 degrees. You're also given the angle ABO is equal to angle CBO. So on the diagram, I've let the angle ABO to be theta and therefore the angle CBO is also theta.

Right, so we are asked to find the gradients of AB and CB in terms of P. Okay, what is P? Right, P is the y coordinates of point B.

So finding gradient is not difficult. So let's go on to find the gradient of AB. Okay, before we go on to find the gradient of AB, let's see what happens if I extend the line BC longer and extend the line BA longer.

Right, so in this case here, you see that we actually have two congruent triangles there. Right. So the next thing is that let's go on to find the gradient of BC, oh sorry, the gradient of AB first.

So basically it's just y2 minus y1 that is p minus 1 over 0 minus bracket negative 2. So that gives you p minus 1 over 2 and let's go on to find the gradient of BC. Alright, so now the next thing is that we need to find a connection between the gradients of AB and the gradients of BC. So let's go back to the diagram.

Alright, so I'm going to let, I've already explained earlier on, I've let the angle ABO to be theta and therefore CBO is also theta. Right, so using the concepts of m equals to tangent theta, we know that the gradient of AB would just be tangent alpha. And the gradient of BC would be this angle here.

Because if you were to, let's recap, m equals to tangent theta again. So theta is the angle that makes with the x-axis measured in an anticlockwise direction. So therefore, the gradient of BC is this angle theta here, which is 180 minus alpha. And tangent alpha is equal to negative tangent 180 minus alpha. when your alpha is between 0 to 180 degrees.

So therefore your tangent bracket 180 minus alpha is just basically minus tangent alpha. Right, so from there we can conclude that the gradient of AB is actually the negative of the gradient of BC. Right, so from there we can form an equation, right, where we let the gradients of AB equals to negative the gradients of BC, and we can find that P is equals to 5. So in this example here, this is the most challenging.

problems involving the concepts of m equals to tangent theta. So the three examples I've just shown you basically will cover most of the things that needed to apply this concept m equals to tangent theta. So it's useful to learn these three different methods well.

Right, what's the next concept that are needed? Right, so if we're given two lines, y equals to m1x plus c1, and another line, y equals to m2x plus c2, and if these two lines are parallel to each other, then the gradients must be equal to each other. Right, so I have an example there to show you how we can actually apply this concept.

So supposing I have a line L, and this line L is parallel to another line, which is x plus 2y equals to 6. And given that this line passes through a certain point 3,4, we are asked to find the coordinates of the point where the line L meets the x-axis. Right, so how do we go on to apply the concepts of m1 equals to m2? Yes, so let's start off from x plus 2y equals to 6. From there, we can see that the gradient is negative half.

First of all, I must remind you, whatever equation is given, you have to rewrite it in the format y equals to mx plus c. Only by rewriting it in the format y equals to mx plus c, then you can actually find out what is the gradient. And here, we found that the gradient of this line is negative half. And since...

L is parallel to this line, L will also have gradient negative half. Alright, so we're going to substitute the gradient negative half and the 0.3 comma 4 into this formula, which is very, very useful to find equations of any line. So from here, we find that the equations of this line L is y equals to negative half x plus 11 over 2. And we are asked to find the coordinates of the point where L meets the x-axis. So on the x-axis, this is where your y-coordinate is 0. So we're going to let y equals to 0 to find the x-coordinate. And we found out to be x equals to 11. So therefore the point where the line L meets the x-axis will be 11,0.

Right, so this is how you can actually make use of this concept. M1 equals to M2 when the two lines are parallel to each other. Right, sometimes in the exam questions, they will not... tell you that the two lines are parallel rather they will say that three points are collinear and when they say that it means that they lie on the same straight line and by saying that they lie on the same straight line it means that the gradients of each line segment will be equal to each other so let's look at an example on you how we can actually apply this concept. Right, so in these questions here, we are asked to show that these three points P, Q and R are collinear.

So if these three points are collinear, means that the gradient of this line segment P, Q will be equal to the gradient of the line segment QR. If we can show that they have the same gradient, then It means that the points PQR lies on the same line. So let's start off by finding the gradients of PQ. right which turn up to be 1. Very important thing is that to bracket this one here okay anything that involves more than one term you should always bracket it and let's go on by finding the gradients of qr which also turns up to be 1. So we can conclude that since the gradient of PQ is the same as gradient of QR, and since Q is a common point, therefore the three points are collinear. Right, let's look at example 6. Sometimes the questions can be phrased in such a way that you are told that the points are collinear.

and then from there you are asked to find a certain unknown value. So this unknown value is the value k. So when these three points are collinear, they are all on the same straight line, it means that the gradient of two points of the gradient of the line segment joining first and second point will be the same as the gradient of the line segment joining the second and the third point. So let's start off by naming, giving a name to all these three points.

So we give cov,,,,. So let's go on to find the gradients of and the gradients of. And since the line on the same straight line means the gradient must be equal to each other, And therefore, from there, we can find the values of k to be negative 3. So this is another example on how we can actually apply the concepts of three points being collinear to one another.

Right, more concepts needed. In this chapter, Coordinate Geometry, we also need to know the midpoint formula. So if you're given two points A and B, so the midpoint of A and B will be just x1 plus x2 divided by 2 and y1 plus y2 divided by 2. So this midpoint formula, it's very commonly used to find the fourth vertex of a special chord such as a parallelogram, rhombus, square and rectangle.

This special collinear lateral has the same midpoint of the diagonal. So let's see how we can actually apply this concept. Right, so let's look at example 7. So we're given the parallelogram and if we come up the vertices point A, B, C and D.

and you're given the point M, it's the midpoint of AC. So we ask to find these coordinates first. And then in part 2, we ask to find the coordinates of D. So this is where we apply the midpoint formula. So let's go on to find M first.

which is just negative 8 which is your x1 plus x2 which is 12 divided by 2 and 0 plus 10 divided by 2 so the midpoint of this diagonal ac is 2 comma 5 and the midpoint of the diagonal bd will also be 2 comma 5 so let's go on to find the midpoint of this diagonal bd So p over 2 will be equals to 2. That makes p equals to 4. Negative 6 plus q over 2 equals to 5. That makes q equals to 16. Therefore, your point D would just be 4 comma 16. Right. So do remember that besides parallelogram, rhombus, rectangle, square also share the same property. and you can use this midpoint formula to find the fourth vertex. Right, another popular concepts that are needed in this chapter is that when you have two lines that are perpendicular to each other, then the gradients of these two lines multiplied together will give you negative one.

Right, and sometimes you'll be asked to find the equations of the perpendicular bisector. So what do you understand by this word perpendicular bisector? Right, it means that from the diagram that's shown, your XY, this line XY will be the perpendicular bisector of AB.

So perpendicular means XY. intercept AB at 90 degrees, that's called perpendicular. Bisector means it divides the line AB into 2 equal half. Right, so we would have to make use of concept number 6. If we're given the gradient of AB, then the gradient of the perpendicular bisector will be negative 1 divided by the gradient of AB. meaning from here, m1 is equal to negative 1 divided by m2, or m2 is equal to negative 1 divided by m1, depending on what gradient you're given.

And we also need to know how to find the distance of a line segment joining two points. So I'm going to show you more examples on how we can apply these concepts in the next few slides. Right, so let's look at example 8. So example 8, we're given two points A and B, and we're asked to find the equations of the perpendicular bisector of AB.

We're also given further that M is the midpoint of AB, and this perpendicular bisector meets the y-axis at point C, and we're asked to calculate the length of CM. Right, so first of all, let's have a rough idea where point A, B are. So from the diagram, we can clearly see that we are able to find the gradient of A, B, which comes up to be negative one-third.

And therefore, the gradient of the perpendicular bisector will be negative 1 divided by negative 1 third. That gives you 3. And the next thing is that we need to find M, which is the midpoint of AB. And with all this information, I can just substitute into my y minus y1 equals to M bracket x minus x1 to find the equations of the perpendicular bisector, which come out to be y equals 3x minus 7. And on the next slide, I'll show you how we can find the length cm. Alright, so given all this information here, c actually lies on the y-axis. So therefore, the x coordinates of point c is 0. So we're going to substitute x equals to 0 into the equations that we found earlier on.

and therefore the point C will have coordinate 0 and negative 7. Right, and from there we can apply the distance formula and substitute in the values for x2, x1 and y2, y1 and that comes out to be 15.8 units. Right, so finally we need to know how to find areas of our triangles or collider laterals using this method called the shoe lace method. And in a lot of coordinate geometries questions, you're asked to find areas of the given collider laterals. So it is very important that you learn how to apply this. Shoe Lace Method correctly.

So in the next slide, I'm going to show an example of how we can apply it. All right, let's look at a slightly harder question. So in this case here, I'm given a polygon.

Okay, this polygon have how many sides? One, two, three, four, five, five sides. Is it five sides of the polygon? All right, yes, it's a five-sided polygon.

We are asked to find the area of this polygon and then part two is to show that the area of a c d e is 25.5 units squared all right so first thing first we're going to just sketch the points okay let's sketch the points see where are all the points so let's start off with point a point b c d and e all right next thing we draw join them up draw all these points up using line segment to get the polygon so one two so we have uh let me use a highlighter one two three four five is this is a five-sided polygon all right okay so what do we do yes let's arrange them okay arrange them so i start with point a all right use take it in the anti clockwise direction and the clockwise direction so start with point a which is negative two negative five not to forget the half must be always there some students forget about the half there Okay, start with the half and the vertical line, then you arrange all your coordinates, starting off with the x coordinate first, followed by y coordinate. So negative 2, negative 5, followed by 3, 5, 1, 8, 0, 9, negative 3, 1, go back to negative 2, negative 5. Just remember, when you come out of the house, you always like to go back to the house, right? So same thing, you start with point A, you end off with point A.

You start with point E, you end off with point E. Okay. Right.

So what do we do? So that will be equals to half. All right.

How do I get 38? So we're going to simplify the calculations. OK, sometimes if you if you were to do what I have done earlier on, basically by writing bracket negative 2 times 10 give you negative 10 and then plus 3 times 8, which is 24 and then plus 1 times 9, which is 9. 0 times 1 is 0 and then negative 3 times negative 5 is positive 15. Okay, you see that it's a rather tedious way of doing it. So instead of doing it, we would really, really try to make use of the calculator as much as possible.

Especially in exam, time is very limited. So we should make use of calculator as much as possible. So use a calculator to just key in negative 2 times 5 plus 3 times 8 plus. 1 times 9 plus 0 times 1 plus negative 3 times negative 5. And you will end up getting 38. Yes. Okay.

You will get 38. All right. So then minus away the other direction. So the other direction is, again, use a calculator.

Okay. Negative 5 times 3 plus 5 times 1 plus 8 times 0 plus 9 times negative 3 plus 1 times negative 2. And you end up get negative 39. Okay, and again, use a calculator. Okay, right, use a calculator as much as possible to help you in the calculation.

All right, you get 38.5 units square for the polygon ABCDE. All right, so the next step is that we need to show that the area of ACDE is 25.5. So one way of course is going to find ACDE, right? But ACDE, I think we'll have more points to find.

So instead, I go on to find ABC. Eh? Yeah.

I go on to find ABC or AEC. What did I find? Wait, let me see.

Area of triangle. Oh, yes. So that means I highlight the diagram wrongly.

Okay. Oh, no, no. I didn't highlight the diagram wrongly.

I went on to find the area of ABC. And then I used 38.5 to minus away the area of ABC to get the area of. A, C, D, E. Okay, the area of A, C, D, E, which is the one highlighted in blue, right? So I need to find, show that the area that's highlighted in blue is 25.5.

So what I do, what I've done is I find the area of A, B, C first, okay, which come out to be 13 units square. and then i take 38.5 minus away 13 to get the area of ACDE. This one is ACDE. Alright and that is 25.5 unit square.

Okay. Alright last but not the least, we need to know some properties of this quadrilateral. Alright let's start off with the rectangle. which is quite easy that the rectangle have four right angles and opposite sides are equal and the square has four right angles and four equal sides. Palaeograms do pass a parallel size the opposite side is equal.

Rhombus have all equal sides and the diagonals intersect each other at 90 degrees. Alright trapezium have two sides parallel. Right, and the kite have two pairs of adjacent sides of the same length. And also the diagonals actually intersect each other at 90 degrees. Right, and just to mention again that for rectangle, for square, for parallelograms, and for rhombus, we can use the midpoint formulas to find the fourth vertex.

So finally, In the last few slides, we're going to apply all the concepts we've learned into solving problem sums in this chapter, Coordinate Geometry. Right, instead of looking at these very complicated questions, so I've actually have broken up these questions into different parts so that you will not feel suffocated by it. Usually for coordinate geometries, it's around 9 to 10 marks questions. And usually in this such of questions, there will always be a phrase here saying that solution to these questions by accurate drawing will not be accepted. Meaning you have to apply the concepts we've learned.

You can't actually draw out and say that, oh, Point D is so and so by drawing. You can't do that. In this question here, we are given a trapezium.

Your BC is parallel to AD. If you are given point AB, if you are given a midpoint M, M is a midpoint of AB. And then the line CD is parallel to the x-axis. These are very important information you have to take note. The line CD parallel to the x-axis means the gradient of CD is 0. It's a horizontal line.

Horizontal lines have 0 gradient. Right. C lies on the y-axis.

And then we are asked to find the coordinates of B, C, D. So first thing first, let the point B be x, y. Let point C be 0, B. Why is it 0, b?

Because c lies on the y-axis, so the x-coordinate is 0. And because cd is parallel to the x-axis, that makes cd a horizontal line. So that makes the y-coordinate of d be b. Okay, so let's see which points we can find first. Right.

And quite obviously, we can find point B. Because you're given point A and you're given the midpoint, so we can just apply the midpoint formula to find point B first. So point B is negative 4 comma 1. So now we have to decide.

We are left with point C and D. So we have to decide which point is easier to find. And in this case here, I think it will be easier to find point C because because we are able to find the gradient of AB which turns out to be 2. And having found gradient of AB, we know that the line AB is perpendicular to the line AD.

And therefore we can use the concept of m1 times m2 equals to negative line or m1 times m2 equals to negative 1. for two perpendicular lines. So from there we can actually find the gradients of AD which is actually the same as gradient of BC because your AD is parallel to BC and using interior angles property your angle ABC is also 90 degrees. So therefore your line BC is also perpendicular to the line BA. Right, so you can just take negative 1 divided by 2. So the gradient of BC will be negative half.

And we can use Y2 minus Y1 over X2 minus X1 to find the gradient of BC. and these two values must be the same because they both refers to the gradient of the same line. Right, so here you just take note.

Negative half means negative 1 over 2 or 1 over negative 2. It can only belongs, the negative can only belongs to the numerator or the denominator but not both. So here what I have done is I check the negative with my numerator. So therefore I take negative 1 times negative 4. And hence my denominator will be a positive 2. So from there, I managed to find the values of b to be negative 1. So hence the coordinates of c will be 0, negative 1. and therefore the coordinates, the y coordinates of d will also be negative 1 and we can go on to find the gradients of ad using y2 minus y1 over x2 minus x1 and let it equals to negative half and from there we can find that a is 20. So therefore, the coordinates of D will be 20, negative 1. Right, so we have managed to find the coordinates of all the three points B, C, and D.

Right, so the next thing is to find the area of the trapezium. So we're going to use the shoelace method. So to use the shoelace method, first of all, make sure that you... take the directions of the coordinates in the anti-clockwise direction.

We start off with having a half there, arrange it in the anti-clockwise direction. We start with 0, 9 or end off with 0, 9. And I always like to do it this way, right? So how do I fill in the values in the bracket, right?

It's basically just 0 times 1 plus negative 4 times negative 1 plus 0 times negative 1. plus zero times 20 plus zero times uh sorry plus 20 times nine you just use your calculator to work it out you get 184. likewise for the other bracket it's just the the other the other directions and then from there you can just use a calculator to work out the values to be 120 units square so that would be the area of the trapezium abcd right so let's look at another question again this one is is rather complicated So I've broken it up into parts by parts. So the first part is to find the coordinates of A and of C. So you're given the quadrilateral ABCD.

You're given point B. Point B lies on the y-axis, 0,3. Point D lies on the x-axis, that's 7,0. And you're given the equations of the line AD to be 5y equals 2x minus 14. and we are also told that point C lies on the line y equals to x. Right, so what it means is that if I'm going to let my x coordinate be a, then my y coordinates is also a.

So if my x is 1, my y is 1. If my x is 2, my y is 2. If my x is 3, my y is 3. That means any points that lie on this line y equals to x will have the same x and y coordinates value. Alright, so your CD is parallel to the Y-axis, so that makes CD a vertical line, and your angle CDO is 90 degrees. So that's a very important observation.

Alright, so since CD is a vertical line, C will have the same X coordinate as point D, which is 7. And since the point C lies on y equals to x, that both x and y coordinate will be the same. So if the x coordinate of point C is 7, the y coordinate will also be 7. So therefore, your point C will have 7,7 as the coordinates. Right, so that's point C.

That's quite easy to find. So point A is harder because point A lies on the perpendicular bisector of BD. So we need to find two equations and solve them simultaneously to find point A. So we need to draw the diagram, the perpendicular bisector of BD.

Yes, so we have to find the equations of this perpendicular bisector and solve it simultaneously with 5y equals to 2x minus 14. Okay, so as usual, to find equations of a perpendicular bisector, we need to find the gradients of BD first, and then take negative 1 divided by the gradients of BD to find the gradients of your perpendicular bisector. Then you need to find the midpoint, and then substitute into y minus y1 equals to gradient bracket x minus x1 to get the equations of a perpendicular bisector, which I'm going to call it equations 1 and then the equations of AD is equation 2 and solve these two equations simultaneously and we can find the x coordinates of point A which is 2 then you can substitute into equations of AD or the perpendicular bisector to find the y coordinates which turns out to be negative 2. Therefore your point A is 2 comma negative 2. Right the second part is rather easy. Once we found the coordinates of point C and A, we can just use the shoelace method to find the area of your quadrilateral ABCD. So again arrange it in the anticlockwise direction.

You start with 2 negative 2 and it ends with 2 negative 2. Right so basically you don't really have to write down 2 times 0 plus 7 times 7 plus 7 times 3 plus 0 times negative 2. You just key in this value into a calculator. Make sure you key in correctly. You'll come out to be 70 and then for the other bracket you just do it during the other directions. and come up to be negative 8 and you use a calculator to work out again you come up to be 39 units square right the last part of the questions require you to explain whether the quadrilateral ABCD is a kite so for that we need to know the properties of a kite so if you were to recall if ABCD is a kite right so this are the properties that the kite must have.

AB is equals to AD, BC must be equals to DC and the diagonals BD and AC must intersect each other at 90 degrees. So we're going to find the gradients of BD and find the gradients of AC right and we're going to take multiply these two gradients together and you see that it doesn't gives you negative one right so it does not satisfy the conditions of a kite So therefore, ABCD is not a kite. Right, let's look at the last questions.

So in these last questions here, okay, you're given a triangle PQR. You're given point P, and you're given that PQR passes through the origin, and OQ is perpendicular to PR. Okay, and you're also given that area of triangle POQ is 15 units square. Find the equations of OQ.

So what you need to take note is the origin, the O meaning refers to the origin, and the origin has x coordinate 0 and also y coordinate 0. So that's something very important. And because OQ is actually perpendicular to RP, right, so we know that the gradients of This two perpendicular lines multiplied together gives you negative 1. So these are the concepts that you need to know. So let's start off by finding the gradients of OP. And from there, we will know that the gradients of OQ would be 2. And if you look at the diagrams, the line OQ actually passes through the origin.

So it means that the y-intercept is 0. So therefore, the equations of OQ will just be y equals to 2x plus 0. That gives you y equals to 2x. Okay, let's look at the second part of the questions, where you are asked to find the coordinates of Q. Right. So since the equations of OQ, it's y equals to 2x, what we are saying that any points on the line OQ, if your x coordinate, let's say it's 1, then your y coordinates will be 2. If your x coordinate is 2, your y coordinates will be 2 times 2, 4. So if I let the x coordinate of Q be a, then the y coordinate will be 2 times a, gives you 2a.

Let's explain this part here. and the origin is just 0 comma 0 right so i'll be using this information here the triangles of p o q is 15 unit square to find the unknown a so here i'm going to use the shoelace method make sure you take it in the anti-clockwise direction six times 2a plus a times 0 plus 0 times negative 3 so that gives you 12a do the other directions that gives you negative 3a so you will end up with half 12a plus 3a equals to 15 so that will be 15a equals to 30 and therefore a is equals to 2 so from there we can find the coordinates of q to be 2 comma 4 Let's look at part 3. In part 3, we are told that the length of OP is 3 times the length of OR. So imagine that you are at the point P, you would have to walk 3 units to the left and 3 units up to reach point O right?

And then you would need to walk 1 unit to the left and 1 unit up to reach point R. Because the ratio is 1 is to 3. So these 3 units... actually represent 6 because the x coordinates of p is 6. So if 3 units represent 6, then 1 unit will represent 2 steps. So take the x coordinates of p, take it as 6 steps.

So 3 units is 6 steps, so 1 unit is 2 steps, right? So therefore here, you will have to walk 2 steps. two steps to the left and it's on the negative side of the x-axis so therefore it's negative 2 that means your x coordinates of your r is negative 2 okay and you need to work one step up one step is positive right so one step it's two unit sorry it's a different story just now we're talking about x coordinate so that is six steps y coordinate is not six steps it's three steps here this is three steps for the y coordinate so this should not be two but rather it should be one because three steps represent three unit three step represent three unit so one step is one unit okay let me repeat now here this one here let's look at your x coordinate first So 6 steps represent 3 units. So 1 unit will represent 2 steps. So you're walking 2 steps to the left.

So therefore, it's negative 2, right? Now let's look at your y coordinate. Y coordinate is 3 steps represent 3 units. So 1 unit will be 1 step.

So you're walking 1 step. Therefore that will be 1. So your coordinates of r will be negative 2 comma 1. Okay so that's filled up what we have known so far. We have followed the coordinates of q to be 2 negative 2 comma 4. coordinates of r to be negative 2 1. So we asked to find a point S such that any point on the line PR, that means any point here, will be same distance from Q, also same distance from the point S. That means anywhere here, same distance. This is what it means.

so that means your line RP is like a mirror line it's like a mirror line right so where do you think your point S is yes your point S if you were to join your point S to O and to Q QS is actually your perpendicular or rather RP is actually the perpendicular bisector of QS so that makes all the midpoint of QS So using the midpoint formula, so your point S is negative 2, negative 4. All right, so here we conclude the whole video on how you can apply all the concepts that are needed to solve any problem sums in this chapter, coordinate geometry. Right, so happy studying.

Transcript for:Coordinate Geometry Essentials for Revision

Transcript for:
Coordinate Geometry Essentials for Revision