CXC CSEC Maths- Lesson On Transformational Geometry 1; Translation and Reflection

okay welcome everyone to a lesson on transformational geometry this is a lesson that was covered in the live class so if you’re a student arm or mr class or if you’re a new student this is an opportunity for you to catch up with a class okay so all this information here was covered in the live class the live sessions okay so are the lessons on transformational geometry and I’ve decided to break up the lesson into two sections so you have no transformational German 21 and this and there is going to be another video on transformational geometry too and in transformational geometry 1 i’m looking on two types of transformation on translation and reflection and in the next video which is transformational geometry 2 i’ll be looking at our rotation and enlargement and also provide i will be providing some additional information to as well okay so the first transformation that I’m going to look at is what you call a translation okay and technically there are four types four main types of transformations your translation reflection rotation and enlargement okay so in this video I’ll be looking at translation and reflection so generally speaking a transformation is a one-to-one relation okay it sort of take an object which can be a point or plane and after you apply the transformation you get what we call an image okay so look now let us look at the first type of transformation which is called a translation translation is a transformation in which a plane figure slides along a straight line and changes its position without turning each point moves the same distance and in the same direction okay so let me get my graph paper here and this is something that time that I be using quite frequently in this lesson ok so I have a graph paper here i have my sentir on my origin at 0-0 ok and this is my x-axis and this is my y-axis so let us start off very basic with a point set up our point right here right here ok remember point of this trip is an opposition let’s call this point P it has coordinate X equal to 3 y equal to two x equal the tree y equal to 2 that’s a coordinate for point P now let’s say that I decided I want to move this point to some other position on this arm graph here and its coordinate grid all right now let’s say that I want to move this point here say in this quadrant in the second quadrant so let’s say that I want to move with one two three four five six seven units today left and say 12 units to the right so i want to put point p right here okay now first of all this original point is what they call your object point okay the object okay and this point that i’m going to get what’s called p prime okay it’s called your image the image point ok now I decided that I want to move this point one two three four five six seven eight units to the left and two units up so can I arm represented using what they call a translation vector or a column vector well I wanted you I want

to move this point eight units to the left so my x-coordinate for my column vector or my translation vector is going to be negative it and I want to move this same point to you in stood up I’m sorry two units up so that’s going to be positive 2 okay so what I can simply do is to take my point P take my point p which has coordinated 32 and add my translation vector T which is negative 8 2 and I’m going to get a new point called pint arm p prime which is the image okay after apply the transformation T so now this is just simply matrices we know what to had matrices with us have the corresponding elements so 3 plus negative 8 is negative 5 and two plus two is four so my new position my new position for this point here is going to be here X equal to negative 5 and y equal to four right here okay the coordinate for P prime is going to be X equal negative 5 and y equal to 4 so what I’ve simple done is to move this point based on some translation vector t and I have decided to move this point eight units to the left H and pull units up so you have one two three four five six seven eight units to the left and 12 units up okay and by doing that I can represent my direction based on some translation vector but I apply to the original object point to get your original arm to get your image point and please note that if you take your original 00 any unit to the left on the arm x-axis are any units on to the right of the arm origin okay so this is the right hand side with the positive values of x and any units to the left would be negative values so if I’m over here and I want to go to the left and that simple mean that you’re going to go arm- units to the left likewise if you’re here you want to go right is going to positive units and for the y axis you can think of this is this is going up or upper odd okay going upward are down a downward whereas if you’re going up then it’s going to be positive units if you’re going down it’s going to be negative units that’s how we normally represent our motion okay now you know of course this is just a point and the wall idea is that you can apply this to any plane figure any pain figure you can take a child for example which is your object triangle I can move the car to sort of direction that you want okay and the form that you’re using is that you take your object point X plus your translation gives you X prime which is your image point your new position after you apply the transformation so let’s let us do this using say for example a triangle let us do this using a triangle let’s say that I have a child here ok i will triangle there and let’s call it turn the ABC again it can do this with any plane figure rectangle square a triangle a trapezium some other polygons you can do it exagone epic gun you name it ok and

what you do you apply the translation vector to each vertex of your object arm figure so for example i have this triangle here which I’d coordinate a coordinate a vertex is X equal to negative 5 and y equal to 6 my b vertex is X equal to negative 2 y equal to 3 and X equal to negative 5 and y equal to tree for my c vertex and let’s say that I want to move this triangle say I don’t know five units to the right all right moving this is your x axis and this is your y axis y axis okay and again this is going up this is going down this is going to the right and this is great to the left and let’s say that I want to move this triangle arm say several units are ten units today to the right okay 10 units to the right and say ah two units up okay this is what you call this coordinate here write it properly here ten and two I want to move each vertex of this triangle 10 units to the right and two units up this is what I call a translation vector or in some text book they call it what they call a column vector okay and of course the translation vector the column vector can be anything you want base and how you want to move the figure so this is your object triangle I need to get your arm your image triangle or your image points again I use your formula take a point a you apply the translation vector to each vertex of your figure and that figure can be any clean figure so for example arm the vertex a at a point a as coordinate neg negative X equal negative 5 y equal 6 plus your translation vector that’s all you want to move the the triangle is equal to a prime it is simply add the corresponding elements negative 5 plus 10 is going to give us five and six plus two is eight and you want to find B Prime say take the point be the vertex B negative 2 3 plus your translation vector 10 2 is equal to B Prime and just you know have your corresponding elements here so you have negative 2 plus 10 that’s going to give us eight and three plus two is five and if you want to find ce primary let’s take an original point our object point c which has coordinate negative 5 and 3 plus your translation vector which is 10 too that’s going to give us c prime okay this little dash here are the upper script represent arm Prime okay we should protect your image of armed see so lets you know find this negative 5 plus deadness give us five and three plus three sorry plus 2 is equal to 5 and now you can plot these image points to get your new position of your triangle okay so you know you have X equal to 5 and y equal to 8 X equal to 5 y equal to 8 let’s be a prime and X equal to 8 and y equal to 5 that’s going to be a B Prime and X

equal to 5 and y equal to 5 that’s going to be a C prime it will simply connect the dots to form your new triangle okay and as you can see that you have moved this triangle this object triangle 10 units today right okay to the right and two units up and each and every vertex or each other point that makes up this arm object try and layer is moved in the same direction and in the same distance okay and in fact if you were to connect if you were connect a 2.82 a prime and undo that say connect a the a prime B the B Prime and C the C prime you can see that all of these lines here are actually parallel lines okay which means that all of these points here are going in the same direction and the length of this line and this line at this line segment is in fact equal which means that all of these points are traveling in the same distance okay so I mean the translation vector or the column vector can be anything you want it to be all right so if you want to move there this original charm for example down here or you want to move it over here or you want to keep it in the same position then you just apply the appropriate translation vector to the object point to each of vertex of the object right to get the corresponding period image vertex are you plug the vertices connected points and then that’s how you will get V this is no the image triangle okay fairly easy does applying the translation vector notice that you are adding the same translation vector 10 to 10 to 10 22 each object arm vertex together our image vertex right and reason we’re adding the same translation vector it’s because for a translation transformation all the points as the information says they’re all the points okay moves in the all points under the same translation move the same distance in the same direction okay now in terms of translation being what they call a congruent I can’t go into transformation okay what I simply mean that if you were to compare your object triangle to your image triangle then I can see that this length and this length is the same so AC and a prime C prime is the same AP and a prime B prime is the same and CP and see payments c prime b prime is the same ok so the lengths are the same ok after the transformation the angles are the same this ad is equal to this angle this angle is equal to this angle and this angle is equal to this angle so let meet us right now this angle C is equal to this angle C prime this angle a is equal to this angle a prime and this angle B is equal to this and the beep I’m the lengths are the same the angles are the same ok and the air of the board challenges will be the same ok so LT nurse will be the same except they’d the position right because as you can see that the position changes from here to here and a person changes color to your translation vector and again remember I said that if you’re going to the right let’s come to the positive unit so if you’re going to the left you make your x value negative if you’re going up you make a y-value positive if you’re going

down you make y value negative ok so let us look at our messenger that’s the reason why we call a translation transformation a congruent transformation because everything stata see ok so now let’s go back to the information here and it says that in terms of properties of translation translation preserves the lent the angle the parallel ism the area the way should order of points the orientation ok we’ll talk about orientation for example I’m talking about in terms of clockwise or anti-clockwise so if you look at the the you know I name this triangle ABC and here you can see that ABC is going to is going into a a a clockwise direction okay it bc and if you look at your image triangle a prime B prime C prime is also going into a clockwise direction so the orientation after the transformation call a translation is the same the orientation is the same as well an order of the points a B C a prime B prime C prime on the order of the points are the vertices of Stajan see okay now of course you know in an exam they can give you for example they can give you the object points and image points and actually to find the translation vector or the column vector that moves the object point to the image point again it can use the same formula right so here is for example if we apply this still say point is let’s say that we did not know the translation vector and only that they give you is your object point and an image point and then add you to find the translation vector that moves arm it to a prime then what we can do is a simpler formula here it’s going to rearrange your formula using basic algebra right so in general if you have an appoint X I apply the translation vector I’ll call that translation vector T all right x + t is going to give us X Prime alright so let’s say that we have this point here call it a right so it’s going to be a witch has coordinate negative 5 6 plus your translation vector let’s call it X Y that’s what we don’t know is equal to a prime which we know that was given to us okay a prime was X has coordinate X equal to 5 and y equal to 8 X equal to 5 y equal to 8 now here you are given a and a primary are to find a translation vector we already know what it is but let us see we can let us show that we can calculate it if we did not know what it is using the basic formula here so you can say that T let me arm content this up here so you guys can see clearly let me erase some of this so you have this is a formula here for example X plus T is equal to X prime then now we can apply the same formula given a point eight original point a machine which we know has coordinates negative 5 and 6 plus a translation vector T that’s what we are trying to find X Y is equal to a prime which we know has coordinate X equal to 5 y equal to 8 so we can easily find on the translation vector X Y is equal to a prime 5 8 minus a okay let me just erase some of this get

the space minus a which is negative 56 okay because it’s a plus right here it bring it over today right hand side it becomes a minus in the same way if you’re solving a linear equation okay you can think of this txc as a variable right and you know if you want to find TX is a simple do a chance position here so I mean in the same way if you will fit this original formula x plus t equal to x prime then if you want to find T is that’s going to equal to X prime minus X the same thing we are here a prime minus a okay and of course now we can easily find the translation vector by subtracting the corresponding armed elements they have 5 minus minus 5 that’s going to be 5 minus minus 5 and it’s going to be 8-6 8-6 ok and this equal to 5 minus minus 5 that’s 10 because minus minus is plus and a minus 6 is 2 so the translation vector all right the translation vector now has x coordinate of 10 and y coordinate of 2 that’s what we already know over here but translation vector is going to be 10 20 k this is a way you can rearrange the original formula to find what you want okay so they give you the original arm point at an object point and they give you the image point you can simply find it translation vector t by rearranging this formula transposing the formula to find the translation vector okay so this is the key formula here let me use green key formulator for translation now let’s go back to our information here yeah and now let’s move on so we talked about the translation can be represented by a column vector usually denoted t and that’s what we talked about so there’s a column vector T arm XY now traditionally translation is denoted by the letter T the capital letter T but you know that’s for consistency’s mathematicians are usually used T but when you’re going to the exam just keep in mind that the examiner can use any letter of the alphabet to represent the translation vector add a column vector ok so don’t don’t go inside exam thinking that well and in letter than any you know T letter we will present a translation know in an exam you know it can be a different thing because obviously the examiner is trying to test your knowledge and test your understanding on certain thing and they try to trick you and tell you where the transformation is is T and then you’re going to think well it since since it is T then it has to be a translation or it can be anything technically speaking but in terms of reading your text box and stuff like that normal it is given a letter T but just keep in mind that in an exam they can actually use any letter of the alphabet to represent the translation now let us go back to this idea call arm in verse translation ok and it said in verse translation moves the image back to the object ok so now this is interesting now let’s go back to our graph here and let’s say let me erase some of these information here so I can have some space here so now let’s say that you have a arm image point and you want to get a object point okay let me keep this year for

consistency different color red so let’s say that um you know you have some image point and you want to find the object point in other words you want to move your image back to its original position you want to move this image strand back to the original position what translation vector that will do that it turns out that the negative of the original translation vector will move your image back to its original position okay so if you take your image point for example a prime which is coordinate X equal to 5 and y equal to 8 and you add what you call your in verse translation okay I in verse translation is a negative of this translation so now you have negative 10 and negative 2 once you multiply this out so you can think of this as negative 1 some scalar you multiply this negative 1 year x 10m x 2 i get this so this is not your in verse translation it’s still can represent it does t and now you want to get your original position a so now you have no five plus or minus 10 that’s going to give us minus 5 all right for the x value of a and 8 plus or minus 2 that’s going to give us six so what I’m saying is that if you take your image point and you add what you call your in verse translation this is your in verse translation and the final in verse translation is a simply multiply you always know the translation by negative 1 not represent your in verse translation and notice that if you apply the inverse translation to your image you get a is negative x equals negative 5 and y equals 6 and as you can see here you will get back your original position on your object position which is armed X equal to negative 5 y equal to six same thing you have here you can apply the same principle to each vertex of your image triangle to get your object arm vertices okay in other words what is simple doing is is taking this imager and moving it back to its always null position applying what you call an inverse translation which is a negative of the original translate arm translation okay so you know you can actually in a move the object to the image I can move the image back to the object ok see if we want to move the object to the image then use a translation vector and if you want to move the image back to the object then I use the inverse translation vector which is the negative of the original translation vector ok well that makes sense and if not I’m send me an email on the six seem odd 22 @ gmail.com so we can so I can explain further what I meant by that and probably can do an example okay so let me go back okay and that’s it pretty much for translation the key to remember that translation is a congruent transformation okay it preserves the length the angle it parallels in the air the ratio the audio finds orientation all of those properties are preserved after you apply the translation vector right and also that all points under the chance same translation move the same distance in the same direction okay and it’s very important and of course translations are current grain translation account grain transformation now let’s move on to reflection okay so reflection is a way of transforming a shape as a mirror plane does at all okay

and the properties of reflection preserves lent the angular parallels in the air at Oratia order points and we said that reflection does not preserve the orientation and there are different types of reflection right you can reflect a point or an object in the x-axis the y-axis y equal to x our line the y equal minus x reflection in the origin reflection in the in the line x equal to b and reflection in the line y equal to c and of course we can use a 2 by 2 matrix to represent a reflection okay so let’s go back to my graph paper here and let me just undo all of this just erase all of that and again this is my x-axis this is my y-axis and let’s say that I at this point here right here and let’s call this point a okay now let’s say that you’re in your washroom are your bathroom at home again and you’re standing here okay and you look in the mirror in a washroom and obviously your reflection will be they will take the opposite position here over here something over here it’s not going to be here it’s not going to be here yeah i mean a reflection is not going to be not going to be beside of you here or beside of you here it’s going to be an opposite side of your mirror lines think of the x axis right here as your mirror line or your mirror at home okay so your x axis is a mirror line and just note that let’s let’s freak them to arm makes it make an observation here that are in fact your x axis as equation y equal to 0 so and you know this is your x axis your x axis your x axis over there when I keep writing that so we remember what am I doing here mmm okay yeah so I mix it up this activity sexy so your x-axis is this line here and I’m saying that arm your x axis also have equation y equal to 0 so if you’re standing here your x axis is your mirror line then obviously you’re going to be somewhere over here and to give you the exact position that you’re going to be your reflection is going to be the same distance from your mirror line as your arm your object here because remember your position he represents your object point here and your image point what I’m saying is going to be in the same way this is one two three units from the mirror line you’re up your image is going to be 1 2 3 units from the Muir line so a prime is going to be somewhere there and note that the line segment there’s a line segment that connects a to a prime okay that connects a the a prime and this is always going to be the case a pram cannot be here it cannot be here it cannot be here it cannot be here it has to be on the same line segment as a as an object okay so this again is your image a prime is the image right and note that as I said before that a

and a pram are equal distance from the mirror line so if this is three units from the mirror line then a prime is also going to be treated from the mirror line and it’s always going to be the case also that a and a prime is going to be on the same line segment that’s always going to be be the case regardless of the reflection of your doing for any reflection now in this case we’re doing a reflection in the x axis and therefore your x axis is your mirror line and please note that I could have said that I am doing a reflection in the line y equal to 0 which is the same thing as your x axis okay also note that your mirror line and the line segment that connects a to a prime is perpendicular and since a and a prime has to be equal distance from a mirror line it means that your mirror line in fact is a perpendicular bisector perpendicular please take note of that perpendicular bisector okay because a mirror line bisects the line segment that connects a to a prime okay because this is three units armed going in that direction and three units going down so if x 6 a line segment and as you can see that the x axes are they the line y equal to 0 is perpendicular to the line segment a and a prime so you form a right angle of 90 degrees right there okay and that’s going to be the case for any reflection whether it be the x axis y axis the reflection to the origin our line y equal to X or Y line y equals negative x those are those are the things we’ve been looking at arm looking at shortly as we move on on reflection now let us look at the object coordinated recorded for the object point so X equal to 4 let me erase this a little bit so you can see that it actually line segment actually passes to 4 so X equal to 4 and y equal to tree and for the image point the coordinate is X equal to 4 and y equal to negative tree and if you were to compare the object point which is for entry and the image point which is 4 and negative tree you can see that your X values stay the same but your Y value changes sign from positive to negative and this is always going to be the case for a reflection in the x axis and the reason why the X valid stay the same is because a and a prime is on the same line segment and in and in fact this line segment here let me judge a different color here this line segment that i’m talking about that connects a to a prime in fact has equation x equal to 4 because this is a vertical line that passes through the x value of 4 so any point on this line segment here is going to have a x coordinate value of for any point even if you join a new point here or here any point is going to have a x value of 4 and that is why a and a prime of the x value of 4 and as you can see that under a reflection in the x axis the sine of Y change and the value of the x stay the same in fact we say that the X arm value is invariant okay because this is a reflection in the x axis and therefore we can develop a formula a general formula that says that

if you take if you take some reflection em let me use a red pen here so if you take some reflection mm usually denoted by reflection all right let em and of course this is same thing as a reflection in the line y equal to 0 or reflection in the x axis same thing okay so if you take some general points say P that has coordinate X Y I you reflected I am mapped it to a P prime then p prime will have coordinate X will stay the same and the sine of Y will change so what we’ve been negative Y so Y will change from why the negative Y so if this Y is positive that this is going to be negative and if this Y is negative then this wise when we’re positive okay so the sine of Y changes so this is the formula for a reflection in the arm x axis and of course you can apply this formula here to the point that we have so if you take point a which has coordinate four and three and you want to find a prime which is the image which is the image okay of a then all it do you you’ll keep the value of what arm the actress is for and let’s call it to the formula I a chain the sine of Y so you become negative tree and same thing that we have here okay and this formula can apply to any object it could be a triangular could be a any plane figure it could be a triangular rectangular square armature p ziam apart some arm polygon on you know you can apply to X garden octagon you apply this to the vertex of each other of these object to get the arm the image vertex so always apply to each other vertex of the object figure to get the vertices of the image figure and it turns out that to make it a little bit easier for you turns out that these are two by two matrix that we flip that represent a reflection arm in the x axis or a reflection in the line y equal to 0 which is same thing as the x axis and this is a 2 by 2 matrix so 2 by 2 matrix will be armed what the leading diagonal will be 1 and negative one and the other diagonal will be 0 and 0 so this is a 2 by 2 matrix that you can apply to your object point to get your image point and again you can apply this the same way on any figure rectangles squares all that right and what is super do is just take for example you take the 2 by 2 matrix and multiply it by your object point let’s say for 3 right to get your a prime okay and we know how to multiply matrices here because we did it before our very first lesson we’re going to take roll times column and you see that this will be 4 and Rho times column this would be negative tree and you get a prime which is for negative tree so this is a general 2 by 2 matrix that represent a reflection in the x axis but it can apply see if you don’t want to use a formula you can’t remember eyes a formula at least memorize a 2 by 2 matrix that represent a reflection in the x axis and you can simply multiply these two by two matrix by the vertices of arm each arm the day the vertices of the object so if this area for let’s say that this was a triangle ok I want let’s say the triangle here I want to reflect this triangle in the x axis or do you

just ate the vertices of this chain called ABC take the coordinates right I multiply each of the coordinate by this two-by-two matrix here to get your a prime B prime C prime you plot it connect the dots are a collective collective vertices ok and then you’ll see that that’s actually a reflection in the arm in the x axis ok so let’s move on to another type of reflection so the first type of reflection we look at is a reflection in the x axis now let’s look at a reflection in the y axis ok so let me erase all of this and start off against this is my x-axis this is my roof y axis so my y-axis is going to be the mirror the mirror line or the line of reflection so let’s say take some point here I could take any point all right and again I could take any figure I triangle rectangle is using point just for presentation purposes just take note that you can use any figure okay and for it I’m going to actually do let us let us actually in turn a few is no point you let us use a triangle here okay let’s use a triangle because this something else I want to arm show when it comes to reflection arm that everything that everything preserve except the arm except the orientation of the object okay so now we have arm dis child here and as vertices a B and C and let’s say that the vertices of a is X equal to 3 y equal to 7 X equal to 3 y.o go 27 B is x equal to 6 y equal to 4 and C has coordinates X equal to 3 and y equal to 4 and let’s say I want to reflect this triangle in the y axis I mean the same thing would go for reflection in the x axis apply the appropriate formula or the matrix but not within a reflection in the arm y axis which mean that your y-axis is actually your mirror line okay are a reflection line and the same principle applies each there’s going to be a line segment that connects eight a prime and of course a pair must be over here and see parents to be over here and be parents to be over here ok because it’s represent a reflection reflection is going to be on the opposite side of your reflection line always doesn’t matter what reflection you’re doing a prime cannot be over here or BTM cannot be over here a CPM cannot be over here that’s be an opposite side of your arm reflection line or a mirror line and in fact it has to be on the line segment a perpendicular line segment okay that connects ate a prime and again as I said before the a and a PM has to be equal distance from the mirror line so notice note that your image your object pointer a is one two three units from your mirror line and therefore a prime is going to be 1 2 3 units from your mirror line so this is going to be a prime and b / b is 1 2 3 4 5 6 units from the mirror line and therefore B prime is going to be 6 units from the mirror line 1 2 3 4 5 6 units be primary somewhere here B prime and then C prime is 1 2 or 3 units and therefore so C is what two three units and his CPM is going to be 1 2 3 units okay the simple connect dots here to form your image triangle okay I let me just write the coordinates of

armed a prime which is X equal negative tree and y equal 274 c prime is X equal to negative 3 and y equal to r 4 and 4 B prime is X equal to negative six x equal to negative 6 and y equal to 4 X equal negative 6 y equal to 44 be prior okay and note that um this line segment here that connects a connects a to a prime is perpendicular to the y-axis which is a mirror line with as expected and again as I said before the mirror line would be a perpendicular bisector okay which by said line segment a and a plan and that’s going to be the same for every other points okay there’s a line segment that connects c to c prime and from and B to B prime all right and all of those line segments we but perpendicular to your mirror line and they you know of course Merline is a perpendicular bisector but if you were to take the coordinates of say a and a prime I look at a has coordinated 37 and you look at a primer has called it negative 3 and 7 what you notice that they the sign of exchanges from 3 to negative tree and the arm value of the y-coordinate stays the same seven and seven and that is not a surprise because in fact this line segment here that connects ate a prime is has um an equation arm y equals seven because the past should update the value of seven okay let me just erase their to show that it passes through the value of seven okay and therefore it has equation y equal to seven so any point on this line segment here is going to have a value of seven a y-value of seven so now we can develop a general formula for a reflection in the y axis and of course they are the y axis also have our equation X equal to 0 ok that’s the equation for the y axis so in an exam if they say find the reflection in the in the end the line x equal to 0 you know that you’ll know that that is the same thing as your y axis so the formula but we represent a reflection in the y axis or the line x equal to 0 you can take some point say p from general point p that has coordinate X Y I reflect it in the y axis and you see that the the sign of exchanges but y value stay the same that’s what we have here for each point the sine of X changes from positive to negative and y values of stay the same all right Y values did the same and the sine of X changes from positive 6 and negative 6 okay so this is a formula here that represent a reflection in the x axes or a reflection in line x equal to 0 now in terms of properties of reflection note that if we go back to my notes here in terms of properties of reflection the length is preserve the air is preserved parallelize them is preserved the error racial order of points and this one thing that is not preserved on the reflection and but it’s called orientation

okay so looking at this chart earlier which is your object triangle and this drag which is an image triangle you can see that arm the length of a and it’s a AC and a pelham and CPM the same bc and I’m BPM and c prime and a B and a permanent and B Prime the same length the angles are the same this argues equal to this angle this angle is equal to this angle and this angle is equal to this angle okay so everything is the area of the triangles are the same okay the parallelism of the triangle are the same okay this line is parallel to a prime C prime okay this line BC is part of this line okay and the parallelism of the lines are the same the air are the same okay now what is different in terms of arm a reflection is orientation of the object and your image and when talk about orientation are talking about clockwise or anti-clockwise so note that a b c the object ABC is going into a clockwise direction okay a bc that’s the orientation for the object all right and if arm you look at the orientation of the image a b c a b c is going into what I call an anti-clockwise direction so ABC is going in that direction and ABC ABC is going in that direction this is going into a clockwise direction and this is going to do an anti-clockwise direction so the object is clockwise and the image is art the clockwise so the orientation changes under a reflection arm under any reflection okay so for the transformation called a reflection the orientation changes and this is very important because if for example an exam you’re giving a arm you given an object say say for example a triangle or a rectangular or square or anything any plainer figure all right and you’re given its image are you add to find the transformation that transform the object to the image all right one of the first thing I want one of the things I want to look for right is the orientation and if what I’m saying that the orientation changes between object and the image then the transformation is a reflection because in fact reflection is the only transformation where the orientation changes out of the four main types of transformation translation reflection rotation and enlargement reflection is the only transformation with orientation changes from clockwise to anti-clockwise our vice versa all right and that is a big clue in an exam once you see that the orientation changes between the object and image then no question asked discharge formation is a reflection and then you can you know state in the exam that need transformation is a reflection okay so this is a formula and in fact there is a 2 by 2 matrix that represent a reflection in the y axis and this is your 2 by 2 matrix okay you have negative 1 100 okay and if you were to take each point of this object arm triangular each vertex of the object

triangle all right you can you can write it out like this ABC all right and writing all the values for example 37 37 b is 64 and c is 34 i can multiply those two matches because this is a two by two and this is a 2 by 3 2 rows three columns and therefore these numbers are equal to 12 and therefore you can multiply these matrices to get what you will get what we call a prime B prime C prime which is going to be a 2 by 3 matrix now to represent your image triangle once you plug the a prime B prime C prime connect the dots are connecting vertices to get it or if you’re not comfortable with multiply matrices when multiplying all the points well it can do is to multiply the 2 by 2 matrix to each vertex okay so you can have 100 1 and multiply that by 4 first of all you multiply by a which is trees seven to get a prime I do this for every every single vertex of the arm object to get your arm your image okay so that’s it for a reflection in the x-axis now let’s not move on to a reflection in the say in the line y equal to X now generally speaking the line y equal to X some line that is like this okay it’s actually forms in 45 degrees angle from your x axis and this is the equation y equal to x all right and again you can take any point here say for example here call this point I’m a witch has coordinate X equal to 1 y equal to 4 and again to find a prime is going to be equal distance from your your reflection line so this y equal to x is actually a reflection line or your mirror line okay and as I said before arm and a prep is going to be equal distance so this is one and a half the end this is going to be off and that’s the one they’re all right a prime is going to be somewhere here which has coordinate X equal to 4 and y equal to 1 and if you were to look at these two coordinates here you can see that a is 14 and a prime is 41 so what happened is that the coordinates switch its position okay this wire this this site is x value I’m in the object point this is the object point again okay and this is the image point so the x value here becomes the Y value here and the y value here becomes the x value here and that’s going to be the case for any reflection in the line y equal to X where the coordinates of the object simpler switches position or switches arm values to get the arm the image position and therefore we can write a formula for a reflection in line y equal to x and if we take some general point XY that is no reflected to get p prime which is simply YX ok so what you do is take the point XY and to get P Prime you know simply switch the coordinate of XRF of P so to use my a and a prime example I have my original point a which has coordinated one and four and if I want to find a prime what I do is simply switch the coordinates ok so this for you become my x value and this one year become my Y value so this is a formula that you can use to find a

reflection in the line y equal to x and of course everything stays the same in terms of the line segment that connects eight a prime the line segment it’s perpendicular line segment a and a prime is equal distance from your reflection line or a mirror line ok everything see the same that’s going to be the same thing that’s that’s going to be the same properties for any reflection and of course there’s a two by two matrix that represent a reflection in the line y equal to X or X equal to Y the same thing this is your two by two matrix 0 1 1 number right it’s over clearly so you guys can see it it’s 01 10 okay so this is the two by two majors that represent a reflection in the line y equal to X again if you are planning to pry the point object point a which has coordinate X is 1 Y is 4 and we were to multiply this to get a prime multiply these two matrices 0 to get a prime you will see that you get X equal to 4 and y equal to 1 which is your a prime is this is a general 2 by 2 matrix that represent a reflection align y equal to X now let’s move on to a reflection in the line y equal to negative x which is going to be this line here with gradient negative 1 okay joris properly and again it can use the same idea here if we take some point say for example here okay I wish I this college point q all right I just coordinate X equal to negative 1 and y equal to three and if i want to find Q prime again this is your reflection lining y equal to negative x this is a mirror line and again the point has to be equidistant from the mirror lines if if Q is one unit here well you could say one space here from the mirror line therefore Q prime is going to be 1 space 2 as well as the Q problem is going to be here which has coordinate X equal to negative tree and y equal to 1 and again the line segment p and p prime is perpendicular Tori arm reflection line y equal to negative x and of course the reflection line is perpendicular bisector tool arm is perpendicular to the line segment Q one Q Prime and I can I said before that’s going to be the same for any reflection now if you look at the points arm are the coordinates of the object point here and your image point Q prime right your object point has negative 1 and tree and here you can see that this is negative 3 and 1 so what’s happening here is that the you first of all you switch position I also switch aside okay so a general reflection in the line y equals negative x this is a formula in the line y equal to negative x if you take some point P that has coordinate X Y okay what is simple do to get P Prime you will have negative y 9 negative x so you switch the coordinates and then change the sign ok so this Y value becomes this x value and this x0 a leprechaun why the y value and then you switch the design so if you were to use my arm point here Q which has coordinate negative 1 and 3 then to find Q Prime based on this formula here right to find Q Prime based on a formula here all right you switch the day the values for Q and then it changes I’m set then a chain assign simultaneously do the two things at the same time switch and then to InDesign the switch first in a chain a sign that becomes negative tree and

then then this becomes 1 right so negative tree and one okay so remember you doing two things simultaneously switching and then changing the sign as you as you switch okay and this is of course the formula and of course there’s a 2 by 2 matrix that represent a reflection in a line y equals negative x and this is a 2 by 2 matrix 0 negative 1 negative 1 0 and as you can see if you take your point Q negative 1 and 3 to find Q Prime it is simple multiply these two matrices and you’ll get negative 3 and 1 so now let’s do a a different type of reflection okay call a reflection in the origin so this is called a reflection in the origin okay now if it takes up part here say for example point a that has coordinate X equal to y equal a tree and if you want to find a prime which is a reflection in the origin so you know your origin is now considered your you no reflection on point that align with a point okay then if you want to find a prime all this simple do is just change the sign okay so this is not become negative 2 negative 2 and negative tree so a primer is going to be here which has coordinate negative 2 negative tree and if you read a complete compound compare these two coordinates here X negative 2 and y equal to tree and you see that the reflected point or the image point will be X equal negative 2 and y equal to negative tree or doing a simple change the sign and just for information purposes this same reflection actually represent a rotation arm our clockwise or anti-clockwise rotation of 180 degrees ok but once you get to rotate in the next videos transformation transformational German 22 then you will see that it is in fact the same thing and of course you can you know have a general formula here I represent a reflection in Jin where the origin is actually a point of reflection so if you take the point XY general point XY and if you want to find P Prime okay a general point P that has coordinated XY you want to find P prime then all they do is just simple change the sign of your original arm point ok so this exit become negative x and this wire become negative Y and of course if this was negative x then this would be positive x and if this was negative Y this will be our positive Y and this represent a reflection in the origin where the origin is not your point of reflection and of course there’s a 2 by 2 matrix that represent this type of reflection and this is a 2 by 2 matrix here ok negative 1 0 0 negative 1 and of course you can apply this to your point a and you will see you’ll get the same a premiere which is negative 2 negative 3 once you multiply this matrix these two by two matrix to your a we will get a prime which is negative two and three okay now in notice that i’ve been using just a point again you can use any object arm arm to and apply the same formula to each vertex of the object so if this was a triangle for example you apply the formula to each vertex of the triangle or if it’s a rectangle you’ll apply the formula to each vertex of the rectangle I will get the same reflection arm in

the origin all right so if this was a triangle up here in this first quadrant then you will see that it will be reflected in the origin and will be somewhere down here ok so the same principle applies so now let’s move on to a different type of reflection which is called a reflection reflection in the line x equal to B where B can be some real number some you know constant value okay so let’s say that we have say B say let’s say we have some point here here which has come you know coordinates so it’s coordinate X equal to 5 and y equal to 3 and I want to reflect this point here at this object pointer in the some line some vertical line okay that has that has equation of this vertical line near the equator of this vertical line here will be X equal to 3 where X equal to B where B where B is a constant so a real number okay in this case B is 3 so again the same principle applies arm you know to find a prime is going to be equal distance from your arm your mirror line so this equation this line here X equal a tree is in fact the mirror line or your reflection line so if you want to find a prime then it’s going to be this is 12 units from the mirror line then this a prime is going to be 12 units from the mirror line and therefore it’s going to have coordinate x equal to 1 and y equal a tree okay and of course if you look at if we compare the coordinates here this is 53 this is 13 but notice that the y-values stay the same and that makes sense because there’s a line segment I connects a the a prime or a prime to a and in fact if we’re to extend this line segment here you can see that this line segment their passes to the y value this is your y-axis year pass to the y value of tree which means that this line segment here as an equation of y equal to tree so any point on this line segment here will have a y-value of tree that is why you have this here okay now note that and something that we have always been seen that the distance from here to here is the same and a distance from here to here is the same and in fact they call this the object distance this is the object is the distance from your mirror line is called object distance this this line segment here from the mirror line to the object is called object distance and this line segment here from a prime or your image to your mirror line is called the image distance as you can imagine image distance and as I said the fear before that object distance and image distance is going to be the same as well that’s what I’ve been saying from the very first part of this lesson that object distance and the image distances is the same okay so in fact to find a the arm the object distance it’s actually going to be X minus B okay so it’s simple does take your arm x value here is subtract B and that’s going to give you two so if we take this x value of 5i a sub B which is tree that’s going to give you two and as you

can see this is 12 units from the area aren’t they the mirror line same thing here if you want to find the distance from here from the mirror line to the image then it’s going to be X minus b2 as well all you get a simple take arm your x value here I a subtype b and it’s actually going to be the absolute value of X minus B because in fact you’re we’re talking about distance here that has to be a positive value so if you take one is subtract tree then you’re going to get negative two we cannot have a negative distance so in fact is the absolute value of this which mean the distant has to be too and as expected because if this is too then this has to be too because the two distance has to be equal distance apart okay and if this is X minus B and this is X minus B then if you want to find you know technically you can say that that I mean if this is X minus B distance from the arm the mirror line and this is X minus B distance from the mirror line then technically if you want to find the distance from arm you know from the image point for example from a it from a prime okay or a prime from a is just simple 2 times X minus B so if you’re at a I want to find a distance between a and a prime which is all of this year it is multiply that by 2 times X minus be all right because from here to ear is X minus B and from here T is X minus be there for the distance from here the all of here is our X minus B it’s 2 times X minus B and therefore if you want to find your image x value after they are the reflection it is simple take X minus 2 times X minus B and this will actually give you the x value for your a prime of 1 okay because you know that they we are not worrying about the Y values because under this reflection the Y values will stay the same tree and tree Alton I’m trying to find here is the the X values are after this reflection in the line x equal to B and I’m trying to find a general formula that represented that x value or to calculate an x value and what I’m saying is that the calculate that x value is going to be X minus 2 times X minus b2 finally on the x value for the image point here of one okay so we can just you know expand this out and simplify it and then we can add X minus 2 times x is 2x + 2 x arm negative 2 minus so negative two x negative B that’s going to be positive to be all right then here you have no ex- arm to XS minus X plus 2 P and we can actually rewrite this as to be minus X this and this is the same thing so what I’m saying is that this one here represent 2p minus X in general and let’s check it if we want to find the x value for arm a prime let us take two times B with B is 32 times p all right minus the arm the x value from your object which is five and 22 26 minus five is equal to 1 this xci represent the x value from the object and that makes sense because after all you’re trying to find an x value for the image using this formula and as you can see two times B which is 2 times 3 with 6 minus the x value from your object which is five you get one and that’s the one you get there so this is the general

formula to find the x value of your image arm for a reflection in the line x equal to B where B can be some constant okay so in general in general if you are given a reflection and the line x equal to B and you have some general point here XY I want to find P prime then to find x value of P prime is going to be 2 times B minus X that’s what we calculated here that’s for your x value of the image and then your wife Isla will stay the same same thing that we have here we have a last 53 and be sorry and a prime is 1 & 3 notice that the Y values stay the same tree and tree but if you want to find the x value is simply uses formerly 2 times B minus X will be represent the value of the equation of the vertical line here which is actually perpendicular to the y-axis sorry which is actually parallel to the y-axis and it’s perpendicular today the line segments and connects n that connects our a and a prime because in fact this line segment here this line here X equal to choose a are perpendicular bisectors as we mentioned before the mirror line is always a perpendicular bisector right and so this be in your formula represented this tree here okay the equation for they are the mirror line and this exia comes from the x value of your object point and of course you use this formula to calculate your x value for your p prime and then your Y value stay the same so if why was she here then y will be 3 here okay now unfortunately there is no 2 by 2 matrix that you can use arm to get the same thing as a formula and in fact in general with two by two you only can get two by two matrix if your reflection line is some line that goes through the origin here your reflection line does not go to the origin this is the origin here I you notice that for all the other reflections that we have been doing all of those lines goes through the origin 0 0 for example x axis go to the origin and we have a 2 by 2 matrix for that the y axis go to the origin and we have a 2 by 2 matrix for that the line y equal to X goes to the origin remember two by two matrix for that and the line y equals negative x are negative x equal to Y goes to the origin are there were two by two matrix for that but notice that this mirror line here X equality does not go to the origin and therefore there is no 2 by 2 matrix that you can use to represent this all right formula here so technically you have to memorize this formula if you’re given a reflection in the line X equal to B where B is some constant number some real value so this is a formula that will have to use that you have to use ok again i’m using a point here but in fact you can use again you know any point during the vertex that’s on a polygon for example a rectangular square a triangle right ethical and X goddess apply this formula to the vertex of each of these objects to get the image arm vertex no no no I’m going to look at one last reflection okay and that’s going to be a reflection in the line y equal to C okay so reflection in the line y equal to C is going to come from the same idea arm for a

reflection in the line x equal to be the same idea that we’re going to use here but let me just um you know mate it’s pretty quick to avoid the video being too long again it’s going to be the same idea as X equal to B again there’s no two by two matrix because this line near this mirror lining it does not go through the origin so this nut there’s not going to be any two by two matrix and let’s say that this is our line y equal to tree where see here see again here the line y equal to c where c is a constant any real value i could it could be in a positive or negative okay so this line here could be y equal negative 3 or y equal negative for all right and you know and again if you choose some point here say here let’s call this s and has coordinate X equal to 3 y equal to 6 and this is again 1 2 3 units from the reflection line this is not your reflection line or a mirror line okay reflection line or a mirror line and therefore if you want to find s prime s prime is simply going to be one two three units armed from the mirror line so one two three so as fam is actually here which has coordinates X equal to 3 and y equal to 0 because remember this is your x axis and any point on the x-axis will have a coordinate a y-value coordinate of zero because in fact as I said before this x axis has equation y equal to 0 so any point on this x axis will have a coordinate of 0 a y coordinate of 0 our Y value of zero okay and of course you can connect a line segment that you can see that this line segment is perpendicular to your mirror line which is again before a mere liye teaser you know a perpendicular bisector the same idea as before and as if you were to do the same that we do it for X for a line X equal to B you will see that you can have a general formula for reflection in line y equal to see all right and if you take some general point p which has coordinate X Y right then define p prime it’s going to be X will stay the same and then you’re going to have 2 times C minus y to get your to get your arm y value for P Prime all right I looking at s and s prime you can see the s the the X values stay the same tree and tree and to find a y-value uses formula right here which is two times c minus y where see again is your you know that comes from your equation for this horizontal line here y equal to 32 C values three okay so two times C is three minus y know this wire comes from the y value of your original arm point the object point so your y value is six ok so 2 times 3 is 6 minus 6 is equal to zero let’s say you get at zero right there so this is a form that you can use to find a y-value arm of your your image point ok so this is this is that’s it pretty much arm so we have looked at some you know common types of reflection let me just list them right here so we look at our reflection reflection in the

x axis y axis line y equal to X line y equal to negative x and the line y equal to see that which you just did and in line X equal to be where CNP can be any constant any real number positive or negative okay so these are 1 2 3 4 5 6 types of reflections that we look at and each other these are lines here represent they are some perpendicular bisectors we’ve said before and in fact each other each of these lines here is the mirror line for that reflection so the x axis will be the reflection for the armor so the x axis would have been the mirror line or the reflection line for a reflection in the x-axis the I axis would be the reflection are the mirror line for the reflection in the y-axis and so on so forth okay now these are the common types of reflections that you you know you can expect to see in an exam at the CSEC level for example but I want to keep I want you to keep you guys to keep in mind that the the line of reflection or reflection line in fact can be any line okay any line okay I am we are just using these you know 10 because they feel a common but just keeping just keep in mind that in fact they are reflection line or the mirror line can be any line and the line at all you want to make it to be any light okay what we are just using these six types here because those are the ones that will feel a common arm especially at the arm the sea second level so memorize these type of reflections okay x axis y axis y equal x y equal negative x y equal to C and X equal to be alright and memorize a formula or RB 2 by 2 matrix up until this point this four year has a two by two matrix remember that these two year does not have a two by two matrix ok so that’s it for this video um this is all the information that was covered in class and as you can see they’re very long video and a lots of information arm covered on these two types of transformation now i’m going to do a next video which transformational german 22 that’s going to cover another to another type to other types of armed transformation and ask of course i’m going to give some additional information at the end of that video ok so that’s it for this video and again this is a video for lesson for students will miss the class a live class or new students arm would like to catch up with the class ok so that’s it for this video