# SHSAT Coordinate Geometry

All right so now we`re going to go over a coordinate geometry and this will be very much related to what we`ve been doing before with functions and coordinates x and y points and I`ve even mentioned this before so let`s just jump straight in how do we represent points on a coordinate system points are numbers that are in a pair so this would be a point three comma four five commas six would be a point negative ten commas zero would be a point but how do we represent this graphically well we would use a coordinate system a rectangular coordinate system where we draw two perpendicular lines where perpendicular means they meet at right angles

This is a right angle right angle right angle right angle which is 90 degrees.

This horizontal axis would be the x axis and this vertical axis would be the y axis now all it is is just a sequence of points we talked about the number line the real number line that`s what the x axis is and the y axis is the same except they are perpendicular to each other and even though they have the same points they both are needed to make up a point a coordinate point so the point on the x axis will be all of these points and the same for the y axis however you need them both to define an actual point on the grid so you need two points to define a point on the grid all right so let`s find two points one on the x axis and one on the y axis and we`ll see how it makes this coordinate.

How about this we`ll make that one two three four five and this will make negative one negative two negative three will make this negative one negative two negative three and this one two three.

What did I do here well as you can see some part of the y-axis has positive points and some have negative points and the same for the x-axis now this is why that`s so so if we cut up the coordinate axis into these four quadrants one two a three four and it has to be in that order we see that this is the positive x-axis positive y-axis so that means any point in this section will be positive x positive Y so x and y will both be positive and then for the second coordinate is the second region we`re going to have a negative x and a positive Y so we`re going to have numbers like negative two commas three so the X points are negative all of them over here but the Y points are positive and for quadrant three our X`s and Y`s would be negative x and negative Y so we`ll have something like negative two negative three and four the fourth quadrant we have positive x and negative y so something like three and negative four so one way of looking at it is if we go vertically up that`s the plus y direction if we go to the right it`s a plus X direction which only makes the opposite directions negative so down will be negative y and left will be negative x so if I were to find the point 2 commas 3 then I would go to the X x-axis first because our points are X comma Y always so I would go find 2 here 2 is right here and then I would go up 3 and so that point is 2 comma 3 in other words the x value is 2 which is that point and the Y value is 3 which is this point so we draw the point at 2 comma 3 go across 2 and up 3 now what if I wanted to find the point 3 negative 5 then I would find the x value that is 3 which is here and then I would go down negative 5 and so that would be somewhere here this would be 3 and negative 5.

Great so that`s how we label points on an axis so let`s let`s try some more problems concerning that.

Here I have my positive x-axis positive y-axis and here is the origin or the point zero zero now I want to graph these points P the point P is zero comma one the point Q is zero comma 2 the point Z is 2 comma 0 and the point F is 1 comma 0 ok let`s graph point P point P is 0 comma 1 so where is 0 on the x axis it is here now where is 1 on the x axis well we just go up 1 and then will be here so 0 and then up 1 so our point is here this would be P which is 0 comma 1 now it`s fine Q 0 comma 2 well we`re here at 0 and then we go up 2 so that would be around here so Q would be 0 comma 2 now let`s find Z which is 2 comma 0 so we need to move over to the right 2 and so here is 2 and then we need to go up 0 so we`re right here we`re wherever it needs to be we are already here so 2 comma 0 and now let`s do F which is 1 comma 0 which is 1 1 would be here 1/2 halfway between 0 and 2 and then go up 0 so that`s it so F is 1 comma 0 now these are our points and I`m just eyeballing where 2 is and where 1 is and that`s fine so let me highlight our points so we have 1 here we have one here one here one here and one here great so these are our points now what can we see about these points can we find what the distance is between any two of these points we know how to graph them but what if I want to find the distance between the two points let`s find the distance between how about Q and P P is zero comma one Q is zero comma – well obviously we are looking for this distance here between Q and P well P and Q both have it next coordinate or an X point of zero but it goes up one from P to Q so this distance here is just one so the distance between P and Q is one because we only went up one it`s a vertical distance so we can represent the distance PQ as this so we have PQ this line segment the distance of that would be one.

I`m just going to write one unit and so now we know it`s one because we just went up one now what about the distance between F and Z between F and Z well again we just went across one unit we went to the right one unit so FZ this sense is also one unit now how can we find these distances without looking at the graph well let`s see what PQ is in terms of its points so P is 0 1 and Q is 0 2 so we have 0 1 and and 0 2 well this is the x value this is the Y value and likewise here now I both I have X`s and Y`s on both places but let me label the X`s and Y`s I`m going to call this x1 y1 and x2 y2 they`re both X&Y points but I just want to label them so that I don`t get confused between the X points and the Y points so the way to actually find the distance without counting it on the graph is to use the distance formula and let`s see if we can use this formula to get PQ as one unit so the distance formula that distance formula is equal to D for distance is equal to the square root of Y 2 minus y 1 plus X 2 minus X 1 but then you squared each quantity.

The square root of Y 2 minus y 1 squared plus X 2 minus X 1 squared so if we have two points and I`ll write you need two points if you have two points you plug it into the d this distance formula and you get the distance between the two so let`s use our two points we have 0 1 & 0 2 so 0 1 X 1 is 0 and y 1 is 1 X 2 is 0 Y 2 is 2 so D so D equals the square root of Y 2 minus y 1 which is 2 minus 1 squared plus is 0 minus 0 squared and then we would get the square root of just 1 squared all right because this cancels out and that would just equal 1 so yes we did get a 1 unit a distance of 1 unit and notice that the X values the X distance or the horizontal distance was 0 doesn`t that make sense between P and Q we don`t move any horizontal distance we just move up which is why the only nonzero quantity here would be 2 minus 1 or the y parts of it where the vertical parts y 2 minus y 1 now let`s see if we get the same thing for FZ the F point is let`s see what that point was that point is 1 0 and the Z point is 2 0 so let`s do that here we have 1 0 and 2 0 so let`s find the distance between the two points well this is x1 y1 and x2 y2 that does not change ok so D equals the square root of y2 minus y1 so that 0 minus 0 and see the Y parts are 0 now plus what is 2 minus 1 or X 2 minus X 1.

Then you square it so that equals the square root of 1 squared which is again one unit so we got the units we got the distance as one unit in one unit great but that`s not that interesting.

I want to find the distance now between points that are not vertical or horizontal that are diagonals or that are added at an angle that it`s not particularly horizontal or vertical so let`s try and find the distance between Q and Z how does that sound this distance here all right so let`s try and find that and that distance right now that`s that distance and we can tell it`s going to be larger than one unit it just looks bigger so let`s find it P is 0 Q is 0 2 and Z is 2 0 so those are our two points ok so we have Q as 0 2 and Z as 2 comma 0 so here`s x1 y1 x2 y2 and the distance is the square root of y2 minus y1 so that`s 0 minus 2 what 0 minus 2 it`s negative 2 squared plus x2 minus x1 which is 2 minus 0 which is 2 squared so let`s simplify we get the square root of well negative 2 squared is 4 plus 2 squared is 4 and so we get that it equals the square root of 8 well we can simplify the square root of 8 by just finding what perfect square is a factor well that`s the square root of 4 times the square root of 2 and that equals 2 radical 2 so the distance Q Z is 2 radical 2 units and 2 radical 2 is bigger than 1 so notice how we just found this black distance here by not counting because it would be really hard to count we use the actual distance formula and likewise we can find the distance between between P and F ok so 0 1 and 1 0 now we know that`s going to be less than the distance the distance QZ but let`s find it anyway so P and F so PMF that would be let me just write it like this so we have P as the point 0 1 and F as the point 1 0 and so let`s label our x ones and y ones so we have x1 y1 x2 and y2 and so our distance would be equals to the square roots of y 2 minus y 1 or 0 minus 1 which is negative 1 squared and plus x2 minus x1 squared which is 1 minus 0 squared and that equals the square root of well negative 1 squared is 1 plus 1 minus 0 squared is 1 and that that equals the square root of 2 so the distance PF is a square root of 2 and really all you`re doing is plugging into this this distance formula which which you should write down and memorize so i`ll a box that again that is right over here so here we go this is the distance formula all right so don`t forget that.

Great so we have all the distances now let me write the distances out just so we can see them in one page we have we have let`s see PQ is equal to 1 we have FZ equal to 1 and we have qf or started qz equal to 2 radical 2 and we have PF equals equals to radical 2 now do we did we miss any combination is there any other combination here of lines that we could have used yes of course we could have used how about q qf we could have used this or we could have used pz like that but it`s

Because we have the gist of it now we know how to find distances between two points now i want to show you how to find the distance another way without using the distance the distance formula without using it directly ok we can actually derive the distance formula so let`s try that right now it`s pretty interesting so what if we have the same coordinate axes or rectangular coordinate system or a grid where we have X comma Y and now we have these two points so we have one here called P and one here called Q I`ll make P equal to 1 comma 1 and Q equal to 3 commas 3 so you can you can make these up as long as it makes sense, for example, you`re not going to put a negative number here, for example, this point cannot be negative three because negative three will be in this axis.

Now we have these two points and of course I`m asking us I`m asking you to find the distance between the two but pretend you don`t know the distance formula all you know is the Pythagorean theorem which you should know by now but I`ll go over it so all it says is I`ll do two as an aside if you have a right triangle like this a right triangle which means this angle is ninety degrees then you can find a relationship between this side which at this distance of a this hot the side which has a distance of B and this side which has a distance of C and now these lengths are all related but let me just label it this B is a length that is called the leg and this is also a leg but this C which is the longest length, of course, is called the hypotenuse hypotenuse.

This distance called the hypotenuse is opposite the 90-degree angle and now what is the relationship I promised it is that a squared plus B squared equals C squared so if you have any right triangle this follows for example if we have a right triangle with the side lengths of three and four that means that 3 squared plus 4 squared equals C squared and this is C so let`s solve for C what`s 3 squared plus 4 squared that is 9 plus 16 equals C squared and then plus 16 is 25 equals C squared if C squared is 25 then we know C is 5 so that means this length equals 5 and yes this length of 5 is bigger than both of the legs great so now that we know the Pythagorean theorem which I`ll label here Pythagorean theorem and this is something you should memorize we can now use that to find distances in our rectangular grid so I have these two points Q and P and I want to find the distance PQ or QP the way to do that is to break it apart into with vertical and horizontal distance so this will be the vertical distance and this will be the horizontal distance but both of these distances make up PQ so I`ll call this point L and so this is PL and this is L Q and now to find this this horizontal distance all we need to know is what this point L is and what is that point L well if this point here is three three that means that this is an x coordinate of 3 so we know that that`s an x coordinate of 3 but what`s the y coordinate well we know that this point here is 1 1 which means that this is a y coordinate of 1 which means that over here is a y coordinate of 1 it`s right across from it there`s no it doesn`t go down ok it just goes straight across as a purely horizontal line now we can just say that it`s 3 commas 1 all right it goes up 1 and across 3 or you can just think of it as across 3 and up 1.

The point is 3 comma 1 now that we have all three points let`s actually find the distances we`re not using the formula all we`re using this common sense if this is 1 1 and this is 3 1 how much time or how far is P from L well it goes up just to because we start at 1 and then we go to an x value of 3 so we move two spaces so this is 2 units or you can think of it as 3 minus 1 so this is a distance of 2 units so all right that this has a distance of 2 units now what is this distance here well we start at 1 and then we go up to 3 so that`s also 3 minus 1 which is a distance of 2 units now that we know this is basically a triangle or start a right triangle with a distance of 2 here distance of 2 here and we`re looking for P Q well let`s use the Pythagorean theorem so we have 2 squared plus 2 squared equals P Q squared and I`ll call PQ just PQ we can leave it like it`s a quantity so PQ squared what`s two squared plus two squared that`s eight.

That equals P Q squared and so we would take the square root of both sides and we get that PQ is the square root of eight or two radical two all right so great we learned two ways to find the distances between points and where do those two ways we have the distance formula if we use this the distance formula for the previous problem we would still get two radical two and the Pythagorean theorem and then we can think of a whole bunch of different problems with these these two methods to find distances between points there are so many different fields that use this concept and so let me give you another problem that uses both of these concepts so let me go and find where I had that problem and so that is going to be similar across yep right here great so we have points here this point is the school imagine you`re you`re living somewhere and there`s a school and then you walk some distance and then there`s your house and then you walk some distance and then there`s your work place and then you can go all the way to school my question to you is if you start at school how far do you have to walk from school to home to work back to school in other words if you`re at school how far do you have to walk to get back to school by covering every single location so if you do that problem all that`s saying is find the lengths between all the two points and then add them up so find the distance between school and home then home to work and then work to school and add them up so let`s use the distance formula why because we`re not so entirely sure that this is a 90-degree angle if it`s not then we cannot use the Pythagorean theorem so we have to use the distance formula so first thing we do is label the points this would be x1 comma y1 this would be x2 comma y2 and this we can make X 3 comma y3 great so which one should we do first let`s do school-to-home so all to home how far do you have to walk to go from school to home well we would use the distance formula so D equals and I`ll write the distance formula here is for reference since we`re learning it for the first time so y2 minus y1 squared plus x2 minus x1 squared so that would be x2 minus x1 squared and all of that`s under the radical sign ok so what is that distance that would be 5 minus 7 y2 minus y1 5 minus 7 is negative 2 we square that plus x2 minus x1 which is 3 minus negative 4 3 minus negative 4 is 7 so that`s 7 squared and that equals the square root of 4 plus 7 or 11 so the distance here is square root of 11 now what about the distance of home to work home to work all right so then that would be these two points here so we have D equals the square root of in this case y3 minus y2 and it doesn`t matter which order you do it in you could do negative 3 minus 5 or 5 minus negative 3 so y3 minus y I wanted to be negative 3 minus 5 or negative 8 squared plus what is now x3 minus x1 which is 3 minus 3 or 0 and then you squared it and so that will give you a distance of the square root of 64 which is 8 so this is a distance of 8 from home to work so let`s start labeling what we have we have this distance as a square root of 11 and we`ll call this feet and then this has a distance of 8 feet and now let`s find the distance between school and work so I`ll do that here school and work or school to work which is the same as work to school and what are the two points well we have school as negative 4/7 so I`ll write that here since we can`t see it and work as three negative three so three negative three.

Now we use the distance formula so we take the Y components and subtract them so that`s negative three minus 7 which is negative 10 and square that plus then 3 minus negative 4 which is 7 and square that and so we get the square root of 100 plus 49 and so the answer is the square root of 149 all right so now we have all three distances so let`s write that the square root of 149 and so since we have all three distances we just need to add them so the total distance the total distance is and we don`t have to find the simplified version but the total distance is D all right D total equals 8 C 8 plus the square root of 11 plus the square roots of 149 and so that`s how far you need to work to go all the way around to come back to school from school to go back to school you need to walk that distance.

Wonderful now I want to try one over one more problem that we can make up

We have the coordinate system again and this time not going to draw just one line I`m going to actually draw a shape so we have a let`s see we have this line up here this line across and this point is this point is B and this point is C now I want us to find the area of this rectangle.

We know that this is the origin of course and I`m going to tell you that this a point here is 2 comma 1 all right and we need to find the area is this possible it seems like we don`t have enough information but this is all we`re given so let`s think how can we find the area of any rectangle well we just multiply the length by the width so the length times the width gives us the area of a rectangle so I`ll write that up here if you didn`t know area of a rectangle that`s on the bottom is equal to length times width or L times W awesome so how can we find Ellen how can we find w well I don`t know what this point is or this point but I know these two points so using these two points I`m thinking I can draw a diagonal here and I know that distance if I use the distance formula and then I can find this length and this length or I can draw a line here and just use the Pythagorean theorem so let`s do the ladder so let me draw the line here ok great and now if this is 0 0 and this is 2 1 what does that make this distance that distance is just 2 because we went over to to get an x coordinate of 2 and now we know that this distance must be 1 because we go up 1 to get to a from B so now we know that this point here for B would be 1 comma 1 comma 0 and this point C now what is that well here we`re at 0 0 and then we go up 2 so that would be 0 comma 2 great so we have all three points actually we have a as to 1 we have B as 1 0 and we have C as 0 comma 2 and of course the origin is 0 0 now our original idea was to find the area so can we find the length now the length must be the length must be the line segment that is ca well see a we can use the distance formula or just use logic if this is 0 2 and this is 2 1 that means to go from C to a we must travel a certain horizontal distance and that horizontal distance is the 2 minus the 0.

That means that this is where we start and this is where we end we take the X components and subtract them so the final x value minus the final wive out on the finest the final x value minus the initial x value sorry so that would be 2 minus 0 which is 2 units ok and then the width would be the final Y value minus the initial y value so this is the final y value and this is the initial Y value if you jumped from here to here

That would be 0-1 but distances can never be negative so instead of negative one we`ll just make it a 1 so a B equals 1 and yes you can do that because all we`re looking for is a distance not an actual negative or positive value we`re just looking for a distance now we have the length and width so getting the area is just a matter of multiplying it so area equals L times W or lunk times width so the answer is 2 times 1 which is 2 units squared area is in units squared so that is our answer the area is 2 that`s that`s pretty interesting and since this diagonal cuts the rectangle in half this is 1 unit squared and this is one unit squared and we can even figure that out because this distance is 2 this distance is 1 so this is a triangle area of a triangle is 1/2 base times height so that`s just I`ll write this as an aside area of a triangle is 1/2 base times height and the base is this horizontal distance the height is this vertical so that`s 1/2 1 times 2 or 2 times 1

1/2 2 times 1 is just 1 which is interesting and makes sense because this is one unit and do the same logic here and this is one unit and so it makes an area of 2 units squared so actually this is one unit squared and this is one unit squared don`t forget the squared or else it`s just say distance not an area alright so in the next video we`ll do some more problems regarding this topic all right so let`s do some problems involving coordinate geometry here for number one we have a rectangular grid system or coordinate system with the quadrants 1 2 3 & 4 and in Roman numerals a is asking which quadrant is the point 3 comma negative 100 n B is asking which quadrant is 4 comma 5 in so positive video and we`ll go over it alright great so let`s go for part a that you did so 3 comma negative 100 so 3 is a positive x value so it`ll be around here negative 101 negative 100 is a negative Y value so it would be somewhere down here which means it would be in the 4th quadrant so the answer is fourth quadrant B which quadrant is 4 comma 5 in 4 would be on the positive x axis and 5 would be on the positive y axis so somewhere here or the first quadrant great now let`s look at number 2 here we have a coordinate grid system with a point D and the D is 3 1 ok so it`s asking to complete a square with the points ABC and D ok so you have a coordinate system and we I want you to make a square with the points a B C and D like like this so this would be a B C and D for example and then find the points a B and C since you already have D so basically make a square on this grid and then find all of the four corner points so pause the video and take a moment to try that problem and then come back alright so let`s go over number two so if we have D as 3 comma 1 and then we know that that from D to this x-axis would be 1 so if this is 1 and then this must be 1 because it`s a square ok so that`s 1 plus 1 this whole distance is 2 so that we know the side length of the square is 2 so all the other side lengths must be 2 so I`m going to extend this point over here until some other point and the point that we`ll find I`ll call it I`ll call this point C and this point B so we already know what C is that would be 3 3 as this point here negative one so we can go down one so that would be three comma negative one and now we need this distance to be two as well if this point is three on the x axis we can move over two and this would be one comma negative one and so then a would be up here all right so I realize this does not look like a square the reason is because I drew this length way longer than this length but all that matters are the points.

If you have the points correctly if you write them correctly then it should have side lengths that are equal therefore it`s a square so that a point would be well you`re on the one x coordinate and so we need to go up but how many points well from here to here is one and we know this whole length this two so we need to go up one point so this is one comma one great and so we made the square I`ll write it over again so it looks neater so we have we have this coordinate axis and x and y don`t forget to draw the arrows meaning it goes on forever and here we have 1 comma 1 for a and then for C we have 1 comma negative 1 for C and then or rather this is B and then for C we have 3 commas negative 1 and then for D we have 3 commas 1 and now this looks more like a square and so all the side lengths are 2 and the areas for if you wanted to do that four units squared thanks for watching the video please be sure to check out our book on Amazon to specialized high school test prep book this book comes with five full-length exams with answer explanations along with a manual to go ahead and better prepare you for the exam you will find the link on the video or in the description above have fun studying