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CCSS.Math:

two points of a two of the points that define a certain quadrilateral are negative four comma negative two so let's plot that so it's negative four comma negative 2 and 0 comma 5 and 0 comma 5 so that's 0 comma 5 right over there the quadrilateral is left unchanged by a reflection over the line y is equal to x over 2 so what does that line look like y is equal to x over 2 I'll do that in blue y is equal to x over 2 so when X is equal to 0 Y is 0 the y intercept is 0 here and the slope is 1/2 every time x increases by 1 Y will increase by 1/2 or when x increases by 2 y will increase by 1 so x increases by 2 y increases by 1 X is increasing bite increases by 2 y is equal it Y increases by 1 another way to think about it Y is always 1/2 of X when X is 4 y is 2 when X is 6 Y is 3 when X is 8 y is 4 so we could connect these let me try my best attempt to draw these in a relatively straight line and then I could keep going when X is negative 2 y is negative 1 when X is negative 4 y is negative 2 so it actually goes through that point right there and it just keeps going with a slope of 1/2 so this line I can draw it a little bit thicker now now that I've dotted it out this is the line this is the line y is equal to x over 2 and they also say that the quadrilateral is left unchanged by reflection over the line over the line y is equal to negative 2x plus 5 so the y-intercept here is 5 when x is 0 Y is 5 so actually goes through that point and the slope is negative 2 every time we increase by one our every time we increase X by one we decrease Y by 2 so that's so we go there we go there and we keep going at a slope of negative 2 so it's going to look something like this it's going to look something something like this it actually goes to that point and just keeps going on so this is my best attempt at drawing that line so that is y is equal to negative 2x plus 5 now let's think about let's see if we can draw this quadrilateral so let's first reflect the quadrilateral or let's reflect the points we have over the line y is equal to x over 2 so this is the line y is equal to x over 2 this magenta point the point negative 4 2 is already on that line so if you it's its own reflection I guess you could say it so it's on the mirror one way one way is to think about it but we can easily reflect this line over here this line if we were to drop a perpendicular if we were to drop a perpendicular we drop a perpendicular and actually this line right over here Y is equal to negative 2 X plus 5 is perpendicular to Y is equal to x over 2 how do we know well if one law if you have a if one line has a slope of M then the line that's perpendicular would be the negative reciprocal of this it would be negative 1 over m so this first line has a slope of 1/2 but what's the negative reciprocal of 1/2 well the reciprocal of 1/2 is 2 over 1 and you make that negative so it is equal to negative 2 so this slope is a negative reciprocal of this slope so these lines are indeed these lines are indeed I'm trying to reset these lines are indeed perpendicular so we literally could drop a perpendicular literally go along this line right over here in our attempt to reflect and we see that we're going kind of down down 2 over 1 down 2 over 1 twice so let's go down 2 over 1 down 2 over 1 twice again the reflection we do that in that same color the reflection of this point across Y is equal to x over 2 is this point right over there so now we have three points of our quadrilateral let's see if we can get 1/4 so let's go to the magenta point the magenta point we've already seen it's kind of it's it's it's sitting on top of y equals x over 2 so trying to reflect it doesn't help us much but we could try to reflect it across Y is equal to negative 2x plus 5 so once again these lines are perpendicular to each other actually let me mark that off these lines are perpendicular so we can drop a perpendicular and try to find its reflection so we're going to the right 2 and up 1 doing that once twice three times on the left side so let's do that once twice three times on the right side so the reflection is right there we essentially want to go to that line and however far we were to kind of the left of it we want to go that same that bottom left direction we do want to go in the same direction to the top right the same distance to get the reflection so there you have it there is our there is our other point so now we have the four points of four points of this quadrilateral four points of this quadrilateral are they are or the four side let me actually just draw the quadrilateral we have our four points so this is one side right over here this is one side right over here this is another side right over here and you can verify that these are parallel how would you verify that they're parallel well they have the same slope if you get from this point to that point you have to go over so your run has to be four and you have to rise one two three four five six seven so the slope here is 7/4 so slope here rise over run or change in Y over change in X is 7/4 it's and over here you go one two three four so you run four and then you rise one two three four five six seven so the slope here is also seven over four so these two lines these two lines are going to be these two lines are going to be parallel and then we could draw these lines over here these lines over here so this one at the top the one at the top right over there and what's its slope well let's see we go from x equals 0 to x equals 8 so we go down our change in Y is negative 1 every time we increase X by 8 so this is the slope of slope is equal to negative 1 over 8 and that's the exact same slope that we have right over here negative 1 over 8 so these two lines are parallel as well this line is parallel to this line as well so at minimum we're dealing with the parallelogram but let's see if we can go even more specific because this kind of looks like a rhombus it looks like a parallelogram where all four sides have the same length so there's a couple of ways that you could verify that this parallelogram is a rhombus one ways you could actually find the distance between the points you could use that we know the coordinates so you could use the distance formula which really comes straight out of the Pythagorean theorem or even better you can look at the diagonals of this rhombus or you can look at the diagonals of this parallelogram we're trying to figure out if it's rhombus and if the diagonals are perpendicular then you're dealing with the rhombus and we've already shown that these diagonals at this diagonal this diagonal and this diagonal are perpendicular are perpendicular they intersect at right angles and so this must be a rhombus