high school co-ordinate geometry with a quadrilateral and two kites
- Thread starter davee
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I'm a middle school student and I want to solve this
I can solve the first question but I don't know how to solve the second and last one.. help me please
So the exercises with which you are having difficulties are the following:
2. A kite, EBCD, is formed by joining the vertices B(3, 2), C(8, 3), and D(9, 8) to the point E(1, 10).
(a) Find the equation of the axis of symmetry of the kite EBCD.
(b) Show that this axis of symmetry goes through the midpoint of the diagonal BD.
3. Consider any point F(x, y) on the line EC. Assuming that F and C are two distinct points on the line, show that FBCD is always a kite.
For (2), have you drawn the picture? (a) Which line segment represents the axis of symmetry? (b) What is the equation of the line that contains the segment BD? Where does this lead?
For (3), what have you drawn? What are your thoughts? Where are you getting stuck?
Please be complete. Thank you!
2-a. The axis of symmetry: EC,
gradient: 3-10/8-1=-1 so -1, and the equation is y=-x+11
I find the equation of this, check it, please.
2-b. I think because the midpoint of BD and the midpoint of the axis of symmetry EC are the same this axis of symmetry goes through the midpoint of the diagonal BD. I think I can use the midpoint equation to solve it. Is it right..?
3. I'm stuck from the beginning.. I cannot understand the problem itself.
2-a. The axis of symmetry: EC,
gradient: 3-10/8-1=-1 so -1, and the equation is y=-x+11
I find the equation of this, check it, please.
2-b. I think because the midpoint of BD and the midpoint of the axis of symmetry EC are the same this axis of symmetry goes through the midpoint of the diagonal BD. I think I can use the midpoint equation to solve it. Is it right..?
3. I'm stuck from the beginning.. I cannot understand the problem itself.
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I get the same line.
No, they are not.
The Midpoint Formula will be helpful, yes. I'm not sure what you mean by "solving" this question, since you are actually instructed to prove the given relationship. Find the midpoint; show that it lies on the specified line.
Draw the picture. Place a dot on line EC that is not on the endpoint C; name this dot "F". Look at the resulting figure FBCD. Where does this lead?
The Highlander
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Dear @davee,
Two other people have now suggested that you need to draw a sketch/diagram for this kind of problem. In your last thread (see here) I also suggested that a diagram was the way to approach your problem but you failed to respond.
When there was no response from you I added a sketch into the thread and it illustrated how it made the problem much clearer and showed how to solve it quite simply!
Why are you not taking this advice?
Again I will help you (by starting you off with the diagram below).
You don't have to use my diagram, you can draw/sketch the problem on squared paper yourself and then upload a picture of your work but, if you do want to proceed with my diagram then just right-click on it and choose: "Save image as..." to download it to your device.
I constructed the figure in the MS Paint™ application which should be readily available to you for further work on the picture (though any image handling App should be suitable; Paint.net is an excellent (and free) one to use.)
I used Arial, 20 point, (emBOLDened) for the letters (ABCD) and Times New Roman, 48 point, for the dots (full stops). The lines were drawn using the 2nd thickness in the drop-down menu.
Please get back to us with your diagram of the situation described in Q.2 and your attempts to answer (all) the questions; confirmation that your answers are correct or further help (if needed) will then be offered.
Now, please tell us what your answer was to Q.1 and what the "reasons" you gave were. (The dotted red lines in my picture are a strong hint for answering that question properly!)
Have you checked on the definition of a kite?
Please look at this web page and pay particular attention to the TWO sections I have highlighted (in purple) on it for you. (Scroll down to see the second highlighted part.)
The reasoning that you (should have) used in Q.1 can also be used to answer Q.3 when it comes to providing the required 'proof'.
2b. What are the coordinates of the midpoint of the diagonal BD? Consider them further with refence to your equation for the line EC (now that you have constructed it on your diagram ?).
3 Assuming you have studied the "MATH is FUN" web page I directed you to earlier, can you now come up with any arguments that prove what is required? Hint: Look at my red dots (...) again.
Thank you for your reply
I've solved other problems too, do you think this is right?
1. Opposite sides are parallel, diagonals are perpendicular, and the length of the diagonals are equal so the type of quadrilateral that ABCD is a rhombus.
2-a. The axis of symmetry: EC,
gradient: 3-10/8-1=-1 so -1, and the equation is y=-x+11
2-b. The equation of BD: y=x-1
y=y, then -x+11=x-1
12=2x
6=x
substitute for x into equation: y=-6+11
y=5
Two diagonal EC and BD meet at (6,5)
3. I consider F=(2,9)
FBCD is always a kite because:
- Diagonal line FC is the perpendicular bisector of BD (same as EC)
- The intersection of line FC and BD is the midpoint of BD (and also the same as the intersection of EC and BD)
- The distance between line FB and line FD are equal (7.07)
- The distance between line BC and CD are equal (5.10)
This evidances show the FBCD is always a kite.
Thank you for your reply
I've solved other problems too, do you think this is right?
1. Opposite sides are parallel, diagonals are perpendicular, and the length of the diagonals are equal so the type of quadrilateral that ABCD is a rhombus.
2-a. The axis of symmetry: EC,
gradient: 3-10/8-1=-1 so -1, and the equation is y=-x+11
2-b. The equation of BD: y=x-1
y=y, then -x+11=x-1
12=2x
6=x
substitute for x into equation: y=-6+11
y=5
Two diagonal EC and BD meet at (6,5)
3. I consider F=(2,9)
FBCD is always a kite because:
- Diagonal line FC is the perpendicular bisector of BD (same as EC)
- The intersection of line FC and BD is the midpoint of BD (and also the same as the intersection of EC and BD)
- The distance between line FB and line FD is equal (7.07)
- The distance between line BC and CD is equal (5.10)
These evidences show the FBCD is always a kite.
The Highlander
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Hi @davee
Thank you for supplying your answers (though I see you are still resistant to the idea of providing a sketch! ?) You don't need to create fancy pictures a snap of a pencil sketch on squared paper is adequate. ?
I have only just seen your post but I'm afraid I have to go out for a few hours.
There are several points I would like to make about what you have written so I will post a full reply when I get back home.
Oh, I forgot to post a sketch, I drew this graph using a website called Desmos.
no rush thank you!
The Highlander
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Hi @davee,
Oh, I forgot to post a sketch, I drew this graph using a website called Desmos.
Like The Terminator (another movie character, lol) I'm back! ?
Desmos is certainly a very useful website for producing drawings & graphs but I would suggest that it's not the most appropriate for your use in problems like this. I know it allows you to plot points just by typing in their coordinates, etc. but I notice you haven't drawn any of the lines (a wee bit trickier perhaps, eh?) and that was one of the things we all asked you to do (because it helps understanding of the problem and illustration of the reasoning behind the answers).
Here is what I think would be an acceptable sketch for this problem...
If you're not happy just hand drawing (in pencil) on squared paper, then uploading a picture of your work, then I suggest that you stick to MS Paint™ or some other image handling program (the free one I mentioned, "Paint.net", is excellent and much more powerful than MS Paint™, though not as simple to use, of course).
I am also attaching a LARGE sheet of "squared paper" that you could open in any image handling App for the purpose of making sketches. You (or anyone else who fancies a copy of it ?) can right-click on it and choose "Save image as..." to download it as: "Squared Paper.jpg". Once you open it (in MS Paint™ for example) just Select an 'area' of it that has enough boxes for your current purposes and then Crop it to that size; that will make the boxes 'bigger' so you can easily add your (numbered?) axes and plot points on it. (Use the Times New Roman font for plotting points as full stops in that font are round unlike, say, in the Arial font where they look square when you make them big enough to display properly. ?)
Blank sheet of squared paper.
But enough on sketching (regardless of how important it is) we now need to look at your answers in the next post.
The Highlander
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Right, lets go through the answers you've shown us. First of all, thank you for posting them. ? It's important that we see your work so we know that you've understood any help we've offered you and been able to use it to get the right answers. We can only help members properly if they are prepared to help us to help them and that means responding to the questions we ask; so many post problems and then we never hear from them again! ?
So now let's go through each of your answers in turn...
(Anything I've added into your post as a comment or correction is shown in red, below; please look for those additions to your answers as well as reading what else I've written in this post.)
Your answer is actually a very good example of this (I must remember it myself to explain "begging the question" to others, lol). You have listed all those properties of a rhombus (to conclude that that ABCD is a rhombus) but you haven't shown that any of the properties you have listed are actually true!
It's a bit like saying:-
Now that's hardly the "mathematical" reasoning you were asked to provide, is it?
The red dotted lines I added in my earlier sketch were meant to be a strong hint as to how you could have proved it had a property that would define it as a rhombus!
You could have said;-
Each side of the quadrilateral, ABCD, is either 1 unit along and 5 units up or 5 units along and 1 unit up, therefore, by Pythagoras, the length of each side is:-
and now that you have shown that all four sides are the same length (and none of its vertices are 90°) you can rightfully claim that it is a rhombus.
The equation of the line EC is of the form y = mx + c. I have already shown that m = ˉ1 and the point (1, 10) lies on the line EC, therefore, when x = 1, y = 10 and so 10 = ˉ1 × 1 + c ⇒ 10 = ˉ1 + c ⇒ 11 = c.
Therefore, the equation of the line EC is: y = -x + 11. (Which may also be written as: y = 11 - x).
(You haven't shown or addressed what the midpoint of BD actually is!)
I would have been tempted to say: " You told me that EBCD is a kite, therefore, its long diagonal (EC) must bisect the shorter one (BD) and so the axis of symmetry (EC) does go through the midpoint of BD. ?
But I think that would be regarded as a bit of a cheat (or a 'smart-aleck' answer, lol).
You mentioned in your earlier posts that you thought you could use the "midpoint equation" to help solve this part of the question and I think that is exactly what was expected of you. I also gave you a hint in that direction in my first post when I said:-
"2b. What are the coordinates of the midpoint of the diagonal BD? Consider them further with refence to your equation for the line EC (now that you have constructed it on your diagram ?)."
So you should have started this part by using that formula...
The midpoint (xm, ym) between any two points, (x1, y1) and (x2, y2) may be found using the formula:-
and so the midpoint of BD is:-
\(\displaystyle \text{M} = \left(\frac{9+3}{2}, \frac{8+2}{2}\right)\implies\text{ M is the point } (6, 5)\)
and now that you've shown what the midpoint of BD is, you can also prove that it lies on the line EC by showing that it fits the equation of EC: 5 = 11 - 6, therefore, M lies on the line EC and the axis of symmetry (EC) does, therefore, pass through the midpoint of BD. QED. ?
I'm afraid your answer doesn't address this part at all. (Continued in next post...)
So now let's go through each of your answers in turn...
(Anything I've added into your post as a comment or correction is shown in red, below; please look for those additions to your answers as well as reading what else I've written in this post.)
I'm afraid that doesn't really meet the requirement specified in Q.1. What you've done here is what's called "begging the question"! A lot of people misuse that phrase to mean that something demands (or strongly points to) another question (when they really ought to say (or mean) something like: "that raises the further question" or "that raises more questions than it answers") but the phrase "begging the question" has a very specific meaning: it is a logical fallacy (a mistake) where the conclusion reached is actually dependent on the conclusion itself being true (without offering any proof that it is true!).
Your answer is actually a very good example of this (I must remember it myself to explain "begging the question" to others, lol). You have listed all those properties of a rhombus (to conclude that that ABCD is a rhombus) but you haven't shown that any of the properties you have listed are actually true!
It's a bit like saying:-
ABCD is a rhombus because it's a rhombus! (ie: it has a rhombus' properties)
Now that's hardly the "mathematical" reasoning you were asked to provide, is it?
The red dotted lines I added in my earlier sketch were meant to be a strong hint as to how you could have proved it had a property that would define it as a rhombus!
You could have said;-
Each side of the quadrilateral, ABCD, is either 1 unit along and 5 units up or 5 units along and 1 unit up, therefore, by Pythagoras, the length of each side is:-
\(\displaystyle \sqrt{1^2+5^2}=\sqrt{1+25}=\sqrt{26}\approx 5.10\)
and now that you have shown that all four sides are the same length (and none of its vertices are 90°) you can rightfully claim that it is a rhombus.
But, since you haven't drawn any of the lines (as requested! ?), again it may not be enough (to some pedants, lol) to just claim it crosses the y-axis at 11! So to satisfy the most demanding critic you might have to prove that. Like so?...
The equation of the line EC is of the form y = mx + c. I have already shown that m = ˉ1 and the point (1, 10) lies on the line EC, therefore, when x = 1, y = 10 and so 10 = ˉ1 × 1 + c ⇒ 10 = ˉ1 + c ⇒ 11 = c.
Therefore, the equation of the line EC is: y = -x + 11. (Which may also be written as: y = 11 - x).
This is all very "wooly" thinking (although I think I understand what you were trying to do) but, again, it doesn't really answer what the question asked: "Show that this axis of symmetry goes through the midpoint of BD."
(You haven't shown or addressed what the midpoint of BD actually is!)
I would have been tempted to say: " You told me that EBCD is a kite, therefore, its long diagonal (EC) must bisect the shorter one (BD) and so the axis of symmetry (EC) does go through the midpoint of BD. ?
But I think that would be regarded as a bit of a cheat (or a 'smart-aleck' answer, lol).
You mentioned in your earlier posts that you thought you could use the "midpoint equation" to help solve this part of the question and I think that is exactly what was expected of you. I also gave you a hint in that direction in my first post when I said:-
"2b. What are the coordinates of the midpoint of the diagonal BD? Consider them further with refence to your equation for the line EC (now that you have constructed it on your diagram ?)."
So you should have started this part by using that formula...
The midpoint (xm, ym) between any two points, (x1, y1) and (x2, y2) may be found using the formula:-
\(\displaystyle (x_m,~y_m) = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
and so the midpoint of BD is:-
\(\displaystyle \text{M} = \left(\frac{9+3}{2}, \frac{8+2}{2}\right)\implies\text{ M is the point } (6, 5)\)
and now that you've shown what the midpoint of BD is, you can also prove that it lies on the line EC by showing that it fits the equation of EC: 5 = 11 - 6, therefore, M lies on the line EC and the axis of symmetry (EC) does, therefore, pass through the midpoint of BD. QED. ?
Please expand your answer (just above) to see all my additions (in red ink).
I'm afraid your answer doesn't address this part at all. (Continued in next post...)
Last edited:
The Highlander
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Continued from previous post...
I'm afraid your answer doesn't address this part properly at all! ?
What you have done (or tried to do) here is to verify that FBCD is a kite for one, single instance, of a point, F, that has the coordinates (2, 9) but that doesn't prove that FBCD will be a kite when F is any point on the line EC!!!
You state, (almost) quite rightly, that "Thedistance between lengths of the lines BC and CD are equal (5.10)"
(Although I have to point out that "5.10" is just an approximation of their true lengths (\(\displaystyle \sqrt{26}\)) and that should always be noted when approximate results are written into calculations/answers. Like I did when I said that \(\displaystyle \sqrt{26}\approx5.10\) in the last post. ?)
But you don't need to give any measure of the lengths of BC & CD at this point (that was what you had to do back in Part 1). All you need to do now is to re-state that their lengths remain equal (regardless of where the point F is) because they remain unchanged by its location.
So what you really need to do at this point is to prove that the lines FB & FD are also of equal length, no matter where the point F lies along the line EC.
So you want to prove that for any point F (xF, yF) the distance FB = FD.
Then FBCD is always a kite (because adjacent sides are of equal length).
Remember I pointed out (above) that the equation of EC, y = -x + 11, could be re-written as y = 11 - x?
So that means that any point on the line EC, eg: the point F, will have the coordinates: (x, 11-x), won't it?
How does that help?
Well, I believe you know that, the distance (d) between any two points (x1, y1) and (x2, y2) is given by the formula...
So, given that F is the point (x, 11-x) and D is the point (9, 8), what will be the length of the side FD in the figure FBCD?
It will be: \(\displaystyle d = \sqrt{(x-9)^2+((11-x)-8)^2}\), won't it? Do you see that OK?
Now it's taken me quite some time to write out this (lengthy) reply addressing all the issues raised by your attempts to answer the question, not just for your benefit but also for the benefit of any other aspiring young mathematicians (in "middle school" or above) who come across it in the forum, so I would like to leave it up to you to try and complete part 3 of the question (it's simple algebra) and come back with your answer soon as you get a chance, eh?
Can you now prove that FB = FD no matter where F lies along the line EC?
Final Hint:-
Rewrite the length of FD as:-
expanding the brackets under the radix; that's the radical sign: \(\displaystyle \sqrt{ }\)).
Hope that helps. ?
Please expand your answer (just above) to see all my additions (in red ink).
I'm afraid your answer doesn't address this part properly at all! ?
What you have done (or tried to do) here is to verify that FBCD is a kite for one, single instance, of a point, F, that has the coordinates (2, 9) but that doesn't prove that FBCD will be a kite when F is any point on the line EC!!!
You state, (almost) quite rightly, that "The
(Although I have to point out that "5.10" is just an approximation of their true lengths (\(\displaystyle \sqrt{26}\)) and that should always be noted when approximate results are written into calculations/answers. Like I did when I said that \(\displaystyle \sqrt{26}\approx5.10\) in the last post. ?)
But you don't need to give any measure of the lengths of BC & CD at this point (that was what you had to do back in Part 1). All you need to do now is to re-state that their lengths remain equal (regardless of where the point F is) because they remain unchanged by its location.
So what you really need to do at this point is to prove that the lines FB & FD are also of equal length, no matter where the point F lies along the line EC.
So you want to prove that for any point F (xF, yF) the distance FB = FD.
Then FBCD is always a kite (because adjacent sides are of equal length).
Remember I pointed out (above) that the equation of EC, y = -x + 11, could be re-written as y = 11 - x?
So that means that any point on the line EC, eg: the point F, will have the coordinates: (x, 11-x), won't it?
How does that help?
Well, I believe you know that, the distance (d) between any two points (x1, y1) and (x2, y2) is given by the formula...
\(\displaystyle d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
So, given that F is the point (x, 11-x) and D is the point (9, 8), what will be the length of the side FD in the figure FBCD?
It will be: \(\displaystyle d = \sqrt{(x-9)^2+((11-x)-8)^2}\), won't it? Do you see that OK?
Now it's taken me quite some time to write out this (lengthy) reply addressing all the issues raised by your attempts to answer the question, not just for your benefit but also for the benefit of any other aspiring young mathematicians (in "middle school" or above) who come across it in the forum, so I would like to leave it up to you to try and complete part 3 of the question (it's simple algebra) and come back with your answer soon as you get a chance, eh?
Can you now prove that FB = FD no matter where F lies along the line EC?
Final Hint:-
Rewrite the length of FD as:-
\(\displaystyle \sqrt{(x-9)^2+((11-x)-8)^2}=\sqrt{(x-9)^2+(11-8-x)^2}=\sqrt{(x-9)^2+(3-x)^2}\)
Then form a similar expression for the length of FB and show that the two are exactly the same (by
expanding the brackets under the radix; that's the radical sign: \(\displaystyle \sqrt{ }\)).
Hope that helps. ?
Last edited:
The Highlander
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C'mon @davee!
Why haven't you come back to us with the final answer?
I have explained how parts 1 & 2 of the question should be done and given you some very strong hints on how to complete part 3.
If you can prove that FB = FD for any point F on the line EC then FBCD is always a kite (because it has adjacent sides of equal length). Thus, you can say...
Part 3.
|BC| = |CD| because they are unaffected by the position of F.
Therefore, if |FB| = |FD|, then FBCD is always a kite because it then has adjacent sides of equal length.
B has the coordinates (3, 2),
D has the coordinates (9, 8)
and
F has the coordinates (x, 11-x) because it lies anywhere on the line EC (with equation: y = 11 - x).
Therefore...
|FB| = \(\displaystyle \sqrt{(x-3)^2+((11-x)-2)^2}=\sqrt{(x-3)^2+(11-2-x)^2}=\sqrt{(x-3)^2+(9-x)^2}\)
and
|FD| = \(\displaystyle \sqrt{\text{ (?)² + (?)² }}=\) ???
Now, please show that these two expressions are equal to prove that FBCD is always a kite when F lies on the line EC.
NB: If you really can't do that then please let us know so that we can show you how to finish it.
Get back to us soon, huh? ?
Why haven't you come back to us with the final answer?
I have explained how parts 1 & 2 of the question should be done and given you some very strong hints on how to complete part 3.
If you can prove that FB = FD for any point F on the line EC then FBCD is always a kite (because it has adjacent sides of equal length). Thus, you can say...
Part 3.
|BC| = |CD| because they are unaffected by the position of F.
Therefore, if |FB| = |FD|, then FBCD is always a kite because it then has adjacent sides of equal length.
B has the coordinates (3, 2),
D has the coordinates (9, 8)
and
F has the coordinates (x, 11-x) because it lies anywhere on the line EC (with equation: y = 11 - x).
Therefore...
|FB| = \(\displaystyle \sqrt{(x-3)^2+((11-x)-2)^2}=\sqrt{(x-3)^2+(11-2-x)^2}=\sqrt{(x-3)^2+(9-x)^2}\)
and
|FD| = \(\displaystyle \sqrt{\text{ (?)² + (?)² }}=\) ???
Now, please show that these two expressions are equal to prove that FBCD is always a kite when F lies on the line EC.
NB: If you really can't do that then please let us know so that we can show you how to finish it.
Get back to us soon, huh? ?