Problem on co-ordinate geometry

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The discussion centers on finding the coordinates of the remaining vertices of square ABCD given two opposite vertices A(3,4) and C(1,-1). Participants explore various geometric methods, including the distance formula, slope calculations, and the midpoint of the square. They identify the midpoint as (2, 1) and discuss the slopes of the diagonals, leading to the formulation of equations to derive the other vertices. Ultimately, the conversation highlights the complexity of the problem and the need for careful planning in solving geometric equations.
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Homework Statement



A(3,4) and [STRIKE]C(1,-2)[/STRIKE] C(1,-1) are the two opposite angular points of a square ABCD . Find the coordinates of the other two vertices .

Here is the image : http://oi55.tinypic.com/2mq80i8.jpg

Homework Equations



Not sure what is most relevant . I tried applying distance formula , section formula and slopes but failed to crack the question .

The Attempt at a Solution



I tried applying distance formula of two diagonals and equated them but failed . Got four variable equation ! I tried slope formulas , but still failed .
I tried combination of above got a very complicated quadratic equation which yielded wrong answer .

Please help .

(I am 14 years , 10th class )

Thanks in advance .


NOTE: The coordinates of C should be as shown in the image: C = (1,-1), not (1,-2)
 
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Hints: Where's the center of the square? What's the slope of the given diagonal? What's the slope of the diagonal between the other two vertices? Write the equation for the diagonal.
 
Doc Al said:
Hints: Where's the center of the square? What's the slope of the given diagonal? What's the slope of the diagonal between the other two vertices? Write the equation for the diagonal.

I have already tried these equations but equations become very complicated . Well thanks for replying Doc .

Centre by section formula is

(2, 3/2)

Slope of diagonal AC : 5/2
So of BD : -2/5
Equation of AC : 2y-5x+7=0

Now ?

:confused:
 
How far are the vertices from the center? Use that fact.
 
Doc Al said:
How far are the vertices from the center? Use that fact.

Ok .

Distance of each vertex from centre :

sqrt(29/4)
sqrt(29)/2

Now ?
 
Make use of that fact! Assume the vertices are at a point Δx, Δy away from the center. Use the slope and the distance to solve for Δx and Δy.
 
Doc Al said:
Make use of that fact! Assume the vertices are at a point Δx, Δy away from the center. Use the slope and the distance to solve for Δx and Δy.

Tally image post #1

I got 1 equation :

(b-3/2)/(a-2) = -2/5


on applying m x m' = -1 of two vertices intersecting at 90 degrees .
 
Good. Now get an equation for the distance to the center.
 
Doc Al said:
Good. Now get an equation for the distance to the center.

Both the equations are coming same ie 2a + 5b = 23/2 !
 
  • #10
sankalpmittal said:
Both the equations are coming same ie 2a + 5b = 23/2 !
Show me your equation for the distance. This can't be it.
 
  • #11
Doc Al said:
Show me your equation for the distance. This can't be it.

(b-3/2)/(a-2) = -2/5 equation 1
On solving you'll get ,
Solution:
4a+10b=23
= 2a+5b=23/2

Now

1+25/4 + (a-2)2 + (b-3/2)2 = (4-b)2 + (3-a)2 equation 2

On solving this you'll get ,
2a+5b=23/2


Now what can I do ?
 
  • #12
sankalpmittal said:
Now

1+25/4 + (a-2)2 + (b-3/2)2 = (4-b)2 + (3-a)2 equation 2
Not sure what you're doing here. The square of the distance to the center is given by:
(a-2)2 + (b-3/2)2

Set that equal to the square of the distance you found before.
 
  • #13
Doc Al said:
Not sure what you're doing here. The square of the distance to the center is given by:
(a-2)2 + (b-3/2)2

Set that equal to the square of the distance you found before.


(a-2)2 + (b-3/2)2 = 29/4 ?


a2-4 a+b2-3 b-1 = 0

It is giving a degree 2 equation .
 
  • #14
sankalpmittal said:
(a-2)2 + (b-3/2)2 = 29/4 ?


a2-4 a+b2-3 b-1 = 0

It is giving a degree 2 equation .
OK. Now combine it with your other equation.
 
  • #15
Problems like this become much easier if you plan how to tackle them. You should almost already know how to get the answer before you write the first equation. This is like in (classical) geometry when you have to first find the answer before you can write the proof.
 
  • #16
Doc Al said:
OK. Now combine it with your other equation.
2a + 5b = 23/2
=> 4a + 10b = 23
=> a= (23 - 10b) / 4

a2-4a+b2-3b-1 = 0

Ok so on combining
((23 - 10b) / 4)2-4((23 - 10b) / 4)+b2-3b-1 = 0

A very complicated quadratic equation indeed .

Let me try solving it .

29\16 (4 b^2-12 b+5) = 0
b = 1\2
so b = 1/2
and d = 5/2

In the same time ,
a = 9/2
C = -1/2

Any easier method ?

Instead of slope can we also do it like this ?
Tally image post 1

AB2 = BC2 ( where AB and BC are sides of square ) -> equation X

So we get one equation in form of a and b .

Now we can get

AC2 ( diagonal ) = 29

Now we put equation of AC2

(3-a)2 + (4-b)2 + (a-1)2 + (b+1)2 = 29

=2 a2-8 a+2 b2-6 b+27 = 29 --> equation Y

Now we combine equation X and Y ??

Well , anyways I got correct answer so thanks a lot for your help .

:)
verty said:
Problems like this become much easier if you plan how to tackle them. You should almost already know how to get the answer before you write the first equation. This is like in (classical) geometry when you have to first find the answer before you can write the proof.
Do you know any easier method ?

Please explain this also : http://in.answers.yahoo.com/question/index?qid=20070306031100AASHBBs

I just know searched it on the net after finishing my question on advice of Doc .
 
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  • #17
sankalpmittal said:
I have already tried these equations but equations become very complicated . Well thanks for replying Doc .

Centre by section formula is

(2, 3/2)
No, that's incorrect. The given points are A(3, 4) and C(1, -2). The midpoint (center of the square) is ((3+1)/2, (4- 2)/2= (2, 1).

Slope of diagonal AC : 5/2
No, that's incorrect. The slope is (4-(-2))/(3- 1)= 6/2= 3.

So of BD : -2/5
Equation of AC : 2y-5x+7=0

Now ?

:confused:

sankalpmittal said:
Ok .

Distance of each vertex from centre :

sqrt(29/4)
sqrt(29)/2

Now ?
No, that's incorrect. The distance between A and C is \sqrt{(3- 1)^2+ (4-(-2))^2}= \sqrt{4+ 36}= \sqrt{40}= 2\sqrt{10}. The distance from the center to each vertex is \sqrt{10}
 
  • #18
HallsofIvy said:
No, that's incorrect. The given points are A(3, 4) and C(1, -2). The midpoint (center of the square) is ((3+1)/2, (4- 2)/2= (2, 1). No, that's incorrect. The slope is (4-(-2))/(3- 1)= 6/2= 3.

No, that's incorrect. The distance between A and C is \sqrt{(3- 1)^2+ (4-(-2))^2}= \sqrt{4+ 36}= \sqrt{40}= 2\sqrt{10}. The distance from the center to each vertex is \sqrt{10}
Thanks, Halls. (I didn't even bother to check those results as I was focused more on the process than the answers. :redface:)

sankalpmittal, my apologies for not checking your work more carefully. In any case, with Halls' corrections, please redo your analysis following the method I outlined.
 
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  • #19
HallsofIvy said:
No, that's incorrect. The given points are A(3, 4) and C(1, -2). The midpoint (center of the square) is ((3+1)/2, (4- 2)/2= (2, 1). No, that's incorrect. The slope is (4-(-2))/(3- 1)= 6/2= 3.

No, that's incorrect. The distance between A and C is \sqrt{(3- 1)^2+ (4-(-2))^2}= \sqrt{4+ 36}= \sqrt{40}= 2\sqrt{10}. The distance from the center to each vertex is \sqrt{10}

Umm Sir Halls , those two coordinates are (3,4) and (1,-1) .
Please tally that image in post 1.

Please check my post 16 .

Doc Al said:
Thanks, Halls. (I didn't even bother to check those results as I was focused more on the process than the answers. :redface:)

sankalpmittal, my apologies for not checking your work more carefully. In any case, with Halls' corrections, please redo your analysis following the method I outlined.
No you checked my work carefully . By mistake sir hallsofivy typed wrong coordinates .
Please check my post 16 .

Oops I wrote wrong question . Sorry . Tally image that is correct question itself in post 1 .

Edit post 1 : In place of (1,-2) edit it to (1,-1)
(I apologize HallOfIvy )
 
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  • #20
sankalpmittal said:
Oops I wrote wrong question . Sorry . Tally image that is correct question itself in post 1 .

Edit post 1 : In place of (1,-2) edit it to (1,-1)
That's the problem: In post #1 you wrote C = (1,-2), but it should have been what you had in the image, C = (1,-1). Right?
 
  • #21
Doc Al said:
That's the problem: In post #1 you wrote C = (1,-2), but it should have been what you had in the image, C = (1,-1). Right?

Yes , exactly , please edit post #1 and check my post #16 .
 
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  • #22
Your post looks OK.

Here's how I would have solved it:

Δy = (-2/5)Δx

(Δy)2 + (Δx)2 = 29/4

Combine to solve for Δx: Δx = ±5/2.

(Δx & Δy are measured from the center, of course.)
 
  • #23
Doc Al said:
Your post looks OK.

Here's how I would have solved it:

Δy = (-2/5)Δx

(Δy)2 + (Δx)2 = 29/4

Combine to solve for Δx: Δx = ±5/2.

(Δx & Δy are measured from the center, of course.)

Thats ok thanks
 

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