# Thread: Problem on coordinate geometry.

1. ## Problem on coordinate geometry.

A(3,4) and 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

2. ## Re: Problem on coordinate geometry.

Originally Posted by sankalpmittal
A(3,4) and 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
Start at point A. How far horizontally do you have to move to get to point C? How far vertically do you have to move to get to point C? This should help you find the other two points...

3. ## Re: Problem on coordinate geometry.

sqrt(29) is the distance. Now what can I do ?

4. ## Re: Problem on coordinate geometry.

That's not what I asked. I asked you how far HORIZONTALLY do you move, and how far VERTICALLY do you move? I did not ask how far you would move if you went straight from one coordinate to the other.

5. ## Re: Problem on coordinate geometry.

Originally Posted by Prove It
That's not what I asked. I asked you how far HORIZONTALLY do you move, and how far VERTICALLY do you move? I did not ask how far you would move if you went straight from one coordinate to the other.
Horizontal distance : sqrt{(a-3)^2 + (b-4)^2}
Vertical distance : sqrt{(a-1)^2 + (b+1)^2}

So on comparing
Horizontal distance = Vertical distance
I got this equation : 4a+10b = 23
(Tally image in post #1)

Now what can I do ?

6. ## Re: Problem on coordinate geometry.

Originally Posted by sankalpmittal
Horizontal distance : sqrt{(a-3)^2 + (b-4)^2}
Vertical distance : sqrt{(a-1)^2 + (b+1)^2}

So on comparing
Horizontal distance = Vertical distance
I got this equation : 4a+10b = 23
(Tally image in post #1)

Now what can I do ?
I guess you can construct 2 vectors, BA = <3-a,4-b> BC = <1-a,-1-b>

Their dot product should be zero since they are perpendicular.

(3 - a)(1 - a) + (4 - b)(-1 - b) = 0

Then use 4a + 10b = 23 to eliminate one of the variables.

This becomes an algebraic nightmare, if you do it by hand, Anybody got a better idea?

7. ## Re: Problem on coordinate geometry.

The other 2 vertices are (-1/2,5/2) and (9/2,1/2)

I used Wolfram Alpha

a &#61; &#40;23 - 10b&#41;&#47;4, &#40;3 - a&#41;&#40;1 - a&#41; - &#40;4 -b&#41;&#40;1 &#43; b&#41; &#61; 0 - Wolfram|Alpha

The plot is not to scale, should be a circle

Wolfram gave both vertices in 1 shot

You can confirm that these vertices are on a square by using distance formula to check length of sides AND check length of diagonals.

You can also construct vectors on the vertices and take dot products, also checking lenths of sides using distance formula

8. ## Re: Problem on coordinate geometry.

Originally Posted by agentmulder
The other 2 vertices are (-1/2,5/2) and (9/2,1/2)

I used Wolfram Alpha

a &#61; &#40;23 - 10b&#41;&#47;4, &#40;3 - a&#41;&#40;1 - a&#41; - &#40;4 -b&#41;&#40;1 &#43; b&#41; &#61; 0 - Wolfram|Alpha

The plot is not to scale, should be a circle

Wolfram gave both vertices in 1 shot

You can confirm that these vertices are on a square by using distance formula to check length of sides AND check length of diagonals.

You can also construct vectors on the vertices and take dot products, also checking lenths of sides using distance formula

Actually , I did it by a lengthy method of using

m x m' = -1 and distance formula.

I was searching for an easier method. Thanks anyways.

[See PM]

9. ## Re: Problem on coordinate geometry.

Originally Posted by sankalpmittal
Actually , I did it by a lengthy method of using

m x m' = -1 and distance formula.

I was searching for an easier method. Thanks anyways.
Isn't it interesting that your method of elimination gave you the equation of the line that contains the OTHER diagonal of the square you were looking for?

4a + 10b = 23

Is a line that runs along the diagonal of the missing vertices, I find that fascinating.

I have to admit, I like this problem very much.

You used the distance formula, set length = width, and got an equation for a line that is 45º to both.

10. ## Re: Problem on coordinate geometry.

A bit more "geometric": The diagonals of a square are congruent and cross at right angles at their midpoint.

The midpoint of the line between (3,4) and (1,-1) is ((3+1)/2,(4- 1)/2)= (2, 3/2). It has slope (4-(-1))/(3-1)= 5/2 and length $\displaystyle \sqrt{(3-1)^2+ (4-(-1))^2}= \sqrt{4+25}= \sqrt{29}$.

The line through (2,3/2) perpendicular to that diagonal is [tex]y- 3/2= (-2/5)(x- 2)[/itex] or y= (-2/5)x+ 4/5+ 3/2= (-2/5)x+ 23/10. That can also be written as 10x- 4y= 23. Find the intersections of that line with the circle, centered on (2,3/2) of radius $\displaystyle \sqrt{29}$, $\displaystyle (x- 2)^2+ (y- 3/2)^2= 29$.

11. ## Re: Problem on coordinate geometry.

Originally Posted by HallsofIvy
A bit more "geometric": The diagonals of a square are congruent and cross at right angles at their midpoint.

The midpoint of the line between (3,4) and (1,-1) is ((3+1)/2,(4- 1)/2)= (2, 3/2). It has slope (4-(-1))/(3-1)= 5/2 and length $\displaystyle \sqrt{(3-1)^2+ (4-(-1))^2}= \sqrt{4+25}= \sqrt{29}$.

The line through (2,3/2) perpendicular to that diagonal is [tex]y- 3/2= (-2/5)(x- 2)[/itex] or y= (-2/5)x+ 4/5+ 3/2= (-2/5)x+ 23/10. That can also be written as 10x- 4y= 23. Find the intersections of that line with the circle, centered on (2,3/2) of radius $\displaystyle \sqrt{29}$, $\displaystyle (x- 2)^2+ (y- 3/2)^2= 29$.
Are you HallsofIvy from Physics forums ?

12. ## Re: Problem on coordinate geometry.

Originally Posted by HallsofIvy
A bit more "geometric": The diagonals of a square are congruent and cross at right angles at their midpoint.The line through (2,3/2) perpendicular to that diagonal is [tex]y- 3/2= (-2/5)(x- 2)[/itex] or y= (-2/5)x+ 4/5+ 3/2= (-2/5)x+ 23/10. That can also be written as 10x- 4y= 23.
I don't mean to nitpick but you have an error in sign at the end of your post.

Make another pot of coffee Halls...