How to find points on a square given two points

• March 8th 2008, 11:05 PM
Yesme
How to find points on a square given two points
Hi there,

I am currently in high school and was posed this problem as a challenge to solve. At first it sounded simple but i am really stuck.(Worried)

A(7,2) C(1,4)
Are the vertices on the diagonal of a sqaure ABCD, find the co-ordinates of B & D (algebraicly)

I know this can be done using vectors or just simply drawing it on the cartesian plane, however it was requested that it be done algebraicly using the perpendicular rules and other stuff like that.

Thanks
• March 9th 2008, 01:35 AM
earboth
Quote:

Originally Posted by Yesme
Hi there,

I am currently in high school and was posed this problem as a challenge to solve. At first it sounded simple but i am really stuck.(Worried)

A(7,2) C(1,4)
Are the vertices on the diagonal of a sqaure ABCD, find the co-ordinates of B & D (algebraicly)

...

1. The points B, D are the endpoints of the other diagonal. The diagonals intersect at the midpoint of AC: $M_{AC}\left(\frac{7+1}{2}~,~\frac{2+4}{2}\right)~\ implies~M_{AC}(4, 3)$

2. The points B, D are placed on the perpendicular bisector of AC:
Calculate the slope of AC:

$m_{AV} = \frac{4-2}{7-1} = -\frac13$

Therefore the slope of BD is:

$m_{BD} = -\frac1{m_{AC}}=3$

3. The points BD are connected by the straight line:

$y-3 = 3(x-4)~\implies~y = 3x-9$

4. The distances $|\overline{MA}|$ and $|\overline{MB}|$ must be equal. To get the coordinates of B you need a parallel to AC which have the distance $|\overline{MA}| = \sqrt{10}$ from AC.

Or:

B is the point of intersection between the perpendicular bisector of AC and the circle around M with radius $r = \sqrt{10}$:

$c: (x-4)^2 +(y-3)^2=10$

I've got: (5, 6), (3, 0)
• March 9th 2008, 10:46 AM
Yesme
didnt get it
Sorry again...
I had got up to the part where you found the equation of BD but from there you said i needed a parallel to AC to find out the co-ordinates of B, that implies that i had to draw the actual square and find a point by plotting.
am i right? i cant solve it like that, i have to solve it algebraicly.

never mind, ill get the answer from school today and ill post what they told me.
• March 9th 2008, 12:41 PM
earboth
Quote:

Originally Posted by Yesme
Sorry again...
I had got up to the part where you found the equation of BD but from there you said i needed a parallel to AC to find out the co-ordinates of B, that implies that i had to draw the actual square and find a point by plotting.
am i right? i cant solve it like that, i have to solve it algebraicly.

I assume that you know how to calculate the equation of a parallel to a given line. I assume also that you know how to calculate the distance of a point from a given line.

1. We have the line passing through A and C: $y = -\frac13x+\frac{13}3 ~\implies~ x+3y-13=0$

2. We know that the point B is placed on the line BD: $y = 3x-9$ Therefore the point B has the coordinates B(b, 3b-9)

3. If a straight line has the equation $Ax + By + C = 0$ then the distance of a point $P(x_P, y_P)$ from this line is calculated by:

$d = \frac{Ax_P + By_P -C}{\sqrt{A^2+B^2}}$

4. We know that the distance must be $|d| = \sqrt{10}$

5. So we can set up the equation:

$\pm\sqrt{10} = \frac{b+3(3b-9)-13}{\sqrt{1^2+3^2}}$

6. Solve this equation for b. You'll get $b = 5 ~\vee~ b = 3$
• March 9th 2008, 05:12 PM
cokhale
u are correct
Quote:

Originally Posted by earboth
1. The points B, D are the endpoints of the other diagonal. The diagonals intersect at the midpoint of AC: $M_{AC}\left(\frac{7+1}{2}~,~\frac{2+4}{2}\right)~\ implies~M_{AC}(4, 3)$

2. The points B, D are placed on the perpendicular bisector of AC:
Calculate the slope of AC:

$m_{AV} = \frac{4-2}{7-1} = -\frac13$

Therefore the slope of BD is:

$m_{BD} = -\frac1{m_{AC}}=3$

3. The points BD are connected by the straight line:

$y-3 = 3(x-4)~\implies~y = 3x-9$

4. The distances $|\overline{MA}|$ and $|\overline{MB}|$ must be equal. To get the coordinates of B you need a parallel to AC which have the distance $|\overline{MA}| = \sqrt{10}$ from AC.

Or:

B is the point of intersection between the perpendicular bisector of AC and the circle around M with radius $r = \sqrt{10}$:

$c: (x-4)^2 +(y-3)^2=10$

I've got: (5, 6), (3, 0)

thank u are correct. clement
nigeria student.
• March 9th 2008, 07:07 PM
Yesme
thanks
Sorry about that... just had a mind block... i realise your method and it was exactly the same as the one given in class today....thanks again(Rock)