# [SOLVED] Finding Co-Ordinates of a Rectangle

• May 9th 2010, 11:29 AM
unstopabl3
[SOLVED] Finding Co-Ordinates of a Rectangle
http://i41.tinypic.com/jiyg4p.jpg

Here's a question from a past paper which I have successfully attempted. My question is regarding part (iii). I have successfully figured out the co-ordinates by the following method:

Is my method correct, considering I did get the right answer?

But is there another simpler method to do this which would save time during an exam.
• May 9th 2010, 01:07 PM
Plato
The diagonals bisect one another. The midpoint of $\overline{AC}$ is ?
• May 9th 2010, 01:13 PM
unstopabl3
Mid of AC is (6,6), the diagnols do bisect each other at the mid but we don't have the x-co-ordinates of B or D to equate the diagnols, or do we?
• May 9th 2010, 01:21 PM
Plato
Quote:

Originally Posted by unstopabl3
Mid of AC is (6,6), the diagnols do bisect each other at the mid but we don't have the x-co-ordinates of B or D to equate the diagnols, or do we?

But the y-coordinate of $B~\&~D$ is 6.
You are given the x-coordinate of $D$, so $D:(h,6)$
• May 9th 2010, 01:28 PM
unstopabl3
No, I meant real value of the x-co-ordinates has not been given since we have to calculate that ourselves.

I have already gotten the correct values by using the method mentioned in my first post.
As stated I want someone to solve this part of the question with a different, possibly easier method.

I am not after the answer, I am looking for an alternate method.
• May 9th 2010, 02:03 PM
masters
Quote:

Originally Posted by unstopabl3
No, I meant real value of the x-co-ordinates has not been given since we have to calculate that ourselves.

I have already gotten the correct values by using the method mentioned in my first post.
As stated I want someone to solve this part of the question with a different, possibly easier method.

I am not after the answer, I am looking for an alternate method.

Hi unstopabl3,

I really don't see what method you used in your first post, but here's how I'd do it.

Plato already told you that the midpoint of BD is M(6, 6).

This means the y-coordinates of B and D are also 6.

This distance from B to D is 20 (found using distance formula)

Each individual segment of the diagonals measure 10 since they are bisected.

Using the distance formula, it is easy to determine the x-coordinates of B and D.

M(6, 6) -----> D(h, 6) = 10

$10=\sqrt{h-6)^2+(6-6)^2}$
• May 9th 2010, 03:42 PM
Soroban
Hello, unstopabl3!

Quote:

Code:

              |          C(12,14)               |          o               |      *    *               |  *        *               *              *         B o - + - - - - - - - o D           *  |          *       ------*-+-------*------------             *|  *               o             A|(0,-2)               |

The diagram shows a rectangle $ABCD.$
We have: . $A(0,-2),\;C(12,14)$
The diagonal $BD$ is parallel to the $x$-axis.

$(i)$ Explain why the $y$-coordinate of $D$ is 6.

The diagonals of a rectangle bisect each other.
. . Hence, the midpoint of $AC$ is the midpoint of $BD.$

The midpoint of AC is: . $\left(\tfrac{0+12}{2},\;\tfrac{-2+14}{2}\right) \:=\:(6,6)$

Therefore, $B$ and $D$ have a $y$-coordinate of 6.

Quote:

The $x$-coordinate of $D$ is $h.$
$(ii)$ Express the gradients of $AD$ and $CD$ in terms of h,.

We have: . $\begin{Bmatrix}A(0, -2) \\ C(12,14) \\ D(h,\;6) \end{Bmatrix}$

$m_{AD} \;=\;\frac{6(-2)}{h-9} \;=\;\frac{8}{h}$

$m_{CD} \;=\;\frac{6-14}{h-12} \;=\;\frac{-8}{h-12}$

Quote:

$(iii)$ Calculate the $x$-coordinates of $D$ and $B.$

Since $m_{AD} \perp m_{CD}$ we have: . $\frac{8}{h} \;=\;\frac{h-12}{8} \quad\Rightarrow\quad h^2-12h - 64 \:=\:0$

Hence: . $(h+4)(h-16) \:=\:0 \quad\Rightarrow\quad h \:=\:-4,\:16$

Therefore: . $D(16,6),\;B(-4,6)$

• May 9th 2010, 10:57 PM
unstopabl3
Quote:

Originally Posted by masters
Hi unstopabl3,

I really don't see what method you used in your first post, but here's how I'd do it.

$10=\sqrt{h-6)^2+(6-6)^2}$

I use the concept of the product of two perpendicular lines = -1
You can see the working in the Soroban's post. That's exactly how I did it!

Quote:

This distance from B to D is 20 (found using distance formula)
How did you get this with only the Y co-ordinates known for both? Did you get the distance of AC which should be equal to BD?

Soroban, thanks for your post, but I've already used that method to solve this problem and already mentioned it in my first post that I am looking for alternative methods to solve it! Thanks nonetheless!
• May 10th 2010, 05:18 AM
masters
Quote:

Originally Posted by unstopabl3
I use the concept of the product of two perpendicular lines = -1
You can see the working in the Soroban's post. That's exactly how I did it!

How did you get this with only the Y co-ordinates known for both? Did you get the distance of AC which should be equal to BD?

Soroban, thanks for your post, but I've already used that method to solve this problem and already mentioned it in my first post that I am looking for alternative methods to solve it! Thanks nonetheless!

The distance BD is 20, found using the distance formula. BD = AC (diagonals of a rectangle are congruent).
• May 10th 2010, 06:42 AM
sa-ri-ga-ma
Co ordinates of B and D are $(x_1 , 6) and (x_2, 6)$
Diagonal AC = BD
AC = 20 = BD.
Distance $BD^2 = (x_1 - x_2)^2$

So (x_1 - x_2) = 20...........(1)

Mid point point of AC = mid point of BD

$\frac{x_1+x_2}{2} = 6$

(x_1 + x_2) = 12......(2)
Solve Eq (1) ans (2) to find the coordinates of B and D.
• May 10th 2010, 08:58 AM
unstopabl3
Thanks for the responses guys!