# [SOLVED] Finding Co-Ordinates of a Rectangle

#### unstopabl3 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.

#### Plato

MHF Helper
The diagonals bisect one another. The midpoint of $$\displaystyle \overline{AC}$$ is ?

#### 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?

#### Plato

MHF Helper
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 $$\displaystyle B~\&~D$$ is 6.
You are given the x-coordinate of $$\displaystyle D$$, so $$\displaystyle D h,6)$$

#### 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.

#### masters

MHF Helper
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

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

#### Soroban

MHF Hall of Honor
Hello, unstopabl3!

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

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

$$\displaystyle (i)$$ Explain why the $$\displaystyle y$$-coordinate of $$\displaystyle D$$ is 6.
The diagonals of a rectangle bisect each other.
. . Hence, the midpoint of $$\displaystyle AC$$ is the midpoint of $$\displaystyle BD.$$

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

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

The $$\displaystyle x$$-coordinate of $$\displaystyle D$$ is $$\displaystyle h.$$
$$\displaystyle (ii)$$ Express the gradients of $$\displaystyle AD$$ and $$\displaystyle CD$$ in terms of h,.
We have: .$$\displaystyle \begin{Bmatrix}A(0, -2) \\ C(12,14) \\ D(h,\;6) \end{Bmatrix}$$

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

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

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

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

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

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

#### unstopabl3

Hi unstopabl3,

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

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

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!

Last edited:

#### masters

MHF Helper
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).

• unstopabl3

#### sa-ri-ga-ma

Co ordinates of B and D are $$\displaystyle (x_1 , 6) and (x_2, 6)$$
Diagonal AC = BD
AC = 20 = BD.
Distance $$\displaystyle BD^2 = (x_1 - x_2)^2$$

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

Mid point point of AC = mid point of BD

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

(x_1 + x_2) = 12......(2)
Solve Eq (1) ans (2) to find the coordinates of B and D.

• unstopabl3