1. ## Proving a quadrilateral is a rectangle?

I'm having a little trouble with a question I'm doing. I'll write it below.

The vertices of a quadrilateral are C(-2,4), D(3,9), E(6,6), and F(1,1).

a) Show that the quadrilateral is a rectangle.

First I calculated the length of its four sides.

CD = square root of 50
DE = square root of 18
EF = square root of 50
CF = square root of 18

Ok, so far so good. Now I have to show that it has a right angle, which would require two adjacent sides being perpendicular.

mCD = 1, but mDE = -6/0?

Did I mess up somewhere? Any help/advice is appreciated.

2. Originally Posted by mathdonkey
mCD = 1, but mDE = -6/0?
Now, what are you doing here?

If you are familiair with vectors, try to express each side as a vector. For an example, $\bar{CD}=\begin{pmatrix}3-(-2) \\ 9-4\end{pmatrix}=\begin{pmatrix}5 \\ 5\end{pmatrix}$
and $\bar{DE}=\begin{pmatrix}6-3 \\ 6-9\end{pmatrix}=\begin{pmatrix}3 \\ -3\end{pmatrix}$

For $\bar{CD}$ and $\bar{DE}$ to perpendicular we must have that the scalar product is zero.

$\bar{CD} \cdot \bar{DE}=\begin{pmatrix}5 \\ 5\end{pmatrix} \cdot \begin{pmatrix}3 \\ -3\end{pmatrix}= 5(3)+5(-3)=0$, so the vectors are perpendicular.

3. Originally Posted by Aliquantus
Now, what are you doing here?

If you are familiair with vectors, try to express each side as a vector. For an example, $\bar{CD}=\begin{pmatrix}3-(-2) \\ 9-4\end{pmatrix}=\begin{pmatrix}5 \\ 5\end{pmatrix}$
and $\bar{DE}=\begin{pmatrix}6-3 \\ 6-9\end{pmatrix}=\begin{pmatrix}3 \\ -3\end{pmatrix}$

For $\bar{CD}$ and $\bar{DE}$ to perpendicular we must have that the scalar product is zero.

$\bar{CD} \cdot \bar{DE}=\begin{pmatrix}5 \\ 5\end{pmatrix} \cdot \begin{pmatrix}3 \\ -3\end{pmatrix}= 5(3)+5(-3)=0$, so the vectors are perpendicular.

mCD=-2-4/3-9
=-6/-6
=1

mDE=3-9/6-6
=-6/0

I think I'm overlooking something, I just can't figure out what.

4. Hello, mathdonkey!

Your reasoning is absolutely correct, but . . .

$m_{CD} \:= \:1\text{, but }m_{DE} \:= \:{\bf\frac{\text{-}6}{0}}\; ?$ . . . . How?

We have: . $D(3,9)\text{ and }E(6,6)$

Therefore: . $m_{DE} \;=\;\frac{6-9}{6-3} \;=\;\frac{\text{-}3}{3} \;=\;-1$

5. Originally Posted by Soroban
Hello, mathdonkey!

Your reasoning is absolutely correct, but . . .

We have: . $D(3,9)\text{ and }E(6,6)$

Therefore: . $m_{DE} \;=\;\frac{6-9}{6-3} \;=\;\frac{\text{-}3}{3} \;=\;-1$

Hi Soroban,

Thanks for looking over my work!

I see what you did to get an answer of -1, but the method you used is different than the one I used. In yours the vertice is seperated and combined with the other vertice, in mine it isn't. I don't understand why this is? Sorry if I'm not being clear enough, I don't talk math often.

6. Originally Posted by mathdonkey
I'm having a little trouble with a question I'm doing. I'll write it below.

The vertices of a quadrilateral are C(-2,4), D(3,9), E(6,6), and F(1,1).

... two adjacent sides being perpendicular.

mCD = 1, but mDE = -6/0?

Did I mess up somewhere? Any help/advice is appreciated.
Hello,

if you have 2 points $P_1\left(x_1,\ y_1 \right)$ and $P_2\left(x_2,\ y_2 \right)$ then the slope of the line which connect these 2 points is calculated by:

$m_{P_1, P_2} = \frac{y_1 - y_2}{x_1 - x_2}$. .......... That's the formula Soroban used in his post:

$m_{D, E}=\frac{9 - 6}{3 - 6} = \frac3{-3}= -1$

I can't imagine how you got a zero in the denominator

7. Originally Posted by earboth
Hello,

if you have 2 points $P_1\left(x_1,\ y_1 \right)$ and $P_2\left(x_2,\ y_2 \right)$ then the slope of the line which connect these 2 points is calculated by:

$m_{P_1, P_2} = \frac{y_1 - y_2}{x_1 - x_2}$. .......... That's the formula Soroban used in his post:

$m_{D, E}=\frac{9 - 6}{3 - 6} = \frac3{-3}= -1$

I can't imagine how you got a zero in the denominator
I wasn't doing it right.

Thanks for the help guys.

8. Oops, ignore this post.

Thanks again for the help guys!

9. Here’s something you can try. Multiply the points C, D, E, F by the matrix $\left(\begin{array}{rr}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\\\
-\,\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right)$
, which is a clockwise rotation of 45° about the origin.

$\frac{1}{\sqrt{2}}\left(\begin{array}{rr}1&1\\-1&1
\end{array}\right)\left(\begin{array}{rrrr}-2&3&6&1\\4&9&6&1
\end{array}\right)=\left(\begin{array}{rrrr}\sqrt{ 2}&6\sqrt{2}&6\sqrt{2}&\sqrt{2}\\3\sqrt{2}&3\sqrt{ 2}&0&0
\end{array}\right)$

Hence we have rotated the vertices of the quadrilateral to $\mathrm{C}'(\sqrt{2},3\sqrt{2})$, $\mathrm{D}'(6\sqrt{2},3\sqrt{2})$, $\mathrm{E}'(6\sqrt{2},0)$, $\mathrm{F}'(\sqrt{2},0)$. Now it is much easier to see that $\mathrm{C}'\mathrm{D}'\mathrm{E}'\mathrm{F}'$ is a rectangle.