1. ## Proving quadrilateral is a rectangle algebraically

How do you prove algebraically, that the following vertices are a rectangle. I can plot it on a graph, but I can't figure out how to prove using algebra.

Help.

2. Originally Posted by GR9STUDENT
How do you prove algebraically, that the following vertices are a rectangle. I can plot it on a graph, but I can't figure out how to prove using algebra.

Help.
Would you happen to have the vertices in question? We typically look for equal lengths and equal slopes...

3. Sorry, I did not list the vertices, they are (1,0) (-1,5) (4,7)(6,2)

4. Originally Posted by GR9STUDENT
Sorry, I did not list the vertices, they are (1,0) (-1,5) (4,7)(6,2)
1 way to do it:

Let the given points correspond to $A,B,C,\text{ and }D$ respectively.

Then to show that $ABCD$ is a rectangle, we must show that each interior angle measures $90^{\circ}$ . To do this, we must need to only compare the slopes of line segments $AB$ with $BC$, and $BC$ with $CD$ and etc...

Do you get what I'm saying?

5. No, I am lost on this one.

I understand that the rectangle has four 90 angles, and that the slope of each pair of lines will equal each other. But I don't know how to show with forumula.

6. Originally Posted by GR9STUDENT
No, I am lost on this one.

I understand that the rectangle has four 90 angles, and that the slope of each pair of lines will equal each other. But I don't know how to show with forumula.
OK.

Perpindicular lines will have slopes that are the negative recirocal respective to each other.

So $AB$ has slope $m_1=\frac{5-0}{-1-1}=-\frac{5}{2}$

and $BC$ has slope $m_2=\frac{7-5}{4-(-1)}=\frac{2}{5}$

Now, since $-\frac{1}{m_1}=m_2$ it follows that $AB$ is perpindicular to $BC$.

7. Thankyou, that makes sense.