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.

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- Jan 10th 2010, 07:08 AMGR9STUDENTProving 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. - Jan 10th 2010, 07:32 AMVonNemo19
- Jan 10th 2010, 07:34 AMGR9STUDENT
Sorry, I did not list the vertices, they are (1,0) (-1,5) (4,7)(6,2)

- Jan 10th 2010, 07:57 AMVonNemo19
1 way to do it:

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

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

Do you get what I'm saying? - Jan 10th 2010, 08:12 AMGR9STUDENT
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. - Jan 10th 2010, 08:24 AMVonNemo19
OK.

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

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

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

Now, since $\displaystyle -\frac{1}{m_1}=m_2$ it follows that $\displaystyle AB$ is perpindicular to $\displaystyle BC$. - Jan 10th 2010, 08:49 AMGR9STUDENT
Thankyou, that makes sense.