Another geometry question... :/

A square graphed on the coordinate plane has a diagonal with endpoints E(2,3) and F(0,-3). What are the coords of the endpoints of the other diagonal?

1. Diagonal: $E (x_E=2,y_E=3)$ and $F(x_F=0,y_F=-3)$.  

What are the coords of the endpoints of the other diagonal $G(x_G,y_G)$ and $H(x_H,y_H)$.

$\begin{array}{rcll} \binom{x_G}{y_G} &=& \binom{x_F}{y_F} +\frac12 \cdot \binom{x_E-x_F}{y_E-y_F} + \frac12 \cdot \binom{-(y_E-y_F)}{x_E-x_F} \\ \binom{x_G}{y_G} &=& \binom{0}{-3} +\frac12 \cdot \binom{2-0}{3-(-3)} + \frac12 \cdot \binom{-(3-(-3))}{2-0} \\ \binom{x_G}{y_G} &=& \binom{0}{-3} +\frac12 \cdot \binom{2}{6} + \frac12 \cdot \binom{-6}{2} \\ \binom{x_G}{y_G} &=& \binom{0}{-3} +\binom{\frac12\cdot 2}{\frac12\cdot 6} + \binom{\frac12\cdot (-6)}{\frac12\cdot 2} \\ \binom{x_G}{y_G} &=& \binom{0}{-3} +\binom{1}{3} + \binom{-3}{1} \\ \binom{x_G}{y_G} &=& \binom{0+1-3}{-3+3+1} \\ \binom{x_G}{y_G} &=& \binom{-2}{1} \\ \end{array}$

$\begin{array}{rcll} \binom{x_H}{y_H} &=& \binom{x_F}{y_F} +\frac12 \cdot \binom{x_E-x_F}{y_E-y_F} - \frac12 \cdot \binom{-(y_E-y_F)}{x_E-x_F} \\ \binom{x_H}{y_H} &=& \binom{0}{-3} +\frac12 \cdot \binom{2-0}{3-(-3)} - \frac12 \cdot \binom{-(3-(-3))}{2-0} \\ \binom{x_H}{y_H} &=& \binom{0}{-3} +\frac12 \cdot \binom{2}{6} - \frac12 \cdot \binom{-6}{2} \\ \binom{x_H}{y_H} &=& \binom{0}{-3} +\binom{\frac12\cdot 2}{\frac12\cdot 6} - \binom{\frac12\cdot (-6)}{\frac12\cdot 2} \\ \binom{x_H}{y_H} &=& \binom{0}{-3} +\binom{1}{3} - \binom{-3}{1} \\ \binom{x_H}{y_H} &=& \binom{0+1-(-3)}{-3+3-1} \\ \binom{x_H}{y_H} &=& \binom{0+1+3}{-3+3-1} \\ \binom{x_H}{y_H} &=& \binom{4}{-1} \\ \end{array}$

1 and -3

:D

y1-y2

---------

x1-x2

Make sense on how I got it? :)

Yes, but I need both endpoints, not just one. Thx though Hayley! :D

Rarinstraw.....here's one way to do this:

Note that the mid-point of the diagonal is located at [ (0 +2)/2, (-3 +3)/2 ] = [2/2 , 0/2] =  (1,0)

Now.....the diagonal length will be =  sqrt [2^2 + (-3-3)^2]  = sqrt(40)

And 1/2 of this  =  sqrt(40)/2

Now....if we construct  a circle centered at (1,0)  with a radius  of 1/2 of the diagonal length, this circle will pass through the given diagonal endpoints as well as the ones we are looking for.  And the equation  of such a circle becomes :

(x - 1)^2  + y^2  = [sqrt(40)/2]^2 

(x - 1)^2 + y^2  = 40/4

(x - 1)^2 + y^2  = 10

Now....the slope of the given diagonal  is  [-3-3] / [0-2]  = -6/-2  = 3

But the other diagonal will lie on a line with the negative reciprocal slope and it will go through the point (1,0)....so the equation  of this line is:

y = (-1/3)(x -1)

y = (-1/3)x + 1/3

y = (1 - x)/3

And putting this into our equation of the circle for y to solve for x, we can find the x coordinates of the diagonal endpoints we are looking for :

(x - 1)^2  + ( [x - 1]/3)^2  = 10   simplify

x^2 - 2x + 1  + [1/9](x^2 - 2x + 1)  = 10      multiply through by 9

9x^2 - 18x + 9 + x^2 - 2x + 1  = 90   simplify

10x^2 - 20x - 80  = 0   divide through by 10

x^2 - 2x  - 8  = 0        factor

(x - 4) (x + 2)  = 0      and setting each factor to 0, we have that   x = 4  and x = -2

So.....we can find the y coordinates of the diagonal endpoints, thusly :

When x = 4  ....   y =  [1- 4]/ 3  = -3/3 = -1

When x= -2 ....    y  = [ 1 - (-2)]/ 3  = 3/3  = 1

So....the endpoints of the other diagonal are (4, -1) and ( -2, 1)

Here's a pic that verifies the results :

Thank you CPhill!

Thanks, Rarinstraw......that one took me a little while to figure out  how to approach it...!!!!

[ There are  probably easier ways.....but....I couldn't think of any of them....LOL!!!  ]

Thanks, heureka........what method did you use for your solution?????

This looks very involved!     

Thanks Chris and Heureka    :D

What method did you use for your solution?

What are the coords of the endpoints of the other diagonal $G(x_G,y_G)$and $H(x_H,y_H)$.

$\begin{array}{rcll} \vec{E} &=& \binom{2}{3} \\ \vec{F} &=& \binom{0}{-3} \\\\ \vec{G} &=& \vec{F} + \frac12 \cdot ( \vec{E}-\vec{F} ) + \frac12 \cdot ( \vec{E}-\vec{F} )_\perp\\ \vec{H} &=& \vec{F} + \frac12 \cdot ( \vec{E}-\vec{F} ) - \frac12 \cdot ( \vec{E}-\vec{F} )_\perp\\\\ \end{array}$

$\begin{array}{lcll} \text{please note, if }~ \vec{a} = \binom{x_a}{y_a} ~\text{ then }~ \vec{a}_\perp = \binom{-y_a}{x_a}~\text{ or }~ \vec{a}_\perp = \binom{y_a}{-x_a} \\ \text{because }~ \vec{a} \cdot \vec{a}_\perp ~\text{ must be null, if be perpendicular to one another.}\\ \binom{x_a}{y_a} \cdot \binom{-y_a}{x_a} = -x_a\cdot y_a + x_a\cdot y_a = 0\\ \binom{x_a}{y_a} \cdot \binom{y_a}{-x_a} = x_a\cdot y_a - x_a\cdot y_a = 0\\ \end{array}$