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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?

 

WHY GEOMETRY WHY????

rarinstraw1195  Feb 5, 2016

Best Answer 

 #8
avatar+18712 
+45

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 \(H(x_H,y_H)\).

 

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

 

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

 

laugh

heureka  Feb 5, 2016
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10+0 Answers

 #1
avatar+8621 
+10

1 and -3

:D

Hayley1  Feb 5, 2016
 #2
avatar+8621 
0

y1-y2

---------

x1-x2

 

Make sense on how I got it? :)

Hayley1  Feb 5, 2016
 #3
avatar+5197 
0

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

rarinstraw1195  Feb 5, 2016
 #4
avatar+78618 
+15

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 :

 

 

 

 

cool cool cool

CPhill  Feb 5, 2016
edited by CPhill  Feb 5, 2016
edited by CPhill  Feb 5, 2016
edited by CPhill  Feb 5, 2016
edited by CPhill  Feb 5, 2016
 #5
avatar+5197 
+5

Thank you CPhill!

rarinstraw1195  Feb 5, 2016
 #6
avatar+78618 
+5

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!!!  ]

 

 

cool cool cool

CPhill  Feb 5, 2016
 #8
avatar+18712 
+45
Best Answer

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 \(H(x_H,y_H)\).

 

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

 

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

 

laugh

heureka  Feb 5, 2016
 #9
avatar+78618 
+5

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

 

 

cool cool cool

CPhill  Feb 5, 2016
 #10
avatar+91001 
+10

This looks very involved!     surprise

 

Thanks Chris and Heureka    :D

Melody  Feb 7, 2016
 #11
avatar+18712 
+10

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 \(H(x_H,y_H)\).

 

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

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

 

laugh

heureka  Feb 8, 2016

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