+5265 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????
+26388 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 \(H(x_H,y_H)\).
\(\begin{array}{rcll} \binom{x_G}{y_G} &=& \binom{x_F}{y_F} +\frac12 \cdot \binom{x_E-x_F}{y_E-y_F} + \frac12 \cdot \binom{-(y_E-y_F)}{x_E-x_F} \\ \binom{x_G}{y_G} &=& \binom{0}{-3} +\frac12 \cdot \binom{2-0}{3-(-3)} + \frac12 \cdot \binom{-(3-(-3))}{2-0} \\ \binom{x_G}{y_G} &=& \binom{0}{-3} +\frac12 \cdot \binom{2}{6} + \frac12 \cdot \binom{-6}{2} \\ \binom{x_G}{y_G} &=& \binom{0}{-3} +\binom{\frac12\cdot 2}{\frac12\cdot 6} + \binom{\frac12\cdot (-6)}{\frac12\cdot 2} \\ \binom{x_G}{y_G} &=& \binom{0}{-3} +\binom{1}{3} + \binom{-3}{1} \\ \binom{x_G}{y_G} &=& \binom{0+1-3}{-3+3+1} \\ \binom{x_G}{y_G} &=& \binom{-2}{1} \\ \end{array}\)
\(\begin{array}{rcll} \binom{x_H}{y_H} &=& \binom{x_F}{y_F} +\frac12 \cdot \binom{x_E-x_F}{y_E-y_F} - \frac12 \cdot \binom{-(y_E-y_F)}{x_E-x_F} \\ \binom{x_H}{y_H} &=& \binom{0}{-3} +\frac12 \cdot \binom{2-0}{3-(-3)} - \frac12 \cdot \binom{-(3-(-3))}{2-0} \\ \binom{x_H}{y_H} &=& \binom{0}{-3} +\frac12 \cdot \binom{2}{6} - \frac12 \cdot \binom{-6}{2} \\ \binom{x_H}{y_H} &=& \binom{0}{-3} +\binom{\frac12\cdot 2}{\frac12\cdot 6} - \binom{\frac12\cdot (-6)}{\frac12\cdot 2} \\ \binom{x_H}{y_H} &=& \binom{0}{-3} +\binom{1}{3} - \binom{-3}{1} \\ \binom{x_H}{y_H} &=& \binom{0+1-(-3)}{-3+3-1} \\ \binom{x_H}{y_H} &=& \binom{0+1+3}{-3+3-1} \\ \binom{x_H}{y_H} &=& \binom{4}{-1} \\ \end{array}\)
+5265 Yes, but I need both endpoints, not just one. Thx though Hayley! :D
+129850 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 :
+129850 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!!! ]
+26388 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 \(H(x_H,y_H)\).
\(\begin{array}{rcll} \binom{x_G}{y_G} &=& \binom{x_F}{y_F} +\frac12 \cdot \binom{x_E-x_F}{y_E-y_F} + \frac12 \cdot \binom{-(y_E-y_F)}{x_E-x_F} \\ \binom{x_G}{y_G} &=& \binom{0}{-3} +\frac12 \cdot \binom{2-0}{3-(-3)} + \frac12 \cdot \binom{-(3-(-3))}{2-0} \\ \binom{x_G}{y_G} &=& \binom{0}{-3} +\frac12 \cdot \binom{2}{6} + \frac12 \cdot \binom{-6}{2} \\ \binom{x_G}{y_G} &=& \binom{0}{-3} +\binom{\frac12\cdot 2}{\frac12\cdot 6} + \binom{\frac12\cdot (-6)}{\frac12\cdot 2} \\ \binom{x_G}{y_G} &=& \binom{0}{-3} +\binom{1}{3} + \binom{-3}{1} \\ \binom{x_G}{y_G} &=& \binom{0+1-3}{-3+3+1} \\ \binom{x_G}{y_G} &=& \binom{-2}{1} \\ \end{array}\)
\(\begin{array}{rcll} \binom{x_H}{y_H} &=& \binom{x_F}{y_F} +\frac12 \cdot \binom{x_E-x_F}{y_E-y_F} - \frac12 \cdot \binom{-(y_E-y_F)}{x_E-x_F} \\ \binom{x_H}{y_H} &=& \binom{0}{-3} +\frac12 \cdot \binom{2-0}{3-(-3)} - \frac12 \cdot \binom{-(3-(-3))}{2-0} \\ \binom{x_H}{y_H} &=& \binom{0}{-3} +\frac12 \cdot \binom{2}{6} - \frac12 \cdot \binom{-6}{2} \\ \binom{x_H}{y_H} &=& \binom{0}{-3} +\binom{\frac12\cdot 2}{\frac12\cdot 6} - \binom{\frac12\cdot (-6)}{\frac12\cdot 2} \\ \binom{x_H}{y_H} &=& \binom{0}{-3} +\binom{1}{3} - \binom{-3}{1} \\ \binom{x_H}{y_H} &=& \binom{0+1-(-3)}{-3+3-1} \\ \binom{x_H}{y_H} &=& \binom{0+1+3}{-3+3-1} \\ \binom{x_H}{y_H} &=& \binom{4}{-1} \\ \end{array}\)
+26388 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 \(H(x_H,y_H)\).
\(\begin{array}{rcll} \vec{E} &=& \binom{2}{3} \\ \vec{F} &=& \binom{0}{-3} \\\\ \vec{G} &=& \vec{F} + \frac12 \cdot ( \vec{E}-\vec{F} ) + \frac12 \cdot ( \vec{E}-\vec{F} )_\perp\\ \vec{H} &=& \vec{F} + \frac12 \cdot ( \vec{E}-\vec{F} ) - \frac12 \cdot ( \vec{E}-\vec{F} )_\perp\\\\ \end{array}\)
\(\begin{array}{lcll} \text{please note, if }~ \vec{a} = \binom{x_a}{y_a} ~\text{ then }~ \vec{a}_\perp = \binom{-y_a}{x_a}~\text{ or }~ \vec{a}_\perp = \binom{y_a}{-x_a} \\ \text{because }~ \vec{a} \cdot \vec{a}_\perp ~\text{ must be null, if be perpendicular to one another.}\\ \binom{x_a}{y_a} \cdot \binom{-y_a}{x_a} = -x_a\cdot y_a + x_a\cdot y_a = 0\\ \binom{x_a}{y_a} \cdot \binom{y_a}{-x_a} = x_a\cdot y_a - x_a\cdot y_a = 0\\ \end{array}\)
