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The quadrilateral ABCD, where A is (4,5) and C (3,-2) is a square. Find the coordinates of B and D.

Guest Mar 1, 2017

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 #2
avatar+18355 
+5

The quadrilateral ABCD, where A is (4,5) and C (3,-2) is a square. Find the coordinates of B and D.


\(\text{Let}\ \vec{A} = \binom{4}{5}\\ \text{Let}\ \vec{C} = \binom{3}{-2} \)

 

\(\vec{O}\) is the midpoint of \(\vec{A}\) and \(\vec{C}\)

 

\(\begin{array}{|rcll|} \hline \vec{AC} &=& \vec{A}-\vec{C} = \binom{4}{5}-\binom{3}{-2}=\binom{1}{7} \\ \vec{OC} &=& \frac12\vec{AC} = \frac12\binom{1}{7}=\binom{0.5}{3.5} \\ \vec{OD} &=& \vec{OC}_{\perp} \qquad \vec{OC}_{\perp} = \binom{-3.5}{0.5} \qquad \vec{OC}\cdot\vec{OC}_{\perp} = 0\ \checkmark \\ &=& \binom{-3.5}{0.5} \\ \vec{OB} &=& -\vec{OD}=\binom{3.5}{-0.5} \\ \vec{D} &=& \vec{C}+\vec{OC}+\vec{OD}= \binom{3}{-2}+\binom{0.5}{3.5}+\binom{-3.5}{0.5} \\ &=& \binom{0}{2} \\ \vec{B} &=& \vec{C}+\vec{OC}+\vec{OB}= \binom{3}{-2}+\binom{0.5}{3.5}+\binom{3.5}{-0.5} \\ &=& \binom{7}{1} \\ \hline \end{array} \)

 

 

laugh

heureka  Mar 1, 2017
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3+0 Answers

 #1
avatar+75299 
0

Maybe not the fastest [or easiest ] way to do this....but

 

The midpoint of ( 4,5)  and ( 3, -2) is  (3.5, 1.5)

 

And we can let A and C  be the endpoints of a diameter of a circle that has its center at (3.5, 1.5)......and the radius^2  = (4 -3.5)^2 + (5 - 1.5)^2  = 12.5

 

So.....the equation of this circle will be  (x - 3.5)^2 + (y - 1.5)^2  = 12.5

 

And the slope of the line joining AC  = 7.........a perpedicular line to this passing through (3.5,1.5) will have a slope  = (-1/7)...

 

So the equation of this line  is  y = (-1/7)(x - 3.5) + 1.5

 

And "B" and "D" will lie on the intersection of this line and the  circle

 

Sub this linear equation into  the equation of the circle for y  and we have

 

(x - 3.5)^2 + [ ( -1/7)(x - 3.5) + 1.5 - 1.5]^2  = 12.5

 

(x - 3.5)^2 + ( -x/7 + .5)^2  = 12.5

 

x^2 - 7x + 12.25 + x^2/49 - x/7 + .25  =  12.5

 

x^2 - 7x + x^2/49 - (1/7)x   = 0

 

49x^2 - 343x + x^2 - 7x  = 0

 

50x^2 - 350x = 0

 

x^2 - 7x  =  0   factor

 

x(x - 7)  = 0      so x = 0  and x = 7

 

And the associated y coordinates are 

 

y = (-1/7) (0 - 3.5) + 1.5  =  2     and

 

y = (-1/7) (7 - 3.5) + 1.5   =   1

 

So....  "B"   = ( 7, 1)    and "D"   = (0, 2)

 

Here's a pic :

 

 

 

cool cool cool

CPhill  Mar 1, 2017
 #2
avatar+18355 
+5
Best Answer

The quadrilateral ABCD, where A is (4,5) and C (3,-2) is a square. Find the coordinates of B and D.


\(\text{Let}\ \vec{A} = \binom{4}{5}\\ \text{Let}\ \vec{C} = \binom{3}{-2} \)

 

\(\vec{O}\) is the midpoint of \(\vec{A}\) and \(\vec{C}\)

 

\(\begin{array}{|rcll|} \hline \vec{AC} &=& \vec{A}-\vec{C} = \binom{4}{5}-\binom{3}{-2}=\binom{1}{7} \\ \vec{OC} &=& \frac12\vec{AC} = \frac12\binom{1}{7}=\binom{0.5}{3.5} \\ \vec{OD} &=& \vec{OC}_{\perp} \qquad \vec{OC}_{\perp} = \binom{-3.5}{0.5} \qquad \vec{OC}\cdot\vec{OC}_{\perp} = 0\ \checkmark \\ &=& \binom{-3.5}{0.5} \\ \vec{OB} &=& -\vec{OD}=\binom{3.5}{-0.5} \\ \vec{D} &=& \vec{C}+\vec{OC}+\vec{OD}= \binom{3}{-2}+\binom{0.5}{3.5}+\binom{-3.5}{0.5} \\ &=& \binom{0}{2} \\ \vec{B} &=& \vec{C}+\vec{OC}+\vec{OB}= \binom{3}{-2}+\binom{0.5}{3.5}+\binom{3.5}{-0.5} \\ &=& \binom{7}{1} \\ \hline \end{array} \)

 

 

laugh

heureka  Mar 1, 2017
 #3
avatar+8629 
+5

The quadrilateral ABCD, where A is (4,5) and C (3,-2) is a square. Find the coordinates of B and D.

 

laugh

Omi67  Mar 1, 2017

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