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# Coordinate Geomerty

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90
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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

#2
+18369
+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 $$\vec{C}$$

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

heureka  Mar 1, 2017
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#1
+75333
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 :

CPhill  Mar 1, 2017
#2
+18369
+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 $$\vec{C}$$

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

heureka  Mar 1, 2017
#3
+8629
+5

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

Omi67  Mar 1, 2017

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