+0

# Coordinate Geomerty

0
197
3

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
+19083
+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
Sort:

#1
+84168
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
+19083
+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
+9046
+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

### 20 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details