The quadrilateral ABCD, where A is (4,5) and C (3,-2) is a square. Find the coordinates of B and D.
+26367 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 \(\vec{C}\)
\(\begin{array}{|rcll|} \hline \vec{AC} &=& \vec{A}-\vec{C} = \binom{4}{5}-\binom{3}{-2}=\binom{1}{7} \\ \vec{OC} &=& \frac12\vec{AC} = \frac12\binom{1}{7}=\binom{0.5}{3.5} \\ \vec{OD} &=& \vec{OC}_{\perp} \qquad \vec{OC}_{\perp} = \binom{-3.5}{0.5} \qquad \vec{OC}\cdot\vec{OC}_{\perp} = 0\ \checkmark \\ &=& \binom{-3.5}{0.5} \\ \vec{OB} &=& -\vec{OD}=\binom{3.5}{-0.5} \\ \vec{D} &=& \vec{C}+\vec{OC}+\vec{OD}= \binom{3}{-2}+\binom{0.5}{3.5}+\binom{-3.5}{0.5} \\ &=& \binom{0}{2} \\ \vec{B} &=& \vec{C}+\vec{OC}+\vec{OB}= \binom{3}{-2}+\binom{0.5}{3.5}+\binom{3.5}{-0.5} \\ &=& \binom{7}{1} \\ \hline \end{array} \)
+128406 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 :
+26367 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 \(\vec{C}\)
\(\begin{array}{|rcll|} \hline \vec{AC} &=& \vec{A}-\vec{C} = \binom{4}{5}-\binom{3}{-2}=\binom{1}{7} \\ \vec{OC} &=& \frac12\vec{AC} = \frac12\binom{1}{7}=\binom{0.5}{3.5} \\ \vec{OD} &=& \vec{OC}_{\perp} \qquad \vec{OC}_{\perp} = \binom{-3.5}{0.5} \qquad \vec{OC}\cdot\vec{OC}_{\perp} = 0\ \checkmark \\ &=& \binom{-3.5}{0.5} \\ \vec{OB} &=& -\vec{OD}=\binom{3.5}{-0.5} \\ \vec{D} &=& \vec{C}+\vec{OC}+\vec{OD}= \binom{3}{-2}+\binom{0.5}{3.5}+\binom{-3.5}{0.5} \\ &=& \binom{0}{2} \\ \vec{B} &=& \vec{C}+\vec{OC}+\vec{OB}= \binom{3}{-2}+\binom{0.5}{3.5}+\binom{3.5}{-0.5} \\ &=& \binom{7}{1} \\ \hline \end{array} \)
