The quadrilateral ABCD, where A is (4,5) and C (3,-2) is a square. Find the coordinates of B and D.
The quadrilateral ABCD, where A is (4,5) and C (3,-2) is a square. Find the coordinates of B and D.
Let →A=(45)Let →C=(3−2)
→O is the midpoint of →A and →C
→AC=→A−→C=(45)−(3−2)=(17)→OC=12→AC=12(17)=(0.53.5)→OD=→OC⊥→OC⊥=(−3.50.5)→OC⋅→OC⊥=0 ✓=(−3.50.5)→OB=−→OD=(3.5−0.5)→D=→C+→OC+→OD=(3−2)+(0.53.5)+(−3.50.5)=(02)→B=→C+→OC+→OB=(3−2)+(0.53.5)+(3.5−0.5)=(71)
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 :
The quadrilateral ABCD, where A is (4,5) and C (3,-2) is a square. Find the coordinates of B and D.
Let →A=(45)Let →C=(3−2)
→O is the midpoint of →A and →C
→AC=→A−→C=(45)−(3−2)=(17)→OC=12→AC=12(17)=(0.53.5)→OD=→OC⊥→OC⊥=(−3.50.5)→OC⋅→OC⊥=0 ✓=(−3.50.5)→OB=−→OD=(3.5−0.5)→D=→C+→OC+→OD=(3−2)+(0.53.5)+(−3.50.5)=(02)→B=→C+→OC+→OB=(3−2)+(0.53.5)+(3.5−0.5)=(71)