Bring the coordinates of point B and point C into the equation, and the equation of straight line L: Y = BX+B.
∴ His equation of the perpendicular passing through point A: y=(- 1/B)x+( 1/B)
∴ coordinate of intersection d [( 1-b 2)/( 1+b 2), 2b/( 1+b 2)]
∴OD= 1 (the calculation process will not be written in detail)
(2) y=bx+b
Passing through point A is the intersection of AD⊥l and D. kAD*kl=- 1.
kAD=- 1/b
The linear equation of AD: y=( 1-x)/b
The coordinates of point D are ((1-b 2)/( 1+b 2), 2b/( 1+b 2)).
s△bda=bd*da/2=(√4/( 1+b^2)*√4b^2/( 1+b^2))/2=2b/( 1+b^2)
S△BOC=BO*OC/2=b/2
s△bda/s△boc=a,∴2b/( 1+b^2)/(b/2)=a
The functional relationship between a and b is a = 4/( 1+b 2) 0.
(3) When D is the axis of DE⊥x, (1) indicates that OD = OA ∴∠ ODA = ∠ A.
∴ cotangent value of angle A =2=AE/DE, let DE=a, then AE=2a=AO+OE= 1+OE.
∴OE=2a- 1，
Right triangle ode ∴ de 2+OE 2 = OD 2
Bring ∴1= a2+(2a-1) 2a (5a-4) = 0a = 0 and discard.
∴a=4/5 has the premise of 0.
The analytical formula y=2x+2 of ∴L:.
(4) Let the intersection of AD and Y axis be f, and find the area of quadrilateral OBDF.
The analytical formula of AD can be obtained as y = (-1/2) x+1/2 ∴ f (0,1/2).
The ∴ area of trapezoid FDEO =1/2 (of+de) OE = 39/100.
The area of right triangle DEB =( 1/2)BE*DE=4/25.
The ∴ area of quadrilateral BDFO = trapezoid+right triangle = 1 1/20.