(2) To find the area of △BDF, we must first know the ordinate of the bottom and height of △BDF, with the bottom =BD and the height = F.
According to the analytical formula of straight line y=kx+2k and straight line AB, the coordinate of F can be obtained.
(3) Whether there is a straight line EF that makes S△DEC=S△AEF can be based on the hypothesis method. If yes, then S△DEC=S△AEF is a condition. According to this condition, an equation can be listed to find the value of K.
Solution:
(1) Let the coordinate of d be (a, 0).
0=ka+2k
ka=-2k
a=-2
∴D(-2,0)
(2) Let the analytical formula of line segment AB be y = k1x+b.
∫A(0，√3)，B( 1，0)
∴b=√3,k 1=-√3
∴y=-√3x+√3 ①
∫y = kx+2k②
Equation ① and Equation ② are simultaneous equations, and ①-② x=(√3-2k)/(k+√3). Substitute x into ① to get y=(3√3k )/(k+√3).
∴△ Height of BDFBD edge h=(3√3k )/(k+√3)
∫BO = 1，DO=2，
∴BD=BO+DO=3
∴s△bdf= 1/2*bd*h= 1/2*3*(3√3k)/(k+√3)=(9√3k)/(2k+2√3
(3) If there is a straight line EF, then S△DEC=S△AEF.
∫S△DEC = S△AEF
∴S△DEC+S quadrilateral CBFE=S△AEF+S quadrilateral CBFE
∴S△DBF=S△ABC
∴( 1/2)bd*[(3√3k)/(k+√3)]=( 1/2)BC * ao
∴( 1/2)*3*[(3√3k)/(k+√3)]=( 1/2)* 2 *√3。
Solution: k=(2√3)/7.
∴ The relation of straight line DF is y=[(2√3)/7]x+(4√3)/7.