help

Let C  = (0,0)     A  = (0,4)  and  B = (7,0)  F = ( 3,0)   E = (0,2)

The area of triangle ABC  = (1/2) ( 4) (7)  = 14

And the area of triangle ACF  = (1/2)(3)(4)  = 6

Now...

The line through  AF has a slope  of (-4/3)

And the equation of this line is  y = -(4/3)x + 4      (1)

The line through EB has a  slope of (-2/7)

And the equation of this line is  y = -(2/7)x + 2    (2)

Find the x intersection of   (1) and (2)

-(4/3)x + 4 =   -(2/7)x + 2

2 = -(2/7)x + (4/3)x

2= (22/21)x

21/11= x

And the  y coordinate of their intersection  = -(2/7)(21/11) + 2  = 16/11

Call this intersction point G = ( 21/11. 16/11)

So.....the height of triangle  DFB  = 16/11    and the  area of triangle  DFB  = (1/2)(4) (16/11)=  32/11

So....area of  triangle ADB  =  

Area of tiangle ABC  - Area of triangle ACF  - Area of triangle  DFB  =

14 - 6 - 32/11 =

8 - 32/11 = 

(88 - 32) /11 =

56 / 11   units^2