In the figure, what is the area of triangle $ABD$

Maybe an easier way.....but.....

A  = (0, 4)

F = (3, 0)

slope of AF = -4/3

equation of line containing AF    ....   y  = (-4/3)x + 4

E  = ( 0, 2)

B = (7,0)

slope of EB  = -2/7

equation of line containing EB  ...  y = (-2/7)x + 2

Find the x coordinate of D by setting these two functions equal

(-4/3)x + 4  = (-2/7)x + 2

4 - 2 = (-2/7  + 4/3)x

2  = [-6 + 28] /21 x

2 = [22/21]x

42/22   =x

21/11   = x

And the associated y coordinate  is  (-2/7)(21/11) + 2  = 16/11

So  D  = (21/11 , 16,11)

And the equation of the line joining AB  is  y  =   (-4/7)x + 4

Write this is standard form  ....   4x + 7y  - 28   =   0

Now...using the  equation to find the distance from  a point to a line we can find the altitude of triangle ABD

l   4(21/11) + 7(16/11)  - 28  l                   l  -112/11 l              112/11

_______________________   =           __________  =        _____

     √[ 4^2 + 7^2 ]                                         √65                       √65

And considering AB to be the base, its length is  √ [4^2  + 7^2]  = √65

So...the area  of triangle ABD  =  (1/2) √65  * (112/11) / √65  =

(1/2) (112 / 11)  =

112 /22  =

56 /21  units^2