Area of Triangle Given Vertices

The required figure in as follows: 

In the above figure, O is the origin and F is the $$x$$-intercept of the side AB.

Hence, F = (4,0) 

Now, it is given that E is dividing AD in two parts such that DE : AE = 2 : 1

Now, Length of OD = Distance of point D from O = 8 units 

Length of OF = Distance of point F from O = 4 units 

OD : OF = 8 : 4 = 2 : 1

Since, the ratio of DE and AE is the same as the ratio of OD and OF, we can say that E lies on the $$y$$-axis as it divides the side AD in the same ratio as O divides DF.

Let the coordinates of E = (0,$$y$$)

Now, In $$\triangle\ $$ DOE and $$\triangle\ $$ DFA, 

$$\angle\ $$ ODE = $$\angle\ $$ FDA 

$$\angle\ $$ DOE = $$\angle\ $$ DFA = 90$$^{\circ\ }$$

Since, two angles are equal, 

Hence, from Angle-Angle criteria, $$\triangle\ $$ DOE $$\sim$$ $$\triangle\ $$ DFA.

Hence, $$\dfrac{OE}{FA}=\dfrac{OD}{FD}$$

$$\dfrac{y}{5}=\dfrac{8}{12}$$

$$y=\dfrac{10}{3}$$

Hence, E = (0,$$\dfrac{10}{3}$$)

In $$\triangle\ $$ ECB, 

Base = CB = 12 units 

Height = BF + OE = $$9+\dfrac{10}{3}=\dfrac{37}{3}$$

Area of $$\triangle\ $$ ECB = $$\dfrac{1}{2}\times12\times\dfrac{37}{3}$$ = 74 square units

Hence, the area of the triangle ECB is 74 square units.

$$\therefore\ $$ The required answer is C.

ALTERNATE SOLUTION

E lie on the side AD such that DE : AE is 2 : 1. So the coordinates of E by section formula can be given as $$\left(\dfrac{2\times4+1\times\left(-8\right)}{2+1},\ \dfrac{2\times5+0}{2+1}\right)=\left(0,\dfrac{10}{3}\right)$$

Now, the area of the triangle ECB whose coordinates of the vertices are $$\left(0,\dfrac{10}{3}\right)$$ , $$(-8,-9)$$ and $$(4,-9)$$ is given by $$\dfrac{1}{2}\left|0\left(-9+9\right)+4\left(-9-\dfrac{10}{3}\right)-8\left(\dfrac{10}{3}+9\right)\right|=\dfrac{1}{2}\ \left|12\times\dfrac{37}{3}\right|=74\ \text{sq units}$$