Solution:
Area of Quadrilateral ACDO= Area (ΔAOD)+Area(ΔACD)-----(1)
Area of Quadrilateral ABDO= Area (ΔAOD)+Area(ΔABD)------(2)
⇒ (1) = 5 ×(2)→→→→GIven
→Area (ΔAOD)+Area(ΔACD)=5 Area (ΔAOD)+5 Area(ΔABD),
→5 Area (ΔABD)+4 Area (ΔAOD)-Area (ΔACD)=0-----(3)
As, Area of a Triangle [tex]={\text{having vertices}} (x_{1},y_{1}),(x_{2},y_{2}) {\text{and}} (x_{3},y_{3})=\frac{1}{2}\times[x_{1}(y_{2}-y_{3})-x_2(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})][/tex]
Area (ΔACD)
[tex]=\frac{1}{2}[0(y-0)-4(0-3)+4(3-y)]\\\\= \frac{1}{2}[12+12-4y]\\\\=12-2 y[/tex]
Area (ΔABD)
[tex]=\frac{1}{2}[0(4-0)-4(0-3)+4(3-4)]\\\\ =\frac{1}{2}[12-4]=4[/tex]
Area (ΔAOD)[tex]=\frac{1}{2}[0(3-0)-0(0-0)+4(0-3)]=6[/tex]
Putting these values in equation (3)
→20+24-12+2 y=0
→ 2 y= -32
Dividing both sides by , 2 we get
→y= -16
