Question

Points A, B, C, D, and O have coordinates (0, 3), (4, 4), (4, y), (4, 0), and (0, 0), respectively. The area of ACDO is 5 times greater than the area of ABDO. Find y.

Answer 1

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 $={\text{having vertices}} (x_{1},y_{1}),(x_{2},y_{2}) {\text{and}} (x_{3},y_{3})=\frac{1}{2}\times[x_{1}(y_{2}-y_{3})-x_2(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

Area (ΔACD)

$=\frac{1}{2}[0(y-0)-4(0-3)+4(3-y)]\\\\= \frac{1}{2}[12+12-4y]\\\\=12-2 y$

Area (ΔABD)

$=\frac{1}{2}[0(4-0)-4(0-3)+4(3-4)]\\\\ =\frac{1}{2}[12-4]=4$

Area  (ΔAOD)$=\frac{1}{2}[0(3-0)-0(0-0)+4(0-3)]=6$

Putting these values in equation (3)

→20+24-12+2 y=0

→  2 y= -32

Dividing both sides by , 2 we get

→y= -16