Question : The graph of $3x+4y-24=0$ forms a $\triangle OAB$ with the coordinate axes, where $O$ is the origin. Also, the graph of $x+y+4=0$ forms a $\triangle OCD$ with the coordinate axes. Then the area of $\triangle OCD$ is equal to:

Option 1: $\frac{1}{2}\text{ of the area of}\ \triangle OAB$

Option 2: $\frac{1}{3}\text{ of the area of}\ \triangle OAB$

Option 3: $\frac{2}{3}\text{ of the area of}\ \triangle OAB$

Option 4: $\text{ the area of}\ \triangle OAB$

Correct Answer: $\frac{1}{3}\text{ of the area of}\ \triangle OAB$

Solution :
Given: The graph of $3x+4y-24=0$ forms a $\triangle OAB$ with the coordinate axes, where $O$ is the origin. Also, the graph of $x+y+4=0$ forms a $\triangle OCD$ with the coordinate axes.
By substituting the value of $x=0$ in the equation $3x + 4y= 24$, we get,
$3\times0+ 4y= 24$
⇒ $4y=24$
⇒ $y = 6$
Therefore, B's coordinates are $(0, 6)$.
By substituting the value of $y=0$ in the equation $3x + 4y = 24$, we get,
$3x + 4\times 0 = 24$
⇒ $3x=24$
⇒ $x = 8$
Therefore, A's coordinates are $(8,0)$.
By substituting the value of $y=0$ in the equation $x + y = –4$, we get,
$x + 0 = –4$
⇒ $x=–4$
Therefore, C's coordinates are $(–4,0 )$.
By substituting the value of $x=0$ in the equation $x + y = –4$, we get,
$0 + y = –4$
⇒ $y=–4$
Therefore, D's coordinates are $(0,–4)$.
Also, the area of $\triangle OAB=\frac{1}{2}\times OA \times OB$.
⇒ $\frac{1}{2}\times 8 \times 6=24$ sq. units
Similarly, the area of $\triangle OCD=\frac{1}{2}\times OC \times OD$.
⇒ $\frac{1}{2}\times 4 \times 4=8$ sq. units
So, $ \text{ area of}\ \triangle OCD=\frac{1}{3} $ of the area of $\triangle OAB$
Hence, the correct answer is $\frac{1}{3}\text{ of the area of}\ \triangle OAB$.