A line intersects the co- ordinate axes in the points A and B such that area of `Delta` OAB is 48 sq. units. If the line passes through the point (3,6) find its equation.

Answer

To find the equation of the line that intersects the coordinate axes at points A and B such that the area of triangle OAB is 48 square units and the line passes through the point (3, 6), we can follow these steps:

### Step 1: Understand the Area of Triangle OAB
The area of triangle OAB formed by the x-intercept (A) and y-intercept (B) can be expressed as:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \]
Here, the base is the x-intercept (let's denote it as A) and the height is the y-intercept (denote it as B). Given that the area is 48 square units, we have:
\[ \frac{1}{2} \times A \times B = 48 \]
This simplifies to:
\[ A \times B = 96 \quad \text{(1)} \]

### Step 2: Express B in terms of A
From equation (1), we can express B in terms of A:
\[ B = \frac{96}{A} \quad \text{(2)} \]

### Step 3: Use the Point (3, 6) to Find the Equation
The line intersects the axes at points A (A, 0) and B (0, B). The slope of the line can be calculated using the intercepts:
\[ \text{slope} = \frac{B - 0}{0 - A} = -\frac{B}{A} \]
Using the point-slope form of the line equation, we can write:
\[ y - B = -\frac{B}{A}(x - 0) \]
Substituting B from equation (2):
\[ y - \frac{96}{A} = -\frac{96}{A^2} x \]

### Step 4: Substitute the Point (3, 6)
Since the line passes through the point (3, 6), we can substitute x = 3 and y = 6 into the equation:
\[ 6 - \frac{96}{A} = -\frac{96}{A^2} \cdot 3 \]
Multiplying through by \(A^2\) to eliminate the denominator:
\[ 6A^2 - 96A = -288 \]
Rearranging gives:
\[ 6A^2 - 96A + 288 = 0 \]
Dividing the entire equation by 6:
\[ A^2 - 16A + 48 = 0 \quad \text{(3)} \]

### Step 5: Solve the Quadratic Equation
Now we can solve the quadratic equation (3) using the quadratic formula:
\[ A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = -16\), and \(c = 48\):
\[ A = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 48}}{2 \cdot 1} \]
\[ A = \frac{16 \pm \sqrt{256 - 192}}{2} \]
\[ A = \frac{16 \pm \sqrt{64}}{2} \]
\[ A = \frac{16 \pm 8}{2} \]
This gives us two possible values for A:
\[ A = \frac{24}{2} = 12 \quad \text{or} \quad A = \frac{8}{2} = 4 \]

### Step 6: Find Corresponding B Values
Using equation (2):
1. If \(A = 12\):
\[ B = \frac{96}{12} = 8 \]
2. If \(A = 4\):
\[ B = \frac{96}{4} = 24 \]

### Step 7: Write the Equations of the Lines
Now we can write the equations of the lines using the intercepts:
1. For \(A = 12\) and \(B = 8\):
\[ \frac{x}{12} + \frac{y}{8} = 1 \quad \text{or} \quad 2x + 3y - 24 = 0 \]
2. For \(A = 4\) and \(B = 24\):
\[ \frac{x}{4} + \frac{y}{24} = 1 \quad \text{or} \quad 6x + y - 24 = 0 \]

### Final Answer
The equations of the lines that satisfy the conditions are:
1. \(2x + 3y - 24 = 0\)
2. \(6x + y - 24 = 0\)