Find the equations of the straight lines which pass through the origin and trisect the portion of the st.line `(x)/(a)+(y)/(b)=1`, which is intercepted between the axes.

Answer

To find the equations of the straight lines that pass through the origin and trisect the portion of the line \(\frac{x}{a} + \frac{y}{b} = 1\) intercepted between the axes, we can follow these steps:

### Step 1: Identify the Intercepts
The line \(\frac{x}{a} + \frac{y}{b} = 1\) intersects the x-axis and y-axis at specific points:
- For the x-intercept, set \(y = 0\):
\[ \frac{x}{a} + \frac{0}{b} = 1 \implies x = a \quad \text{(Point A: (a, 0))} \]
- For the y-intercept, set \(x = 0\):
\[ \frac{0}{a} + \frac{y}{b} = 1 \implies y = b \quad \text{(Point B: (0, b))} \]

### Step 2: Find the Length of the Segment AB
The segment AB is the line segment between points A and B. We can find the coordinates of points A and B:
- Point A: \((a, 0)\)
- Point B: \((0, b)\)

### Step 3: Trisect the Segment AB
To trisect the segment AB, we need to find two points C and D that divide the segment into three equal parts. The coordinates of points C and D can be found using the section formula.

#### Finding Point C
Point C divides segment AB in the ratio \(2:1\):
\[ C = \left(\frac{2 \cdot 0 + 1 \cdot a}{2 + 1}, \frac{2 \cdot b + 1 \cdot 0}{2 + 1}\right) = \left(\frac{a}{3}, \frac{2b}{3}\right) \]

#### Finding Point D
Point D divides segment AB in the ratio \(1:2\):
\[ D = \left(\frac{1 \cdot 0 + 2 \cdot a}{1 + 2}, \frac{1 \cdot b + 2 \cdot 0}{1 + 2}\right) = \left(\frac{2a}{3}, \frac{b}{3}\right) \]

### Step 4: Find the Equations of Lines OC and OD
Now we need to find the equations of the lines passing through the origin (0, 0) and points C and D.

#### Equation of Line OC
The slope \(m_{OC}\) of line OC is given by:
\[ m_{OC} = \frac{\frac{2b}{3} - 0}{\frac{a}{3} - 0} = \frac{2b}{a} \]
Using the point-slope form of the equation of a line:
\[ y - 0 = \frac{2b}{a}(x - 0) \implies y = \frac{2b}{a}x \]
Rearranging gives:
\[ 2bx - ay = 0 \quad \text{(Equation 1)} \]

#### Equation of Line OD
The slope \(m_{OD}\) of line OD is given by:
\[ m_{OD} = \frac{\frac{b}{3} - 0}{\frac{2a}{3} - 0} = \frac{b}{2a} \]
Using the point-slope form of the equation of a line:
\[ y - 0 = \frac{b}{2a}(x - 0) \implies y = \frac{b}{2a}x \]
Rearranging gives:
\[ bx - 2ay = 0 \quad \text{(Equation 2)} \]

### Final Answer
The equations of the straight lines that pass through the origin and trisect the portion of the line \(\frac{x}{a} + \frac{y}{b} = 1\) are:
1. \(2bx - ay = 0\)
2. \(bx - 2ay = 0\)