Find the equation of the straight line which passes through the point of intersection of the straight lines `3x-4y+1=0 and 5x+y-1=0` and cuts off equal intercepts from the axes.

Answer

To find the equation of the straight line that passes through the point of intersection of the lines \(3x - 4y + 1 = 0\) and \(5x + y - 1 = 0\) and cuts off equal intercepts from the axes, we can follow these steps:

### Step 1: Find the Point of Intersection
We start by solving the two equations to find their point of intersection.

1. The first equation is:
\[ 3x - 4y + 1 = 0 \quad \text{(1)} \]
2. The second equation is:
\[ 5x + y - 1 = 0 \quad \text{(2)} \]

We can express \(y\) from equation (2):
\[ y = 1 - 5x \quad \text{(3)} \]

Now, substitute equation (3) into equation (1):
\[ 3x - 4(1 - 5x) + 1 = 0 \]
\[ 3x - 4 + 20x + 1 = 0 \]
\[ 23x - 3 = 0 \]
\[ x = \frac{3}{23} \]

Now, substitute \(x = \frac{3}{23}\) back into equation (3) to find \(y\):
\[ y = 1 - 5\left(\frac{3}{23}\right) \]
\[ y = 1 - \frac{15}{23} = \frac{23 - 15}{23} = \frac{8}{23} \]

Thus, the point of intersection \(P\) is:
\[ P\left(\frac{3}{23}, \frac{8}{23}\right) \]

### Step 2: Equation of the Line with Equal Intercepts
A line that cuts off equal intercepts from the axes can be represented as:
\[ x + y = a \]
where \(a\) is the length of the intercepts on both axes.

### Step 3: Substitute the Point of Intersection
Since the line passes through the point \(P\left(\frac{3}{23}, \frac{8}{23}\right)\), we substitute these coordinates into the equation \(x + y = a\):
\[ \frac{3}{23} + \frac{8}{23} = a \]
\[ \frac{11}{23} = a \]

### Step 4: Write the Final Equation
Thus, the equation of the line is:
\[ x + y = \frac{11}{23} \]

To express it in standard form, we can multiply through by 23:
\[ 23x + 23y = 11 \]

### Final Answer
The equation of the straight line is:
\[ 23x + 23y - 11 = 0 \]