Locus of the points which are at equal distance from `3x + 4y-11 = 0` and `12x + 5y + 2= 0` and which is near the origin is: (a) `21x - 77y +153 = 0` (b) `99x +77y - 133 = 0` (c) `7x - 11y = 19` (d) None of these

Options

A. `21x - 77y +153 = 0`
B. `99x +77y - 133 = 0`
C. `7x - 11y = 19`
D. None of these

Explanation

To find the locus of points that are equidistant from the lines \(3x + 4y - 11 = 0\) and \(12x + 5y + 2 = 0\), we can follow these steps:

### Step 1: Write the equations of the lines in standard form
We have the two lines:
1. \(3x + 4y - 11 = 0\)
2. \(12x + 5y + 2 = 0\)

### Step 2: Identify the coefficients
For the first line, the coefficients are:
- \(a_1 = 3\)
- \(b_1 = 4\)
- \(c_1 = -11\)

For the second line, the coefficients are:
- \(a_2 = 12\)
- \(b_2 = 5\)
- \(c_2 = 2\)

### Step 3: Set up the distance formula
The distance \(d_1\) from a point \((h, k)\) to the first line is given by:
\[ d_1 = \frac{|3h + 4k - 11|}{\sqrt{3^2 + 4^2}} = \frac{|3h + 4k - 11|}{5} \]

The distance \(d_2\) from the point \((h, k)\) to the second line is given by:
\[ d_2 = \frac{|12h + 5k + 2|}{\sqrt{12^2 + 5^2}} = \frac{|12h + 5k + 2|}{13} \]

### Step 4: Set the distances equal
Since we want the points that are equidistant from both lines, we set \(d_1 = d_2\):
\[ \frac{|3h + 4k - 11|}{5} = \frac{|12h + 5k + 2|}{13} \]

### Step 5: Cross-multiply to eliminate the fractions
Cross-multiplying gives:
\[ 13|3h + 4k - 11| = 5|12h + 5k + 2| \]

### Step 6: Consider the cases for absolute values
We can consider two cases based on the absolute values.

**Case 1:** \(3h + 4k - 11 \geq 0\) and \(12h + 5k + 2 \geq 0\)
\[ 13(3h + 4k - 11) = 5(12h + 5k + 2) \]
Expanding this:
\[ 39h + 52k - 143 = 60h + 25k + 10 \]
Rearranging gives:
\[ 99h + 77k - 153 = 0 \]

**Case 2:** \(3h + 4k - 11 < 0\) and \(12h + 5k + 2 < 0\)
\[ 13(-3h - 4k + 11) = 5(-12h - 5k - 2) \]
Expanding this:
\[ -39h - 52k + 143 = -60h - 25k - 10 \]
Rearranging gives:
\[ 99h + 77k - 133 = 0 \]

### Step 7: Identify the correct locus
The locus of points that are equidistant from the two lines is given by:
\[ 99x + 77y - 133 = 0 \]
This matches option (b).

### Final Answer
The locus of the points which are at equal distance from the given lines and which is near the origin is:
**(b) \(99x + 77y - 133 = 0\)**