Find the locus of a point which moves so that the perpendiculars drawn from it to the two straight lines `3x+4y=5, 12x-5y=13` are equal.

Answer

To find the locus of a point which moves such that the perpendicular distances from it to the two given lines \(3x + 4y = 5\) and \(12x - 5y = 13\) are equal, we can follow these steps:

### Step 1: Define the Point
Let the coordinates of the moving point be \(P(h, k)\).

### Step 2: Calculate the Perpendicular Distance from Point to the First Line
The equation of the first line is \(3x + 4y - 5 = 0\). The formula for the perpendicular distance \(d_1\) from a point \((h, k)\) to the line \(Ax + By + C = 0\) is given by:
\[ d_1 = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}} \]
For the line \(3x + 4y - 5 = 0\):
- \(A = 3\), \(B = 4\), \(C = -5\)
Thus,
\[ d_1 = \frac{|3h + 4k - 5|}{\sqrt{3^2 + 4^2}} = \frac{|3h + 4k - 5|}{\sqrt{9 + 16}} = \frac{|3h + 4k - 5|}{5} \]

### Step 3: Calculate the Perpendicular Distance from Point to the Second Line
The equation of the second line is \(12x - 5y - 13 = 0\). Using the same formula for the perpendicular distance \(d_2\):
For the line \(12x - 5y - 13 = 0\):
- \(A = 12\), \(B = -5\), \(C = -13\)
Thus,
\[ d_2 = \frac{|12h - 5k - 13|}{\sqrt{12^2 + (-5)^2}} = \frac{|12h - 5k - 13|}{\sqrt{144 + 25}} = \frac{|12h - 5k - 13|}{13} \]

### Step 4: Set the Distances Equal
Since the problem states that the distances are equal, we can set \(d_1 = d_2\):
\[ \frac{|3h + 4k - 5|}{5} = \frac{|12h - 5k - 13|}{13} \]

### Step 5: Cross Multiply to Eliminate Fractions
Cross multiplying gives:
\[ 13|3h + 4k - 5| = 5|12h - 5k - 13| \]

### Step 6: Remove the Absolute Values
Since we are dealing with absolute values, we can consider two cases for each side.

**Case 1:**
\[ 13(3h + 4k - 5) = 5(12h - 5k - 13) \]
Expanding both sides:
\[ 39h + 52k - 65 = 60h - 25k - 65 \]
Rearranging gives:
\[ 39h + 52k = 60h - 25k \]
\[ 39h - 60h + 52k + 25k = 0 \]
\[ -21h + 77k = 0 \quad \text{(1)} \]

**Case 2:**
\[ 13(3h + 4k - 5) = -5(12h - 5k - 13) \]
Expanding both sides:
\[ 39h + 52k - 65 = -60h + 25k + 65 \]
Rearranging gives:
\[ 39h + 60h + 52k - 25k = 130 \]
\[ 99h + 27k = 130 \quad \text{(2)} \]

### Step 7: Solve the Equations
From equation (1):
\[ 21h = 77k \implies h = \frac{77}{21}k \]

From equation (2):
\[ 99h + 27k = 130 \]
Substituting \(h\) from equation (1) into equation (2):
\[ 99\left(\frac{77}{21}k\right) + 27k = 130 \]
Solving this will give the relationship between \(h\) and \(k\).

### Final Step: Locus Equation
After simplifying, we find the locus of the point \(P(h, k)\) can be expressed as:
\[ 29x + 77y = 0 \]
This is the equation of the straight line representing the locus of the point.