Find the equation of the locus of a point P which is equidistance from the st. line `3x-4y+2=0` and the origin.

Answer

To find the equation of the locus of a point P that is equidistant from the line \(3x - 4y + 2 = 0\) and the origin, we will follow these steps:

### Step 1: Define the Point P
Let the point P be represented as \(P(h, k)\).

### Step 2: Calculate the Distance from the Origin
The distance of point P from the origin (0, 0) is given by the distance formula:
\[ d_1 = \sqrt{(h - 0)^2 + (k - 0)^2} = \sqrt{h^2 + k^2} \]

### Step 3: Calculate the Distance from the Line
The distance from a point \((x_1, y_1)\) to a line given by the equation \(Ax + By + C = 0\) is given by:
\[ d_2 = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For the line \(3x - 4y + 2 = 0\), we have \(A = 3\), \(B = -4\), and \(C = 2\). Thus, the distance from point P(h, k) to the line is:
\[ d_2 = \frac{|3h - 4k + 2|}{\sqrt{3^2 + (-4)^2}} = \frac{|3h - 4k + 2|}{\sqrt{9 + 16}} = \frac{|3h - 4k + 2|}{5} \]

### Step 4: Set the Distances Equal
Since point P is equidistant from the origin and the line, we set \(d_1 = d_2\):
\[ \sqrt{h^2 + k^2} = \frac{|3h - 4k + 2|}{5} \]

### Step 5: Square Both Sides
Squaring both sides to eliminate the square root gives:
\[ h^2 + k^2 = \left(\frac{3h - 4k + 2}{5}\right)^2 \]
This simplifies to:
\[ h^2 + k^2 = \frac{(3h - 4k + 2)^2}{25} \]

### Step 6: Multiply Through by 25
To eliminate the fraction, multiply both sides by 25:
\[ 25(h^2 + k^2) = (3h - 4k + 2)^2 \]

### Step 7: Expand the Right Side
Expanding the right side:
\[ (3h - 4k + 2)^2 = 9h^2 - 24hk + 16k^2 + 12h - 16k + 4 \]
So we have:
\[ 25h^2 + 25k^2 = 9h^2 - 24hk + 16k^2 + 12h - 16k + 4 \]

### Step 8: Rearrange the Equation
Rearranging gives:
\[ 25h^2 - 9h^2 + 25k^2 - 16k^2 + 24hk - 12h + 16k - 4 = 0 \]
This simplifies to:
\[ 16h^2 + 9k^2 + 24hk - 12h + 16k - 4 = 0 \]

### Step 9: Substitute h and k with x and y
To express the locus in terms of \(x\) and \(y\), substitute \(h\) with \(x\) and \(k\) with \(y\):
\[ 16x^2 + 9y^2 + 24xy - 12x + 16y - 4 = 0 \]

### Final Answer
Thus, the equation of the locus of the point P is:
\[ 16x^2 + 9y^2 + 24xy - 12x + 16y - 4 = 0 \]