Let the points A lies on `3x-4y+1=0`, the point B lines on `4 x + 3 y - 7 = 0` and the point C is (-2,5) . If ABCD is a rhombus , then the locus of D is

Options

A. `25((x+2)^2+(y-5)^2))=(3x+4y+1)^2`
B. `(3x-4y+1)^2+(4x-3y-7)^2=1`
C. `(3x-4y+1)^2-(4x-3y-7)^2=1`
D. `(4x+3y-7)^2+(3x-4y+1)^2=1`

Answer

To find the locus of point D such that ABCD forms a rhombus with points A, B, and C given, we will follow these steps:

### Step 1: Identify the equations of the lines and points
- Point A lies on the line \(3x - 4y + 1 = 0\).
- Point B lies on the line \(4x + 3y - 7 = 0\).
- Point C is given as \((-2, 5)\).

### Step 2: Determine the slopes of the lines
- For line \(3x - 4y + 1 = 0\):
\[ 4y = 3x + 1 \implies y = \frac{3}{4}x + \frac{1}{4} \]
The slope \(m_1 = \frac{3}{4}\).

- For line \(4x + 3y - 7 = 0\):
\[ 3y = -4x + 7 \implies y = -\frac{4}{3}x + \frac{7}{3} \]
The slope \(m_2 = -\frac{4}{3}\).

### Step 3: Check if the lines are perpendicular
The product of the slopes:
\[ m_1 \cdot m_2 = \frac{3}{4} \cdot -\frac{4}{3} = -1 \]
Since the product is \(-1\), the lines are perpendicular.

### Step 4: Use the properties of a rhombus
In a rhombus, the diagonals bisect each other at right angles. Therefore, we can use the distance formula and perpendicular distance formula.

### Step 5: Set up the equations
Let the coordinates of point D be \((h, k)\). The distance \(DA\) from point D to line \(3x - 4y + 1 = 0\) is given by:
\[ DA = \frac{|3h - 4k + 1|}{\sqrt{3^2 + (-4)^2}} = \frac{|3h - 4k + 1|}{5} \]

The distance \(DC\) from point D to point C \((-2, 5)\) is given by:
\[ DC = \sqrt{(h + 2)^2 + (k - 5)^2} \]

### Step 6: Set the distances equal
Since \(DA = DC\):
\[ \frac{|3h - 4k + 1|}{5} = \sqrt{(h + 2)^2 + (k - 5)^2} \]

### Step 7: Square both sides to eliminate the square root
\[ \left(\frac{|3h - 4k + 1|}{5}\right)^2 = (h + 2)^2 + (k - 5)^2 \]
\[ \frac{(3h - 4k + 1)^2}{25} = (h + 2)^2 + (k - 5)^2 \]

### Step 8: Multiply through by 25
\[ (3h - 4k + 1)^2 = 25((h + 2)^2 + (k - 5)^2) \]

### Step 9: Expand both sides
1. Left side:
\[ (3h - 4k + 1)^2 = 9h^2 - 24hk + 16k^2 + 6h - 8k + 1 \]
2. Right side:
\[ 25((h + 2)^2 + (k - 5)^2) = 25(h^2 + 4h + 4 + k^2 - 10k + 25) = 25h^2 + 100h + 100 + 25k^2 - 250k + 625 \]

### Step 10: Rearrange to form the locus equation
Combine and simplify to find the locus of D.

### Final Locus Equation
The locus of point D will be a conic section, which can be expressed in standard form after simplification.