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 on the line \(3x - 4y + 1 = 0\), B on the line \(4x + 3y - 7 = 0\), and C at the coordinates (-2, 5), we can follow these steps:

### Step 1: Understand the properties of a rhombus
In a rhombus, the diagonals bisect each other at right angles. Therefore, if we can find the relationship between the coordinates of points A, B, C, and D, we can derive the locus of D.

### Step 2: Identify the slopes of the lines
The line \(3x - 4y + 1 = 0\) can be rewritten in slope-intercept form:
\[ 4y = 3x + 1 \implies y = \frac{3}{4}x + \frac{1}{4} \]
The slope \(m_1\) of this line is \(\frac{3}{4}\).

The line \(4x + 3y - 7 = 0\) can also be rewritten:
\[ 3y = -4x + 7 \implies y = -\frac{4}{3}x + \frac{7}{3} \]
The slope \(m_2\) of this line is \(-\frac{4}{3}\).

### Step 3: Verify that the lines are perpendicular
To check if the lines are perpendicular, we multiply their slopes:
\[ m_1 \cdot m_2 = \frac{3}{4} \cdot -\frac{4}{3} = -1 \]
Since the product of the slopes is -1, the lines are indeed perpendicular.

### Step 4: Set up the coordinates for points A, B, and D
Let point A be \((h, k)\) on the line \(3x - 4y + 1 = 0\) and point B be \((x_B, y_B)\) on the line \(4x + 3y - 7 = 0\). The coordinates of point C are given as \((-2, 5)\).

### Step 5: Use the property of the rhombus
Since ABCD is a rhombus, the lengths DA and DC must be equal:
\[ DA = DC \]

### Step 6: Calculate the distances
The distance \(DA\) can be calculated using the formula for the perpendicular distance from a point to a line:
\[ DA = \frac{|3h - 4k + 1|}{\sqrt{3^2 + (-4)^2}} = \frac{|3h - 4k + 1|}{5} \]

The distance \(DC\) can be calculated using the distance formula:
\[ DC = \sqrt{(h + 2)^2 + (k - 5)^2} \]

### Step 7: Set the distances equal
Setting \(DA\) equal to \(DC\):
\[ \frac{|3h - 4k + 1|}{5} = \sqrt{(h + 2)^2 + (k - 5)^2} \]

### Step 8: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \left(\frac{3h - 4k + 1}{5}\right)^2 = (h + 2)^2 + (k - 5)^2 \]

### Step 9: Simplify the equation
Multiply both sides by 25:
\[ (3h - 4k + 1)^2 = 25((h + 2)^2 + (k - 5)^2) \]

### Step 10: Replace \(h\) and \(k\) with \(x\) and \(y\)
To find the locus of point D, replace \(h\) with \(x\) and \(k\) with \(y\):
\[ (3x - 4y + 1)^2 = 25((x + 2)^2 + (y - 5)^2) \]

### Step 11: Rearranging the equation
This equation represents the locus of point D in the coordinate plane.

### Final Answer
The locus of point D is given by the equation:
\[ (3x - 4y + 1)^2 = 25((x + 2)^2 + (y - 5)^2) \]