For what value of k are the two straight lines 3x + 4y = 1 and 4x + 3y + 2k = 0 , equidistant from the point (1 , 1) ?

Options

A. `1//2`
B. 2
C. `-2`
D. `-1//2`

Answer

To find the value of \( k \) for which the two lines \( 3x + 4y = 1 \) and \( 4x + 3y + 2k = 0 \) are equidistant from the point \( (1, 1) \), we can follow these steps:

### Step 1: Find the perpendicular distance from the point to the first line
The formula for the perpendicular distance \( d \) from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is given by:

\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

For the line \( 3x + 4y = 1 \), we can rewrite it in the form \( Ax + By + C = 0 \):

\[ 3x + 4y - 1 = 0 \]

Here, \( A = 3 \), \( B = 4 \), and \( C = -1 \). The point \( (x_1, y_1) = (1, 1) \).

Substituting these values into the distance formula:

\[ d_1 = \frac{|3(1) + 4(1) - 1|}{\sqrt{3^2 + 4^2}} = \frac{|3 + 4 - 1|}{\sqrt{9 + 16}} = \frac{|6|}{5} = \frac{6}{5} \]

### Step 2: Find the perpendicular distance from the point to the second line
For the second line \( 4x + 3y + 2k = 0 \), we have \( A = 4 \), \( B = 3 \), and \( C = 2k \).

Using the distance formula again:

\[ d_2 = \frac{|4(1) + 3(1) + 2k|}{\sqrt{4^2 + 3^2}} = \frac{|4 + 3 + 2k|}{\sqrt{16 + 9}} = \frac{|7 + 2k|}{5} \]

### Step 3: Set the distances equal to each other
According to the problem, the distances \( d_1 \) and \( d_2 \) are equal:

\[ d_1 = d_2 \]

Substituting the values we found:

\[ \frac{6}{5} = \frac{|7 + 2k|}{5} \]

### Step 4: Solve for \( k \)
Multiplying both sides by 5 to eliminate the denominator:

\[ 6 = |7 + 2k| \]

This gives us two cases to consider:

**Case 1:**
\[ 7 + 2k = 6 \]
\[ 2k = 6 - 7 \]
\[ 2k = -1 \implies k = -\frac{1}{2} \]

**Case 2:**
\[ 7 + 2k = -6 \]
\[ 2k = -6 - 7 \]
\[ 2k = -13 \implies k = -\frac{13}{2} \]

### Step 5: Conclusion
The value of \( k \) that makes the two lines equidistant from the point \( (1, 1) \) is:

\[ k = -\frac{1}{2} \]