Consider a family of lines `(4a+3)x-(a+1)y-(2a+1)=0` where `a in R` a member of this family with positive gradient making an angle of 45 with the line `3x-4y=2`, is

Options

A. `7x-y-5=0`
B. `4x-3y+2=0`
C. `x+7y=15`
D. `5x-3y-4=0`

Explanation

To solve the problem, we need to find the value of \( a \) such that the line represented by the equation \( (4a + 3)x - (a + 1)y - (2a + 1) = 0 \) has a positive gradient and makes an angle of \( 45^\circ \) with the line \( 3x - 4y = 2 \).

### Step 1: Find the slope of the given family of lines

The equation of the family of lines is given as:

\[ (4a + 3)x - (a + 1)y - (2a + 1) = 0 \]

We can rearrange this into the slope-intercept form \( y = mx + c \):

\[ (a + 1)y = (4a + 3)x - (2a + 1) \]

Dividing through by \( (a + 1) \):

\[ y = \frac{4a + 3}{a + 1} x - \frac{2a + 1}{a + 1} \]

Thus, the slope \( m_1 \) of this line is:

\[ m_1 = \frac{4a + 3}{a + 1} \]

### Step 2: Find the slope of the line \( 3x - 4y = 2 \)

Rearranging the equation \( 3x - 4y = 2 \) into slope-intercept form:

\[ 4y = 3x - 2 \implies y = \frac{3}{4}x - \frac{1}{2} \]

Thus, the slope \( m_2 \) of this line is:

\[ m_2 = \frac{3}{4} \]

### Step 3: Use the angle formula for slopes

The angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:

\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

Since we want the angle to be \( 45^\circ \), we have \( \tan 45^\circ = 1 \). Thus:

\[ 1 = \left| \frac{\frac{4a + 3}{a + 1} - \frac{3}{4}}{1 + \frac{4a + 3}{a + 1} \cdot \frac{3}{4}} \right| \]

### Step 4: Solve for \( a \)

Setting up the equation:

\[ 1 = \frac{\frac{4a + 3}{a + 1} - \frac{3}{4}}{1 + \frac{4a + 3}{a + 1} \cdot \frac{3}{4}} \]

Cross-multiplying gives:

\[ 1 + \frac{4a + 3}{a + 1} \cdot \frac{3}{4} = \frac{4a + 3}{a + 1} - \frac{3}{4} \]

To simplify, we need to find a common denominator and solve for \( a \). After simplification, we will get two equations:

1. \( 13a + 9 = 16a + 13 \)
2. \( 13a + 9 = -16a - 13 \)

Solving these equations:

1. From \( 13a + 9 = 16a + 13 \):
\[ -3a = 4 \implies a = -\frac{4}{3} \]

2. From \( 13a + 9 = -16a - 13 \):
\[ 29a = -22 \implies a = -\frac{22}{29} \]

### Step 5: Substitute \( a \) back into the line equation

We will take \( a = -\frac{4}{3} \) (the one with a positive gradient) and substitute it back into the original line equation:

\[ (4(-\frac{4}{3}) + 3)x - (-\frac{4}{3} + 1)y - (2(-\frac{4}{3}) + 1) = 0 \]

Calculating this will yield the specific line equation.

### Final Result

After substituting and simplifying, we find the equation of the line. The final answer is:

\[ 7x - y - 5 = 0 \]