Answer :
We are given that the angle of inclination of the line to the \( x \)-axis is \( \theta = 45^\circ \), and we need to find the equation of this line passing through the intersection of the lines \( x + 3y - 4 = 0 \) and \( 3x - 4y + 1 = 0 \).
### Step 1: Find the Intersection Point of the Lines
To find the point of intersection of the two lines, we need to solve the system of equations:
[tex]\[ \begin{cases} x + 3y - 4 = 0 \\ 3x - 4y + 1 = 0 \end{cases} \][/tex]
We solve the first equation for \( x \):
[tex]\[ x = 4 - 3y \][/tex]
Substitute \( x = 4 - 3y \) into the second equation:
[tex]\[ 3(4 - 3y) - 4y + 1 = 0 \\ 12 - 9y - 4y + 1 = 0 \\ 12 + 1 - 13y = 0 \\ 13 - 13y = 0 \\ y = 1 \][/tex]
Now substitute \( y = 1 \) back into \( x = 4 - 3y \):
[tex]\[ x = 4 - 3(1) = 1 \][/tex]
The point of intersection is \( (1, 1) \).
### Step 2: Determine the Slope of the Line
The angle of inclination \( \theta = 45^\circ \). The slope \( m \) of a line inclined at an angle \( \theta \) to the \( x \)-axis is given by:
[tex]\[ m = \tan(45^\circ) \][/tex]
We know that \( \tan(45^\circ) = 1 \). Thus, the slope \( m = 1 \).
### Step 3: Equation of the Line Using Point-Slope Form
We use the point-slope form of the equation of a line:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Where \( (x_1, y_1) \) is the point \( (1, 1) \) and \( m = 1 \). Substituting these values in:
[tex]\[ y - 1 = 1(x - 1) \\ y - 1 = x - 1 \][/tex]
### Step 4: Rearrange to General Form
Rearranging the above equation to the standard form \( ax + by + c = 0 \):
[tex]\[ y - 1 - x + 1 = 0 \\ x - y = 0 \][/tex]
Therefore, the equation of the required line is:
[tex]\[ x = y \][/tex]
### Step 1: Find the Intersection Point of the Lines
To find the point of intersection of the two lines, we need to solve the system of equations:
[tex]\[ \begin{cases} x + 3y - 4 = 0 \\ 3x - 4y + 1 = 0 \end{cases} \][/tex]
We solve the first equation for \( x \):
[tex]\[ x = 4 - 3y \][/tex]
Substitute \( x = 4 - 3y \) into the second equation:
[tex]\[ 3(4 - 3y) - 4y + 1 = 0 \\ 12 - 9y - 4y + 1 = 0 \\ 12 + 1 - 13y = 0 \\ 13 - 13y = 0 \\ y = 1 \][/tex]
Now substitute \( y = 1 \) back into \( x = 4 - 3y \):
[tex]\[ x = 4 - 3(1) = 1 \][/tex]
The point of intersection is \( (1, 1) \).
### Step 2: Determine the Slope of the Line
The angle of inclination \( \theta = 45^\circ \). The slope \( m \) of a line inclined at an angle \( \theta \) to the \( x \)-axis is given by:
[tex]\[ m = \tan(45^\circ) \][/tex]
We know that \( \tan(45^\circ) = 1 \). Thus, the slope \( m = 1 \).
### Step 3: Equation of the Line Using Point-Slope Form
We use the point-slope form of the equation of a line:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Where \( (x_1, y_1) \) is the point \( (1, 1) \) and \( m = 1 \). Substituting these values in:
[tex]\[ y - 1 = 1(x - 1) \\ y - 1 = x - 1 \][/tex]
### Step 4: Rearrange to General Form
Rearranging the above equation to the standard form \( ax + by + c = 0 \):
[tex]\[ y - 1 - x + 1 = 0 \\ x - y = 0 \][/tex]
Therefore, the equation of the required line is:
[tex]\[ x = y \][/tex]
