To solve the problem, we need to find the equation of a line that passes through the midpoints of the sides AB and AD of rhombus ABCD, given that one diagonal is represented by the equation \(3x - 4y + 5 = 0\) and one vertex \(A(3, 1)\).
### Step 1: Identify the Diagonal
The equation of the diagonal is given as:
\[
3x - 4y + 5 = 0
\]
We can rewrite this in slope-intercept form (y = mx + b):
\[
4y = 3x + 5 \implies y = \frac{3}{4}x + \frac{5}{4}
\]
The slope of this diagonal (let's call it BD) is \(m_{BD} = \frac{3}{4}\).
### Step 2: Find the Slope of the Other Diagonal
Since the diagonals of a rhombus are perpendicular to each other, the slope of the other diagonal (let's call it AC) will be:
\[
m_{AC} = -\frac{1}{m_{BD}} = -\frac{1}{\frac{3}{4}} = -\frac{4}{3}
\]
### Step 3: Find the Equation of Diagonal AC
Using point-slope form of the line equation, and knowing point A(3, 1):
\[
y - y_1 = m(x - x_1) \implies y - 1 = -\frac{4}{3}(x - 3)
\]
Expanding this:
\[
y - 1 = -\frac{4}{3}x + 4 \implies y = -\frac{4}{3}x + 5
\]
Rearranging gives us:
\[
4x + 3y - 15 = 0
\]
### Step 4: Find the Intersection of the Diagonals
To find the intersection point O of the diagonals, we solve the system of equations:
1. \(3x - 4y + 5 = 0\)
2. \(4x + 3y - 15 = 0\)
We can solve these equations simultaneously. We can express \(y\) from the first equation:
\[
4y = 3x + 5 \implies y = \frac{3}{4}x + \frac{5}{4}
\]
Substituting this into the second equation:
\[
4x + 3\left(\frac{3}{4}x + \frac{5}{4}\right) - 15 = 0
\]
This simplifies to:
\[
4x + \frac{9}{4}x + \frac{15}{4} - 15 = 0
\]
Multiplying through by 4 to eliminate the fraction:
\[
16x + 9x + 15 - 60 = 0 \implies 25x - 45 = 0 \implies x = \frac{45}{25} = \frac{9}{5}
\]
Now substituting \(x = \frac{9}{5}\) back into the first equation to find \(y\):
\[
3\left(\frac{9}{5}\right) - 4y + 5 = 0 \implies \frac{27}{5} - 4y + 5 = 0
\]
Converting 5 to a fraction:
\[
\frac{27}{5} - 4y + \frac{25}{5} = 0 \implies \frac{52}{5} - 4y = 0 \implies 4y = \frac{52}{5} \implies y = \frac{13}{5}
\]
Thus, the intersection point O is:
\[
O\left(\frac{9}{5}, \frac{13}{5}\right)
\]
### Step 5: Find Midpoint R of AO
The midpoint R of segment AO where A(3, 1) and O\(\left(\frac{9}{5}, \frac{13}{5}\right)\) is given by:
\[
R\left(\frac{3 + \frac{9}{5}}{2}, \frac{1 + \frac{13}{5}}{2}\right) = R\left(\frac{\frac{15}{5} + \frac{9}{5}}{2}, \frac{\frac{5}{5} + \frac{13}{5}}{2}\right) = R\left(\frac{24/5}{2}, \frac{18/5}{2}\right) = R\left(\frac{12}{5}, \frac{9}{5}\right)
\]
### Step 6: Find the Equation of Line PQ
Since line PQ is parallel to BD, it has the same slope \(m_{BD} = \frac{3}{4}\). Using point-slope form again:
\[
y - \frac{9}{5} = \frac{3}{4}\left(x - \frac{12}{5}\right)
\]
Expanding this:
\[
y - \frac{9}{5} = \frac{3}{4}x - \frac{36}{20} \implies y = \frac{3}{4}x - \frac{36}{20} + \frac{9}{5}
\]
Converting \(\frac{9}{5}\) to a fraction with a denominator of 20:
\[
y = \frac{3}{4}x - \frac{36}{20} + \frac{36}{20} = \frac{3}{4}x
\]
Rearranging gives:
\[
3x - 4y = 0
\]
### Step 7: Express in Standard Form
The equation can be expressed as:
\[
3x - 4y + 0 = 0
\]
Thus, \(a = 3\), \(b = -4\), and \(c = 0\).
### Step 8: Calculate \(|a + b + c|\)
Now we calculate:
\[
|a + b + c| = |3 - 4 + 0| = |-1| = 1
\]
### Final Answer
The absolute value of \(a + b + c\) is:
\[
\boxed{1}
\]
