To solve the problem step by step, we will follow these instructions:
### Step 1: Identify the Diagonal
The equation of one diagonal of the rhombus is given as \(3x - 4y + 5 = 0\). We need to check if point A(3, 1) lies on this diagonal.
**Calculation:**
Substituting \(x = 3\) and \(y = 1\) into the equation:
\[
3(3) - 4(1) + 5 = 9 - 4 + 5 = 10 \neq 0
\]
Since the result is not equal to zero, point A does not lie on this diagonal. Therefore, this diagonal must be BD.
### Step 2: Find the Slope of BD
To find the slope of the line \(3x - 4y + 5 = 0\), we can rearrange it into slope-intercept form \(y = mx + c\).
**Calculation:**
\[
4y = 3x + 5 \implies y = \frac{3}{4}x + \frac{5}{4}
\]
Thus, the slope of line BD is \(m_{BD} = \frac{3}{4}\).
### Step 3: Find the Slope of AC
The diagonals of a rhombus are perpendicular to each other. Therefore, the slope of diagonal AC will be the negative reciprocal of the slope of BD.
**Calculation:**
\[
m_{AC} = -\frac{1}{m_{BD}} = -\frac{1}{\frac{3}{4}} = -\frac{4}{3}
\]
### Step 4: Find the Equation of AC
Now we will use point A(3, 1) and the slope of AC to find the equation of line AC using the point-slope form \(y - y_1 = m(x - x_1)\).
**Calculation:**
\[
y - 1 = -\frac{4}{3}(x - 3)
\]
Expanding this:
\[
y - 1 = -\frac{4}{3}x + 4 \implies y = -\frac{4}{3}x + 5
\]
Rearranging to standard form:
\[
4x + 3y - 15 = 0
\]
### Step 5: Find the Intersection Point O of AC and BD
We need to find the intersection of lines AC and BD. We have:
1. \(3x - 4y + 5 = 0\) (BD)
2. \(4x + 3y - 15 = 0\) (AC)
**Calculation:**
From the first equation, express \(y\) in terms of \(x\):
\[
4y = 3x + 5 \implies y = \frac{3}{4}x + \frac{5}{4}
\]
Substituting this into the second equation:
\[
4x + 3\left(\frac{3}{4}x + \frac{5}{4}\right) - 15 = 0
\]
Multiplying through by 4 to eliminate fractions:
\[
16x + 9x + 15 - 60 = 0 \implies 25x - 45 = 0 \implies x = \frac{9}{5}
\]
Now substituting \(x = \frac{9}{5}\) back into the equation for \(y\):
\[
y = \frac{3}{4}\left(\frac{9}{5}\right) + \frac{5}{4} = \frac{27}{20} + \frac{25}{20} = \frac{52}{20} = \frac{13}{5}
\]
Thus, the intersection point O is \(\left(\frac{9}{5}, \frac{13}{5}\right)\).
### Step 6: Find the Midpoint R of AO
Now we need to find the midpoint R of segment AO, where A is (3, 1) and O is \(\left(\frac{9}{5}, \frac{13}{5}\right)\).
**Calculation:**
\[
R_x = \frac{3 + \frac{9}{5}}{2} = \frac{\frac{15}{5} + \frac{9}{5}}{2} = \frac{\frac{24}{5}}{2} = \frac{12}{5}
\]
\[
R_y = \frac{1 + \frac{13}{5}}{2} = \frac{\frac{5}{5} + \frac{13}{5}}{2} = \frac{\frac{18}{5}}{2} = \frac{9}{5}
\]
So, \(R\) is \(\left(\frac{12}{5}, \frac{9}{5}\right)\).
### Step 7: Find the Equation of Line PQ
Since PQ is parallel to BD, it will have the same slope \(m = \frac{3}{4}\). Using point R to find the equation:
\[
y - \frac{9}{5} = \frac{3}{4}\left(x - \frac{12}{5}\right)
\]
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 to standard form:
\[
3x - 4y = 0
\]
### Step 8: Find the values of a, b, c
The equation can be written as \(3x - 4y + 0 = 0\), where \(a = 3\), \(b = -4\), and \(c = 0\).
### Step 9: Calculate \(|a + b + c|\)
\[
|a + b + c| = |3 - 4 + 0| = |-1| = 1
\]
### Final Answer
The absolute value of \(a + b + c\) is \(1\).
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