To find the lengths of the altitudes of the triangle formed by the lines \(3x - 4y = 5\), \(4x + 3y = 5\), and \(x + y = 1\), we will follow these steps:
### Step 1: Find the intersection points of the lines
We need to find the vertices of the triangle formed by the intersection of the three lines.
1. **Intersection of \(3x - 4y = 5\) and \(4x + 3y = 5\)**:
- Solve the equations simultaneously.
\[
3x - 4y = 5 \quad \text{(1)}
\]
\[
4x + 3y = 5 \quad \text{(2)}
\]
From (1), express \(y\) in terms of \(x\):
\[
4y = 3x - 5 \implies y = \frac{3x - 5}{4}
\]
Substitute \(y\) in (2):
\[
4x + 3\left(\frac{3x - 5}{4}\right) = 5
\]
\[
4x + \frac{9x - 15}{4} = 5
\]
Multiply through by 4 to eliminate the fraction:
\[
16x + 9x - 15 = 20
\]
\[
25x = 35 \implies x = \frac{7}{5}
\]
Substitute \(x\) back to find \(y\):
\[
y = \frac{3\left(\frac{7}{5}\right) - 5}{4} = \frac{\frac{21}{5} - 5}{4} = \frac{\frac{21}{5} - \frac{25}{5}}{4} = \frac{-4/5}{4} = -\frac{1}{5}
\]
So, the intersection point \(A\) is \(\left(\frac{7}{5}, -\frac{1}{5}\right)\).
2. **Intersection of \(4x + 3y = 5\) and \(x + y = 1\)**:
- Solve similarly.
From \(x + y = 1\), express \(y\):
\[
y = 1 - x
\]
Substitute in \(4x + 3(1 - x) = 5\):
\[
4x + 3 - 3x = 5 \implies x + 3 = 5 \implies x = 2
\]
\[
y = 1 - 2 = -1
\]
So, the intersection point \(B\) is \((2, -1)\).
3. **Intersection of \(3x - 4y = 5\) and \(x + y = 1\)**:
- Solve similarly.
From \(x + y = 1\), express \(y\):
\[
y = 1 - x
\]
Substitute in \(3x - 4(1 - x) = 5\):
\[
3x - 4 + 4x = 5 \implies 7x - 4 = 5 \implies 7x = 9 \implies x = \frac{9}{7}
\]
\[
y = 1 - \frac{9}{7} = -\frac{2}{7}
\]
So, the intersection point \(C\) is \(\left(\frac{9}{7}, -\frac{2}{7}\right)\).
### Step 2: Calculate the lengths of the sides of the triangle
Now we have the vertices \(A\), \(B\), and \(C\):
- \(A = \left(\frac{7}{5}, -\frac{1}{5}\right)\)
- \(B = (2, -1)\)
- \(C = \left(\frac{9}{7}, -\frac{2}{7}\right)\)
We can find the lengths of the sides \(AB\), \(BC\), and \(CA\) using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
1. **Length of \(AB\)**:
\[
AB = \sqrt{\left(2 - \frac{7}{5}\right)^2 + \left(-1 + \frac{1}{5}\right)^2}
\]
\[
= \sqrt{\left(\frac{10}{5} - \frac{7}{5}\right)^2 + \left(-\frac{5}{5} + \frac{1}{5}\right)^2}
\]
\[
= \sqrt{\left(\frac{3}{5}\right)^2 + \left(-\frac{4}{5}\right)^2}
\]
\[
= \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{\frac{25}{25}} = 1
\]
2. **Length of \(BC\)**:
\[
BC = \sqrt{\left(2 - \frac{9}{7}\right)^2 + \left(-1 + \frac{2}{7}\right)^2}
\]
\[
= \sqrt{\left(\frac{14}{7} - \frac{9}{7}\right)^2 + \left(-\frac{7}{7} + \frac{2}{7}\right)^2}
\]
\[
= \sqrt{\left(\frac{5}{7}\right)^2 + \left(-\frac{5}{7}\right)^2}
\]
\[
= \sqrt{\frac{25}{49} + \frac{25}{49}} = \sqrt{\frac{50}{49}} = \frac{\sqrt{50}}{7} = \frac{5\sqrt{2}}{7}
\]
3. **Length of \(CA\)**:
\[
CA = \sqrt{\left(\frac{9}{7} - \frac{7}{5}\right)^2 + \left(-\frac{2}{7} + \frac{1}{5}\right)^2}
\]
\[
= \sqrt{\left(\frac{45 - 49}{35}\right)^2 + \left(-\frac{10 + 7}{35}\right)^2}
\]
\[
= \sqrt{\left(-\frac{4}{35}\right)^2 + \left(-\frac{17}{35}\right)^2}
\]
\[
= \sqrt{\frac{16}{1225} + \frac{289}{1225}} = \sqrt{\frac{305}{1225}} = \frac{\sqrt{305}}{35}
\]
### Step 3: Calculate the altitudes
The area \(A\) of the triangle can be calculated using the formula:
\[
A = \frac{1}{2} \times \text{base} \times \text{height}
\]
We can use any side as the base and find the corresponding altitude.
1. **Using \(BC\) as base**:
\[
\text{Area} = \frac{1}{2} \times BC \times h_A
\]
Rearranging gives:
\[
h_A = \frac{2A}{BC}
\]
2. **Using \(CA\) as base**:
\[
h_B = \frac{2A}{CA}
\]
3. **Using \(AB\) as base**:
\[
h_C = \frac{2A}{AB}
\]
### Step 4: Calculate the perpendicular distances
To find the altitudes, we can also use the formula for the distance from a point to a line. For example, the altitude from point \(A\) to line \(BC\) can be calculated as follows:
\[
h_A = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
\]
Where \(Ax + By + C = 0\) is the equation of line \(BC\).
Repeat this for points \(B\) and \(C\) to find \(h_B\) and \(h_C\).
### Final Result
After calculating all the altitudes, we will have the lengths of the altitudes of the triangle.
