To find the area of the triangle formed by the lines given by the equations \(2x - y + 1 = 0\), \(3x - y + 5 = 0\), and \(2x - 5y + 11 = 0\), we will follow these steps:
### Step 1: Find the intersection points of the lines
We need to find the intersection points of the three lines, which will be the vertices of the triangle.
#### Intersection of Line 1 and Line 2:
1. **Line 1**: \(2x - y + 1 = 0\) → \(y = 2x + 1\)
2. **Line 2**: \(3x - y + 5 = 0\) → \(y = 3x + 5\)
Set the equations for \(y\) equal to each other:
\[
2x + 1 = 3x + 5
\]
Rearranging gives:
\[
2x - 3x = 5 - 1 \implies -x = 4 \implies x = -4
\]
Substituting \(x = -4\) back into Line 1:
\[
y = 2(-4) + 1 = -8 + 1 = -7
\]
Thus, the intersection point is \((-4, -7)\).
#### Intersection of Line 2 and Line 3:
1. **Line 2**: \(3x - y + 5 = 0\) → \(y = 3x + 5\)
2. **Line 3**: \(2x - 5y + 11 = 0\) → \(5y = 2x + 11 \implies y = \frac{2}{5}x + \frac{11}{5}\)
Set the equations for \(y\) equal to each other:
\[
3x + 5 = \frac{2}{5}x + \frac{11}{5}
\]
Multiplying through by 5 to eliminate the fraction:
\[
15x + 25 = 2x + 11
\]
Rearranging gives:
\[
15x - 2x = 11 - 25 \implies 13x = -14 \implies x = -\frac{14}{13}
\]
Substituting \(x = -\frac{14}{13}\) back into Line 2:
\[
y = 3\left(-\frac{14}{13}\right) + 5 = -\frac{42}{13} + \frac{65}{13} = \frac{23}{13}
\]
Thus, the intersection point is \(\left(-\frac{14}{13}, \frac{23}{13}\right)\).
#### Intersection of Line 1 and Line 3:
1. **Line 1**: \(2x - y + 1 = 0\) → \(y = 2x + 1\)
2. **Line 3**: \(2x - 5y + 11 = 0\) → \(5y = 2x + 11 \implies y = \frac{2}{5}x + \frac{11}{5}\)
Set the equations for \(y\) equal to each other:
\[
2x + 1 = \frac{2}{5}x + \frac{11}{5}
\]
Multiplying through by 5:
\[
10x + 5 = 2x + 11
\]
Rearranging gives:
\[
10x - 2x = 11 - 5 \implies 8x = 6 \implies x = \frac{3}{4}
\]
Substituting \(x = \frac{3}{4}\) back into Line 1:
\[
y = 2\left(\frac{3}{4}\right) + 1 = \frac{3}{2} + 1 = \frac{5}{2}
\]
Thus, the intersection point is \(\left(\frac{3}{4}, \frac{5}{2}\right)\).
### Step 2: Area of the triangle
Now we have the three vertices of the triangle:
1. \(A(-4, -7)\)
2. \(B\left(-\frac{14}{13}, \frac{23}{13}\right)\)
3. \(C\left(\frac{3}{4}, \frac{5}{2}\right)\)
The area \(A\) of the triangle formed by the points \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\) is given by the formula:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the coordinates:
\[
\text{Area} = \frac{1}{2} \left| -4\left(\frac{23}{13} - \frac{5}{2}\right) + \left(-\frac{14}{13}\right)\left(\frac{5}{2} + 7\right) + \frac{3}{4}\left(-7 - \frac{23}{13}\right) \right|
\]
Calculating each term:
1. For the first term:
\[
\frac{23}{13} - \frac{5}{2} = \frac{23}{13} - \frac{65}{26} = \frac{46 - 65}{26} = -\frac{19}{26}
\]
Thus, the first term becomes:
\[
-4 \left(-\frac{19}{26}\right) = \frac{76}{26} = \frac{38}{13}
\]
2. For the second term:
\[
\frac{5}{2} + 7 = \frac{5}{2} + \frac{14}{2} = \frac{19}{2}
\]
Thus, the second term becomes:
\[
-\frac{14}{13} \cdot \frac{19}{2} = -\frac{266}{26} = -\frac{133}{13}
\]
3. For the third term:
\[
-7 - \frac{23}{13} = -\frac{91}{13} - \frac{23}{13} = -\frac{114}{13}
\]
Thus, the third term becomes:
\[
\frac{3}{4} \cdot \left(-\frac{114}{13}\right) = -\frac{342}{52} = -\frac{171}{26}
\]
Combining all terms:
\[
\text{Area} = \frac{1}{2} \left| \frac{38}{13} - \frac{133}{13} - \frac{171}{26} \right|
\]
Finding a common denominator (26):
\[
\text{Area} = \frac{1}{2} \left| \frac{76 - 266 - 171}{26} \right| = \frac{1}{2} \left| \frac{-361}{26} \right| = \frac{361}{52}
\]
### Final Answer
Thus, the area of the triangle formed by the given lines is:
\[
\text{Area} = \frac{361}{52}
\]
