To find the area of the triangle formed by the lines given by the equations \(x - y = 0\), \(2x + y = 6\), and \(x + 2y = 3\), we will first determine the points of intersection of these lines, which will serve as the vertices of the triangle.
### Step 1: Find the intersection of the first two lines
We start with the equations:
1. \(x - y = 0\) (or \(x = y\))
2. \(2x + y = 6\)
Substituting \(y\) from the first equation into the second equation:
\[
2x + x = 6 \implies 3x = 6 \implies x = 2
\]
Now substituting \(x = 2\) back into \(y = x\):
\[
y = 2
\]
Thus, the point of intersection \(A\) is \((2, 2)\).
### Step 2: Find the intersection of the second and third lines
Next, we consider the equations:
1. \(2x + y = 6\)
2. \(x + 2y = 3\)
We can solve these equations simultaneously. First, we can express \(y\) from the first equation:
\[
y = 6 - 2x
\]
Now substituting into the second equation:
\[
x + 2(6 - 2x) = 3 \implies x + 12 - 4x = 3 \implies -3x + 12 = 3 \implies -3x = -9 \implies x = 3
\]
Substituting \(x = 3\) back into \(y = 6 - 2x\):
\[
y = 6 - 2(3) = 0
\]
Thus, the point of intersection \(B\) is \((3, 0)\).
### Step 3: Find the intersection of the first and third lines
Now we consider the equations:
1. \(x - y = 0\) (or \(x = y\))
2. \(x + 2y = 3\)
Substituting \(y = x\) into the second equation:
\[
x + 2x = 3 \implies 3x = 3 \implies x = 1
\]
Now substituting \(x = 1\) back into \(y = x\):
\[
y = 1
\]
Thus, the point of intersection \(C\) is \((1, 1)\).
### Step 4: Calculate the area of the triangle
Now we have the vertices of the triangle:
- \(A(2, 2)\)
- \(B(3, 0)\)
- \(C(1, 1)\)
We can use the formula for the area of a triangle given by vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\):
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the coordinates:
\[
\text{Area} = \frac{1}{2} \left| 2(0 - 1) + 3(1 - 2) + 1(2 - 0) \right|
\]
Calculating each term:
\[
= \frac{1}{2} \left| 2(-1) + 3(-1) + 1(2) \right| = \frac{1}{2} \left| -2 - 3 + 2 \right| = \frac{1}{2} \left| -3 \right| = \frac{3}{2}
\]
Thus, the area of the triangle formed by the lines is \(\frac{3}{2}\).
