Find the area of the triangle formed by the lines `y+x-6=0, 3y-x+2=0 and 3y=5x+2`.

Answer

To find the area of the triangle formed by the lines \( y + x - 6 = 0 \), \( 3y - x + 2 = 0 \), and \( 3y = 5x + 2 \), we will follow these steps:

### Step 1: Rewrite the equations in standard form
1. The first equation is already in standard form:
\[ x + y = 6 \]

2. The second equation can be rearranged:
\[ 3y - x + 2 = 0 \implies x - 3y = 2 \]

3. The third equation can also be rearranged:
\[ 3y = 5x + 2 \implies 5x - 3y = -2 \]

Now we have the three equations:
- \( x + y = 6 \) (Equation 1)
- \( x - 3y = 2 \) (Equation 2)
- \( 5x - 3y = -2 \) (Equation 3)

### Step 2: Find the intersection points of the lines
#### Intersection of Equation 1 and Equation 2
To find the intersection of \( x + y = 6 \) and \( x - 3y = 2 \):
1. From Equation 1, express \( y \):
\[ y = 6 - x \]

2. Substitute \( y \) in Equation 2:
\[ x - 3(6 - x) = 2 \]
\[ x - 18 + 3x = 2 \]
\[ 4x - 18 = 2 \]
\[ 4x = 20 \]
\[ x = 5 \]

3. Substitute \( x = 5 \) back into Equation 1 to find \( y \):
\[ y = 6 - 5 = 1 \]

So, the intersection point \( A \) is \( (5, 1) \).

#### Intersection of Equation 2 and Equation 3
To find the intersection of \( x - 3y = 2 \) and \( 5x - 3y = -2 \):
1. Subtract Equation 2 from Equation 3:
\[ (5x - 3y) - (x - 3y) = -2 - 2 \]
\[ 4x = -4 \]
\[ x = -1 \]

2. Substitute \( x = -1 \) back into Equation 2:
\[ -1 - 3y = 2 \]
\[ -3y = 3 \]
\[ y = -1 \]

So, the intersection point \( B \) is \( (-1, -1) \).

#### Intersection of Equation 1 and Equation 3
To find the intersection of \( x + y = 6 \) and \( 5x - 3y = -2 \):
1. From Equation 1, express \( y \):
\[ y = 6 - x \]

2. Substitute \( y \) in Equation 3:
\[ 5x - 3(6 - x) = -2 \]
\[ 5x - 18 + 3x = -2 \]
\[ 8x - 18 = -2 \]
\[ 8x = 16 \]
\[ x = 2 \]

3. Substitute \( x = 2 \) back into Equation 1 to find \( y \):
\[ y = 6 - 2 = 4 \]

So, the intersection point \( C \) is \( (2, 4) \).

### Step 3: Calculate the area of the triangle
The vertices of the triangle are:
- \( A(5, 1) \)
- \( B(-1, -1) \)
- \( C(2, 4) \)

Using the formula for the area of a triangle given 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:
\[ \text{Area} = \frac{1}{2} \left| 5(-1 - 4) + (-1)(4 - 1) + 2(1 - (-1)) \right| \]
\[ = \frac{1}{2} \left| 5(-5) - 1(3) + 2(2) \right| \]
\[ = \frac{1}{2} \left| -25 - 3 + 4 \right| \]
\[ = \frac{1}{2} \left| -24 \right| = \frac{24}{2} = 12 \]

Thus, the area of the triangle is \( 12 \) square units.