Area of triangle formed by straight lines `2x - 3y + 6 =0, 2x + 3y - 18 = 0 and y -1 = 0` is

Options

A. 27 sq. unit
B. `(27)/(2)` sq. unit
C. 9 sq. unit
D. None of these

Answer

To find the area of the triangle formed by the lines \(2x - 3y + 6 = 0\), \(2x + 3y - 18 = 0\), and \(y - 1 = 0\), we will follow these steps:

### Step 1: Find the intersection points of the lines

1. **Line 1**: \(2x - 3y + 6 = 0\)
- To find the x-intercept, set \(y = 0\):
\[ 2x + 6 = 0 \implies x = -3 \implies (-3, 0) \]
- To find the y-intercept, set \(x = 0\):
\[ -3y + 6 = 0 \implies y = 2 \implies (0, 2) \]

2. **Line 2**: \(2x + 3y - 18 = 0\)
- To find the x-intercept, set \(y = 0\):
\[ 2x - 18 = 0 \implies x = 9 \implies (9, 0) \]
- To find the y-intercept, set \(x = 0\):
\[ 3y - 18 = 0 \implies y = 6 \implies (0, 6) \]

3. **Line 3**: \(y - 1 = 0\)
- This line is horizontal at \(y = 1\).

### Step 2: Find the intersection points of the lines

1. **Intersection of Line 1 and Line 2**:
- Solve the equations:
\[ 2x - 3y + 6 = 0 \quad (1) \]
\[ 2x + 3y - 18 = 0 \quad (2) \]
- Adding (1) and (2):
\[ 4x - 12 = 0 \implies x = 3 \]
- Substitute \(x = 3\) into (1):
\[ 2(3) - 3y + 6 = 0 \implies 6 - 3y + 6 = 0 \implies 12 - 3y = 0 \implies y = 4 \]
- Intersection point: \((3, 4)\).

2. **Intersection of Line 1 and Line 3**:
- Substitute \(y = 1\) into Line 1:
\[ 2x - 3(1) + 6 = 0 \implies 2x + 3 = 0 \implies x = -\frac{3}{2} \]
- Intersection point: \(\left(-\frac{3}{2}, 1\right)\).

3. **Intersection of Line 2 and Line 3**:
- Substitute \(y = 1\) into Line 2:
\[ 2x + 3(1) - 18 = 0 \implies 2x + 3 - 18 = 0 \implies 2x - 15 = 0 \implies x = \frac{15}{2} \]
- Intersection point: \(\left(\frac{15}{2}, 1\right)\).

### Step 3: Calculate the area of the triangle

The vertices of the triangle are:
- \(A(3, 4)\)
- \(B\left(-\frac{3}{2}, 1\right)\)
- \(C\left(\frac{15}{2}, 1\right)\)

1. **Base**: The distance between points \(B\) and \(C\):
\[ BC = \left(\frac{15}{2} - \left(-\frac{3}{2}\right)\right) = \frac{15}{2} + \frac{3}{2} = \frac{18}{2} = 9 \]

2. **Height**: The height from point \(A\) to line \(y = 1\):
\[ \text{Height} = 4 - 1 = 3 \]

3. **Area of the triangle**:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 9 \times 3 = \frac{27}{2} \text{ square units} \]

### Final Answer
The area of the triangle is \(\frac{27}{2}\) square units.

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