Find the area of the triangle formed by the straight lines `2 x + 3 y = 12 , x - y = 1` and the y - axis

Options

A. 15 sq. units
B. `(15)/( 2)` sq. units
C. 14 sq.units
D. 1 sq. units

Answer

To find the area of the triangle formed by the lines \(2x + 3y = 12\), \(x - y = 1\), and the y-axis, we will follow these steps:

### Step 1: Find the intersection points with the y-axis
1. For the line \(2x + 3y = 12\):
- Set \(x = 0\):
\[ 2(0) + 3y = 12 \implies 3y = 12 \implies y = 4 \]
- This gives the point \(A(0, 4)\).

2. For the line \(x - y = 1\):
- Set \(y = 0\):
\[ x - 0 = 1 \implies x = 1 \]
- This gives the point \(B(1, 0)\).

### Step 2: Find the intersection point of the two lines
To find the intersection point \(C\) of the lines \(2x + 3y = 12\) and \(x - y = 1\), we will solve the equations simultaneously.

1. Rearranging the second equation:
\[ x = y + 1 \]

2. Substitute \(x\) in the first equation:
\[ 2(y + 1) + 3y = 12 \implies 2y + 2 + 3y = 12 \implies 5y + 2 = 12 \]
\[ 5y = 10 \implies y = 2 \]

3. Substitute \(y\) back to find \(x\):
\[ x = 2 + 1 = 3 \]
- This gives the point \(C(3, 2)\).

### Step 3: Determine the vertices of the triangle
The vertices of the triangle formed by the lines and the y-axis are:
- \(A(0, 4)\)
- \(B(1, 0)\)
- \(C(3, 2)\)

### Step 4: Calculate the area of the triangle
The area \(A\) of a triangle given vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using 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 of points \(A(0, 4)\), \(B(1, 0)\), and \(C(3, 2)\):
\[ \text{Area} = \frac{1}{2} \left| 0(0 - 2) + 1(2 - 4) + 3(4 - 0) \right| \]
\[ = \frac{1}{2} \left| 0 + 1(-2) + 3(4) \right| \]
\[ = \frac{1}{2} \left| -2 + 12 \right| = \frac{1}{2} \left| 10 \right| = \frac{10}{2} = 5 \]

Thus, the area of the triangle is \(5\) square units.