What is the area (in sq. units) of the triangle formed by the graph of the equation 2x + 5y -12 = 0, x + y = 3 and y = 0?
समीकरण 2x + 5y -12 = 0, x + y = 3 और y = 0 के आरेख द्वारा निर्मित त्रिभुज का क्षेत्रफल (वर्ग इकाइयों में) क्या है?

Options

A. 6
B. 5
C. 3
D. 2

Answer

To find the area of the triangle formed by the equations \(2x + 5y - 12 = 0\), \(x + y = 3\), and \(y = 0\), we will follow these steps:

### Step 1: Find the intersection points of the lines
1. **Find the intersection of \(2x + 5y - 12 = 0\) and \(x + y = 3\)**:
- From \(x + y = 3\), we can express \(y\) in terms of \(x\):
\[ y = 3 - x \]
- Substitute \(y\) into the first equation:
\[ 2x + 5(3 - x) - 12 = 0 \]
\[ 2x + 15 - 5x - 12 = 0 \]
\[ -3x + 3 = 0 \implies 3x = 3 \implies x = 1 \]
- Substitute \(x = 1\) back into \(y = 3 - x\):
\[ y = 3 - 1 = 2 \]
- So, the intersection point is \((1, 2)\).

2. **Find the intersection of \(2x + 5y - 12 = 0\) and \(y = 0\)**:
- Substitute \(y = 0\) into the first equation:
\[ 2x + 5(0) - 12 = 0 \implies 2x - 12 = 0 \implies 2x = 12 \implies x = 6 \]
- So, the intersection point is \((6, 0)\).

3. **Find the intersection of \(x + y = 3\) and \(y = 0\)**:
- Substitute \(y = 0\) into the second equation:
\[ x + 0 = 3 \implies x = 3 \]
- So, the intersection point is \((3, 0)\).

### Step 2: Identify the vertices of the triangle
The vertices of the triangle formed by the lines are:
- \(A(1, 2)\)
- \(B(6, 0)\)
- \(C(3, 0)\)

### Step 3: 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:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates of the vertices:
\[ A = \frac{1}{2} \left| 1(0 - 0) + 6(0 - 2) + 3(2 - 0) \right| \]
\[ = \frac{1}{2} \left| 0 - 12 + 6 \right| \]
\[ = \frac{1}{2} \left| -6 \right| = \frac{1}{2} \times 6 = 3 \]

### Final Answer
The area of the triangle is \(3\) square units.