Area of quadrilaeral formed by straight lines `x + y = 2, 3x + 4y = 24` and coordinate axes is

Options

A. 22 sq. unit
B. 26 sq. unit
C. 44 sq. unit
D. 11 sq. unit

Answer

To find the area of the quadrilateral formed by the lines \(x + y = 2\), \(3x + 4y = 24\), and the coordinate axes, we will follow these steps:

### Step 1: Find the intercepts of the first line \(x + y = 2\).
- Set \(x = 0\):
\[ 0 + y = 2 \implies y = 2 \quad \text{(Point: (0, 2))} \]
- Set \(y = 0\):
\[ x + 0 = 2 \implies x = 2 \quad \text{(Point: (2, 0))} \]

### Step 2: Find the intercepts of the second line \(3x + 4y = 24\).
- Set \(x = 0\):
\[ 3(0) + 4y = 24 \implies 4y = 24 \implies y = 6 \quad \text{(Point: (0, 6))} \]
- Set \(y = 0\):
\[ 3x + 4(0) = 24 \implies 3x = 24 \implies x = 8 \quad \text{(Point: (8, 0))} \]

### Step 3: Identify the vertices of the quadrilateral.
The vertices of the quadrilateral formed by the lines and the axes are:
- \(A(0, 2)\) from the first line
- \(B(2, 0)\) from the first line
- \(C(0, 6)\) from the second line
- \(D(8, 0)\) from the second line

### Step 4: Calculate the area of the quadrilateral.
The area of the quadrilateral can be found by subtracting the area of triangle \(AOB\) from the area of triangle \(COD\).

#### Area of triangle \(COD\):
- Base \(CD = 8\) (from \(C(0, 6)\) to \(D(8, 0)\))
- Height = 6 (the y-coordinate of point \(C\))
\[ \text{Area}_{COD} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 8 \times 6 = 24 \]

#### Area of triangle \(AOB\):
- Base \(AB = 2\) (from \(A(0, 2)\) to \(B(2, 0)\))
- Height = 2 (the y-coordinate of point \(A\))
\[ \text{Area}_{AOB} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 2 \times 2 = 2 \]

### Step 5: Calculate the area of the quadrilateral \(ABCD\).
\[ \text{Area}_{ABCD} = \text{Area}_{COD} - \text{Area}_{AOB} = 24 - 2 = 22 \text{ square units} \]

### Final Answer:
The area of the quadrilateral formed by the lines and the coordinate axes is **22 square units**.

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