Two straight lines pass through the origin (x_0, y_0) = (0,0). One of them passes
through the point (x1, y1) = (1,3) and the other passes through the point
(x2, y2) = (1,2).
What is the area enclosed between the straight lines in the interval [0, 1] on
the x-axis?
Options:
(A) 0.5
(B) 1.0
(C) 1.5
(D) 2.0
Official Answer Key:
A
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.2)**
**1. Mathematical Formulation**:
- **Given Lines**:
- Line 1 passes through $(0,0)$ and $(1,3)$: slope $m_1 = \frac{3 - 0}{1 - 0} = 3 \implies y_1(x) = 3x$.
- Line 2 passes through $(0,0)$ and $(1,2)$: slope $m_2 = \frac{2 - 0}{1 - 0} = 2 \implies y_2(x) = 2x$.
- **Interval**: $x \in [0, 1]$.
- **Objective**: Compute the area enclosed between $y_1(x)$ and $y_2(x)$ on $[0, 1]$.
**2. Detailed Step-by-Step Derivation**:
- For any $x \in [0, 1]$, $y_1(x) \ge y_2(x)$ since $3x \ge 2x$.
- The enclosed area $A$ is given by the definite integral:
$A = \int_{0}^{1} [y_1(x) - y_2(x)]\, dx = \int_{0}^{1} (3x - 2x)\, dx = \int_{0}^{1} x\, dx$
- Evaluating the integral:
$A = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - 0 = \frac{1}{2} = 0.5$
- Alternatively, geometrically this forms a triangle with vertices at $(0,0)$, $(1,2)$, and $(1,3)$:
$\text{Base} = 3 - 2 = 1, \quad \text{Height} = 1 - 0 = 1 \implies \text{Area} = \frac{1}{2} \times 1 \times 1 = 0.5$
**3. Official Verification**:
The verified official IIT Master Answer is **Option (A)** (`0.5`).
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