Q.2 Two straight lines pass through the origin (𝑥଴, 𝑦଴) = (0,0). One of them passes through the point (𝑥ଵ, 𝑦ଵ) = (1,3) and the other passes through the point (𝑥ଶ, 𝑦ଶ) = (1,2). What is the area enclosed between the straight lines in the interval [0, 1] on the 𝑥 -axis? (A) 0.5 (B) 1.0 (C) 1.5 (D) 2.0

Q.2 Two straight lines pass through the origin (x0, y0) = (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?

  1. 0.5
  2. 1.0
  3. 1.5
  4. 2.0

Area Enclosed Between Straight Lines Through Origin and Points (1,3) (1,2) in [0,1]

Answer: 0.5

Two straight lines pass through the origin (0,0), one through (1,3) and the other through (1,2). The area enclosed between them over x from 0 to 1 is 0.5 square units.

Line Equations

The first line has slope m1 = 3/1 = 3, so y = 3x. The second line has slope m2 = 2/1 = 2, so y = 2x. For x ≥ 0, y = 3x lies above y = 2x.

Area Calculation

The area A between the curves from x = 0 to x = 1 uses the integral A = ∫01 (3x – 2x) dx = ∫01 x dx = [ (1/2)x² ] from 0 to 1 = 1/2 = 0.5.

Alternatively, subtract triangle areas: upper triangle base 1 height 3 has area (1/2)*1*3 = 1.5; lower has (1/2)*1*2 = 1; difference 1.5 – 1 = 0.5.

Option Analysis

  • 0.5: Correct, matches integral (1/2)x² from 0 to 1.
  • 1.0: Wrong; equals full lower triangle area, ignores subtraction.
  • 1.5: Wrong; upper triangle only, excludes lower line.
  • 2.0: Wrong; sums heights without halving or subtracting.

Introduction to Area Enclosed Between Straight Lines Origin (1,3) (1,2)

Finding the area enclosed between straight lines origin (1,3) (1,2) in interval [0,1] tests definite integration basics for competitive exams like GATE or IIT JAM. Two lines from (0,0) through (1,3) and (1,2) form y=3x and y=2x. The region between them from x=0 to x=1 has area 0.5, key for geometry practice.

Step-by-Step Solution: Equations and Integral

Find line equations through origin: Slope for first line is Δy/Δx = 3/1 = 3, so y₁ = 3x. Second is 2/1 = 2, so y₂ = 2x.

Set up area integral: Upper curve minus lower: |y₁ – y₂| = x. Thus, A = ∫01 x dx = [½x²]01 = ½(1)² – ½(0)² = 0.5.

Geometrically, it’s difference of triangular areas: (½*1*3) – (½*1*2) = 1.5 – 1 = 0.5.

Why Other Options Fail

Option Calculation Error Correct Value
0.5 Matches ∫x dx = ½ 0.5
1.0 Full lower triangle only 1.0
1.5 Upper triangle ignores subtraction 1.5
2.0 Sum of heights, no area formula N/A

Exam Tips for Similar Problems

  • Always derive slopes for lines through origin: m = y/x.
  • Confirm upper/lower curves in interval.
  • Use integration for curved regions, triangles for linear.
  • Practice verifies 0.5 as GATE 2022 answer.

 

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