Linear InequalitiesMCQPYQ Jan 21Question 1125 of 146
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The common region in the graph of the inequalities x+y≤4\displaystyle x + y \le 4, x−y≤4\displaystyle x - y \le 4, x≥2\displaystyle x \ge 2 is

Options

AEquilateral triangle
BIsosceles triangle
CQuadrilateral
DSquare
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Correct Answer

✅ Option b — Isosceles triangle

All Options:

  • AEquilateral triangle
  • BIsosceles triangle
  • CQuadrilateral
  • DSquare

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Detailed Solution & Explanation

Let's identify the boundary lines and find their intersection points to determine the vertices of the common region:
1) x+y=4\displaystyle x + y = 4
2) x−y=4\displaystyle x - y = 4
3) x=2\displaystyle x = 2

**Step 1: Intercepts & Origin Testing**
- Line 1 (x+y=4\displaystyle x+y=4): intercepts (0,4),(4,0)\displaystyle (0,4), (4,0). Testing (0,0)\displaystyle (0,0) gives 0≤4\displaystyle 0 \le 4 (True). Shaded region is below the line.
- Line 2 (x−y=4\displaystyle x-y=4): intercepts (0,−4),(4,0)\displaystyle (0,-4), (4,0). Testing (0,0)\displaystyle (0,0) gives 0≤4\displaystyle 0 \le 4 (True). Shaded region is above the line.
- Line 3 (x=2\displaystyle x=2): vertical line. Shaded region lies to the right (x≥2\displaystyle x \ge 2).

**Step 2: Intersection Points (Vertices)**
- **Intersection of x+y=4\displaystyle x + y = 4 and x−y=4\displaystyle x - y = 4**:
Adding both equations:
2x=8  ⟹  x=4  ⟹  y=0  ⟹  P1(4,0)2x = 8 \implies x = 4 \implies y = 0 \implies P_1(4, 0)
- **Intersection of x+y=4\displaystyle x + y = 4 and x=2\displaystyle x = 2**:
2+y=4  ⟹  y=2  ⟹  P2(2,2)2 + y = 4 \implies y = 2 \implies P_2(2, 2)
- **Intersection of x−y=4\displaystyle x - y = 4 and x=2\displaystyle x = 2**:
2−y=4  ⟹  y=−2  ⟹  P3(2,−2)2 - y = 4 \implies y = -2 \implies P_3(2, -2)

The three intersection points are the vertices of the common region:
P1(4,0),P2(2,2),P3(2,−2)P_1(4, 0), \quad P_2(2, 2), \quad P_3(2, -2)

**Step 3: Distance Calculations**
Let's calculate the lengths of the sides of this triangle using the distance formula d=(x2−x1)2+(y2−y1)2\displaystyle d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}:
- **Length of side P1P2\displaystyle P_1 P_2**:
d(P1,P2)=(2−4)2+(2−0)2=(−2)2+22=4+4=22d(P_1, P_2) = \sqrt{(2 - 4)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = 2\sqrt{2}
- **Length of side P1P3\displaystyle P_1 P_3**:
d(P1,P3)=(2−4)2+(−2−0)2=(−2)2+(−2)2=4+4=22d(P_1, P_3) = \sqrt{(2 - 4)^2 + (-2 - 0)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = 2\sqrt{2}
- **Length of side P2P3\displaystyle P_2 P_3**:
d(P2,P3)=(2−2)2+(−2−2)2=02+(−4)2=16=4d(P_2, P_3) = \sqrt{(2 - 2)^2 + (-2 - 2)^2} = \sqrt{0^2 + (-4)^2} = \sqrt{16} = 4

Since two sides are equal in length (P1P2=P1P3=22\displaystyle P_1 P_2 = P_1 P_3 = 2\sqrt{2}) and the third side is different (4\displaystyle 4), the triangle is an **isosceles triangle**.

Hence, **Option B** is the correct answer.

About This Chapter: Linear Inequalities

Paper

Paper 3: Quantitative Aptitude

Weightage

1-3 Marks

Key Topics

Linear Inequalities in one & two variables

This chapter covers Linear Inequalities in one & two variables and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 1-3 Marks weightage. Focus on understanding core concepts rather than memorizing.

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