Question

Suppose \(A\) and \(B\) are the points at which the line \[ x+y-\lambda=0 \] meets the pair of straight lines \[ x^2+y^2-2x-4y+2=0. \] If \(\angle AOB=90^\circ\), then the value of \(\lambda\) is:

• A. \(2\)
• B. \(3\)
• C. \(4\)
• D. \(0\)

Hint:

If two points \(A(x_1,y_1)\) and \(B(x_2,y_2)\) subtend a right angle at the origin, then use \[ \overrightarrow{OA}\cdot\overrightarrow{OB}=x_1x_2+y_1y_2=0. \]

Answer 1

The Correct Option is A

Step 1: Express y from the line equation.
From x + y - λ = 0, we obtain y = λ - x.

Step 2: Substitute into the pair of lines equation.
Plugging y = λ - x into x² + y² - 2x - 4y + 2 = 0 and expanding gives 2x² + (2 - 2λ)x + (λ² - 4λ + 2) = 0. Let the roots be x₁ and x₂, corresponding to the intersection points A and B.

Step 3: Extract sum and product of roots.
x₁ + x₂ = λ - 1, and x₁x₂ = (λ² - 4λ + 2)/2.

Step 4: Impose the perpendicularity condition ∠AOB = 90°.
OA⃗·OB⃗ = 0 gives x₁x₂ + (λ - x₁)(λ - x₂) = 0. Expanding: 2x₁x₂ + λ² - λ(x₁ + x₂) = 0.

Step 5: Substitute the sum and product and solve for λ.
2[(λ² - 4λ + 2)/2] + λ² - λ(λ - 1) = 0 → λ² - 4λ + 2 + λ² - λ² + λ = 0 → λ² - 3λ + 2 = 0. Factoring: (λ - 1)(λ - 2) = 0, so λ = 1 or λ = 2. Among the options, only λ = 2 appears.

Step 6: Final conclusion.
The required value is λ = 2.