2y - 1 = 0 are prepedicular, then the..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep."> 2y - 1 = 0 are prepedicular, then the..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep." /> 2y - 1 = 0 are prepedicular, then the..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep." />

Question 22

Mathematics Line Easy

If the lines x + (a - 1)y + 1 = 0 and 2x + a<sup>2</sup>y - 1 = 0 are prepedicular, then the condition satisfied by a is

(A) |a| = 2
(B) 0 &lt; a &lt; 1
(C) -1 &lt; a &lt; 0
(D) a = -1
✓ Correct Answer: Option D Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

\begin{aligned} &\text{Given, the lines } x + (a - 1)y + 1 = 0 \text{ and } 2x + a^2 y - 1 = 0 \text{ are perpendicular.} \\ &\text{The condition for perpendicularity of two lines } Ax + By + C = 0 \text{ and } A'x + B'y + C' = 0 \text{ is:} \\ &A \cdot A' + B \cdot B' = 0. \\ &\text{Here, comparing with the standard form:} \\ &A = 1, \quad B = (a - 1), \quad A' = 2, \quad B' = a^2. \\ &\text{Applying the perpendicularity condition:} \\ &1 \times 2 + (a - 1) \times a^2 = 0. \\ &\Rightarrow 2 + a^3 - a^2 = 0. \\ &\Rightarrow a^3 - a^2 + 2 = 0. \\ &\text{Rewriting the equation:} \\ & a^3 + a^2 - 2a^2 - 2a + 2a + 2 = 0. \\ &\Rightarrow a^2 (a + 1) - 2a (a + 1) + 2 (a + 1) = 0. \\ &\Rightarrow (a + 1)(a^2 - 2a + 2) = 0. \\ &\Rightarrow (a + 1) = 0. \\ &\Rightarrow a = -1. \end{aligned}

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