Question

Let $a \ne 1$ be a non-zero real number. If the lines $2x+ay=1$ and $x+2y=1$ are perpendicular, then the value of $a$ is equal to ________.

• A. 1
• B. -2
• C. 2
• D. -1
• E. $-\frac{1}{2}$

Hint:

For perpendicular lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, $a_1a_2 + b_1b_2 = 0$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
This problem deals with the condition for two lines to be perpendicular in a 2D Cartesian coordinate system. The condition is based on the relationship between their slopes.
Step 2: Key Formula or Approach
Two non-vertical lines with slopes \(m_1\) and \(m_2\) are perpendicular if and only if the product of their slopes is -1.
\[ m_1 \cdot m_2 = -1 \] The slope of a line given in the general form \(Ax + By + C = 0\) is \(m = -\frac{A}{B}\).
Step 3: Detailed Explanation
1. Find the slope of the first line.
The equation of the first line is \(2x + ay = 1\), which can be written as \(2x + ay - 1 = 0\).
Here, \(A=2\) and \(B=a\). The slope \(m_1\) is:
\[ m_1 = -\frac{2}{a} \] (Note: The problem states \(a\) is non-zero, so the slope is well-defined).
2. Find the slope of the second line.
The equation of the second line is \(x + 2y = 1\), which can be written as \(x + 2y - 1 = 0\).
Here, \(A=1\) and \(B=2\). The slope \(m_2\) is:
\[ m_2 = -\frac{1}{2} \] 3. Apply the perpendicularity condition.
Since the lines are perpendicular, we must have \(m_1 \cdot m_2 = -1\).
\[ \left(-\frac{2}{a}\right) \cdot \left(-\frac{1}{2}\right) = -1 \] 4. Solve for a.
\[ \frac{2}{2a} = -1 \] \[ \frac{1}{a} = -1 \] Multiplying both sides by \(a\) gives:
\[ 1 = -a \] \[ a = -1 \] The condition \(a \neq 1\) is satisfied.
Step 4: Final Answer
The value of a is -1.