Question

A straight line perpendicular to the line \( 2x + y = 3 \) is passing through \( (1,1) \). Its y-intercept is:

• A. \( 1 \)
• B. \( 2 \)
• C. \( 3 \)
• D. \( \frac{1}{2} \)
• E. \( \frac{1}{3} \)

Hint:

The slope of the given line is \(-2\). The perpendicular slope is \(+1/2\). Using \( y - y_1 = m(x - x_1) \): \[ y - 1 = \frac{1}{2}(x - 1) \] At \( x = 0 \), \( y - 1 = -1/2 \), so \( y = 1/2 \).

Answer 1

The Correct Option is D

Concept: If a line has slope \( m \), any line perpendicular to it has slope \( -1/m \). The general equation of a line perpendicular to \( Ax + By + C = 0 \) is \( Bx - Ay + \lambda = 0 \).

Step 1: Determine the equation of the perpendicular line.
The given line is \( 2x + 1y - 3 = 0 \). A line perpendicular to it will be of the form: \[ 1x - 2y + \lambda = 0 \]

Step 2: Solve for the constant using the point (1, 1).
Substitute \( x = 1 \) and \( y = 1 \): \[ 1(1) - 2(1) + \lambda = 0 \] \[ 1 - 2 + \lambda = 0 \quad \Rightarrow \quad \lambda = 1 \] The equation is \( x - 2y + 1 = 0 \).

Step 3: Find the y-intercept.
To find the y-intercept, set \( x = 0 \): \[ 0 - 2y + 1 = 0 \] \[ 2y = 1 \quad \Rightarrow \quad y = \frac{1}{2} \]