Question

The equation of the line passing through the point \((1, 2)\) and perpendicular to the line \(3x + 4y - 12 = 0\) is:

• A. \(4x - 3y + 1 = 0\)
• B. \(4x + 3y - 11 = 0\)
• C. \(4x - 3y + 2 = 0\)
• D. \(3x + 4y - 10 = 0\)

Hint:

For a line perpendicular to another, use the negative reciprocal of the given line's slope. Use the point-slope form and verify the point lies on the final line.

Answer 1

The Correct Option is C

Step 1: Find the slope of the given line \(3x + 4y - 12 = 0\). 
Rewrite in slope-intercept form \(y = mx + c\): \[ 3x + 4y - 12 = 0 \implies 4y = -3x + 12 \implies y = -\frac{3}{4}x + 3. \] So, the slope of the given line is: \[ m_1 = -\frac{3}{4}. \] Step 2: Find the slope of the perpendicular line. 
The slope of the line perpendicular to the given line is the negative reciprocal of \(m_1\): \[ m_2 = \frac{4}{3}. \] Step 3: Use point-slope form with point \((1, 2)\) and slope \(m_2 = \frac{4}{3}\). 
\[ y - 2 = \frac{4}{3}(x - 1). \] Multiply both sides by 3: \[ 3(y - 2) = 4(x - 1) \implies 3y - 6 = 4x - 4. \] Bring all terms to one side: \[ 4x - 3y - 2 = 0. \] Step 4: Check if the point \((1, 2)\) lies on the line. 
Substituting \(x=1, y=2\): \[ 4(1) - 3(2) - 2 = 4 - 6 - 2 = -4 \neq 0. \] To ensure the line passes through \((1,2)\), adjust the constant term \(c\) in: \[ 4x - 3y + c = 0, \] Substitute \((1, 2)\): \[ 4(1) - 3(2) + c = 0 \implies 4 - 6 + c = 0 \implies c = 2. \] Hence, the correct equation is: \[ 4x - 3y + 2 = 0. \]