3 Write the equation at a line perpendicular to \( x-2 y+6=0 \) and Passing, through \( (3,-2) \)

To find the equation of a line that is perpendicular to the given line \( x - 2y + 6 = 0 \) and passes through the point \( (3, -2) \), follow these steps: ### 1. **Find the Slope of the Given Line** First, rewrite the given equation in **slope-intercept form** (\( y = mx + b \)) to identify its slope. \[ x - 2y + 6 = 0 \\ -2y = -x - 6 \\ y = \frac{1}{2}x + 3 \] **Slope of the given line (\( m_1 \))**: \( \frac{1}{2} \) ### 2. **Determine the Slope of the Perpendicular Line** The slope of a line perpendicular to another is the **negative reciprocal** of the original line's slope. \[ m_2 = -\frac{1}{m_1} = -\frac{1}{\frac{1}{2}} = -2 \] **Slope of the perpendicular line (\( m_2 \))**: \( -2 \) ### 3. **Use the Point-Slope Form to Find the Equation** With the slope \( m_2 = -2 \) and the point \( (3, -2) \), apply the **point-slope formula**: \[ y - y_1 = m(x - x_1) \] Substituting the known values: \[ y - (-2) = -2(x - 3) \\ y + 2 = -2x + 6 \\ y = -2x + 4 \] ### **Final Equation** The equation of the line perpendicular to \( x - 2y + 6 = 0 \) and passing through \( (3, -2) \) is: \[ y = -2x + 4 \]