Question

find the equation of the line that is perpendicular to $y = \frac{1}{3}x - 6$ and contains the point $(5,-6)$.
$y = ?x + \quad$

Explanation:

Step1: Find the slope of the perpendicular line

The slope of the given line \( y = \frac{1}{3}x - 6 \) is \( m_1=\frac{1}{3} \). For two perpendicular lines, the product of their slopes is \( - 1 \), i.e., \( m_1\times m_2=-1 \). Let the slope of the perpendicular line be \( m_2 \). Then \( \frac{1}{3}\times m_2=-1 \), so \( m_2=-3 \).

Step2: Use point - slope form to find the equation

The point - slope form of a line is \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(5,-6) \) and \( m = m_2=-3 \). Substitute these values into the formula: \( y-(-6)=-3(x - 5) \).
Simplify the left - hand side: \( y + 6=-3(x - 5) \).
Expand the right - hand side: \( y+6=-3x + 15 \).
Subtract 6 from both sides to get the slope - intercept form \( y=-3x+15 - 6 \), so \( y=-3x + 9 \).

Answer:

\( y=-3x + 9 \)