Question

write the equation of a line that is perpendicular to $y = 3x - 2$ and that passes through the point $(-9, 5).$

Explanation:

Step1: Find the slope of the perpendicular line

The slope of the given line \( y = 3x - 2 \) is \( m_1 = 3 \). For two perpendicular lines, the product of their slopes is \( -1 \), so \( m_1 \times m_2=-1 \). Substituting \( m_1 = 3 \), we get \( 3\times m_2=-1 \), so \( m_2 = -\frac{1}{3} \).

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 \( (x_1,y_1)=(-9,5) \) and \( m = -\frac{1}{3} \). Substituting these values, we have \( y - 5=-\frac{1}{3}(x + 9) \).

Step3: Simplify the equation

Expand the right - hand side: \( y - 5=-\frac{1}{3}x-3 \). Then add 5 to both sides: \( y=-\frac{1}{3}x - 3 + 5 \), so \( y=-\frac{1}{3}x+2 \).

Answer:

\( y = -\frac{1}{3}x + 2 \)