Question

12 write the equation of a line in slope-intercept form that is perpendicular to $y = \frac{6}{5}x - 3$ and goes through the point $(12, -3)$. equation of the line: \boxed{}

Explanation:

Step1: Find the slope of the perpendicular line

The slope of the given line \( y = \frac{6}{5}x - 3 \) is \( m_1=\frac{6}{5} \). For two perpendicular lines, the product of their slopes is -1, so \( m_1\times m_2=-1 \). Substituting \( m_1 \), we get \( \frac{6}{5}\times m_2=-1 \), so \( m_2 = -\frac{5}{6} \).

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 \( (x_1,y_1)=(12,-3) \) and \( m = -\frac{5}{6} \). Substituting these values, we have \( y - (-3)=-\frac{5}{6}(x - 12) \), which simplifies to \( y + 3=-\frac{5}{6}x+10 \).

Step3: Convert to slope - intercept form

Subtract 3 from both sides of the equation \( y + 3=-\frac{5}{6}x + 10 \). So \( y=-\frac{5}{6}x+10 - 3 \), which simplifies to \( y=-\frac{5}{6}x + 7 \).

Answer:

\( y = -\frac{5}{6}x + 7 \)