Question

line c1 is represented by the equation $y = \frac{2}{3}x + 16$. determine the equation, in slope - intercept form, of the line c2 that is perpendicular to line c1 and passes through the point (12, 18).
there is a table with columns: slope of line c1 ($m_1$), slope of line c2 ($m_2$), point - slope form of line c2 ($y - y_1 = m(x - x_1)$).
there is also a coordinate grid with the graph of line c1 and some marked points.

Explanation:

Step1: Find slope of line \( l_1 \)

The equation of line \( l_1 \) is \( y = \frac{2}{3}x + 18 \). In slope - intercept form \( y=mx + b \), the slope \( m_1=\frac{2}{3} \).

Step2: Find slope of perpendicular line \( l_2 \)

If two lines are perpendicular, the product of their slopes is - 1. Let the slope of \( l_2 \) be \( m_2 \). Then \( m_1\times m_2=- 1 \). Substituting \( m_1 = \frac{2}{3} \), we get \( \frac{2}{3}\times m_2=-1 \). Solving for \( m_2 \), we have \( m_2=-\frac{3}{2} \).

Step3: Use point - slope form

The point - slope form of a line is \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(12,18) \) and \( m = m_2=-\frac{3}{2} \). Substituting these values, we get \( y - 18=-\frac{3}{2}(x - 12) \).

Answer:

• Slope of line \( l_1 \): \( \frac{2}{3} \)
• Slope of line \( l_2 \): \( -\frac{3}{2} \)
• Point - Slope form of line \( l_2 \): \( y - 18=-\frac{3}{2}(x - 12) \)