Question

line rv is perpendicular to line ac. line ac is represented by the equation $y = \frac{3}{2}x + 5$. line rv passes through the point $r(0, 3)$. determine the equation of line rv in slope - intercept form. $y = -\frac{2}{3}x + 3$ (table: slope of line ac $m_1$, slope of line rv $m_2$, point - slope form of line rv $y - y_1 = m(x - x_1)$)

Explanation:

Step1: Find slope of AC

The equation of line \( AC \) is \( y = \frac{3}{2}x + 5 \). In slope - intercept form \( y=mx + b \), the slope \( m_1 \) of line \( AC \) is \( \frac{3}{2} \).

Step2: Find slope of RV

If two lines are perpendicular, the product of their slopes is \( - 1 \), i.e., \( m_1\times m_2=-1 \). Let the slope of line \( RV \) be \( m_2 \). We know \( m_1=\frac{3}{2} \), so \( \frac{3}{2}\times m_2=-1 \). Solving for \( m_2 \), we get \( m_2=-\frac{2}{3} \).

Step3: Use point - slope form

The line \( RV \) passes through the point \( R(0,3) \). The point - slope form of a line is \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(0,3) \) and \( m = m_2=-\frac{2}{3} \). Substituting these values, we have \( y - 3=-\frac{2}{3}(x - 0) \).

Step4: Convert to slope - intercept form

Simplify the equation \( y - 3=-\frac{2}{3}x \). Add 3 to both sides of the equation: \( y=-\frac{2}{3}x + 3 \).

For the table:

• Slope of Line \( AC \) (\( m_1 \)): \( \frac{3}{2} \)
• Slope of Line \( RV \) (\( m_2 \)): \( -\frac{2}{3} \)
• Point - Slope Form of Line \( RV \): Substitute \( x_1 = 0,y_1 = 3 \) and \( m=-\frac{2}{3} \) into \( y - y_1=m(x - x_1) \), we get \( y - 3=-\frac{2}{3}(x - 0) \) or \( y - 3=-\frac{2}{3}x \)

Answer:

• Slope of Line \( AC \): \( \frac{3}{2} \)
• Slope of Line \( RV \): \( -\frac{2}{3} \)
• Point - Slope Form of Line \( RV \): \( y - 3=-\frac{2}{3}(x - 0) \) (or \( y - 3=-\frac{2}{3}x \))
• Equation of line \( RV \) in slope - intercept form: \( y = -\frac{2}{3}x+3 \)