Question

write the equation of a line perpendicular to $y = -\frac{3}{8}x - 2$ that passes through the point (-3, -3).
$\bigcirc$ $y = \frac{8}{3}x - 5$
$\bigcirc$ $y = \frac{8}{3}x + 5$
$\bigcirc$ $y = -\frac{3}{8}x - 5$
$\bigcirc$ $y = -\frac{3}{8}x + 5$

Explanation:

Step1: Find the slope of the perpendicular line

The slope of the given line \( y = -\frac{3}{8}x - 2 \) is \( m_1 = -\frac{3}{8} \). For two perpendicular lines, the product of their slopes is -1, so \( m_1 \times m_2 = -1 \). Solving for \( m_2 \): \( m_2 = \frac{-1}{m_1} = \frac{-1}{-\frac{3}{8}} = \frac{8}{3} \).

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 \( (x_1,y_1)=(-3,-3) \) and \( m = \frac{8}{3} \). Substitute these values: \( y - (-3)=\frac{8}{3}(x - (-3)) \), which simplifies to \( y + 3=\frac{8}{3}(x + 3) \).

Step3: Simplify the equation

Expand the right - hand side: \( y+3=\frac{8}{3}x + 8 \). Subtract 3 from both sides: \( y=\frac{8}{3}x+8 - 3=\frac{8}{3}x + 5 \).

Answer:

B. \( y=\frac{8}{3}x + 5 \)