Question

write the equation of a line perpendicular to $y = -\frac{2}{3}x + 7$ and passing through the point $(6, -10)$. select the correct answer $y = -\frac{5}{2}x - 19$ $y = -\frac{2}{3}x - 19$ $y = \frac{2}{3}x - 19$ $y = \frac{3}{2}x - 19$

Explanation:

Step1: Find the slope of the perpendicular line

The slope of the given line \( y = -\frac{2}{3}x + 7 \) is \( m_1 = -\frac{2}{3} \). For two perpendicular lines, the product of their slopes is \( -1 \), so \( m_1 \times m_2 = -1 \). Solving for \( m_2 \): \( m_2 = \frac{-1}{m_1} = \frac{-1}{-\frac{2}{3}} = \frac{3}{2} \).

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 \( (x_1,y_1)=(6, - 10) \) and \( m=\frac{3}{2} \). Substitute these values: \( y - (-10)=\frac{3}{2}(x - 6) \). Simplify: \( y + 10=\frac{3}{2}x-9 \). Then \( y=\frac{3}{2}x-9 - 10=\frac{3}{2}x-19 \).

Answer:

\( y=\frac{3}{2}x - 19 \) (the fourth option: \( y=\frac{3}{2}x - 19 \))