Question

1. write an equation of the line that is perpendicular to $y = \frac{1}{3}x + 5$ and passes through the point $(-4,1)$. $y = \underline{\quad\quad}x + \underline{\quad\quad}$

Explanation:

Step1: Find the slope of the perpendicular line

The slope of the given line \( y = \frac{1}{3}x + 5 \) is \( m_1=\frac{1}{3} \). For two perpendicular lines, the product of their slopes is \( - 1 \), i.e., \( m_1\times m_2=-1 \). Let the slope of the perpendicular line be \( m_2 \). Then \( \frac{1}{3}\times m_2=-1 \), solving for \( m_2 \), we get \( m_2=-3 \).

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 \( (x_1,y_1)=(-4,1) \) and \( m = - 3 \). Substitute these values into the formula:
\( y - 1=-3(x - (-4)) \)
\( y - 1=-3(x + 4) \)
Expand the right - hand side: \( y - 1=-3x-12 \)
Add 1 to both sides to get the slope - intercept form \( y=mx + b \): \( y=-3x-12 + 1=-3x-11 \)

Answer:

\( y=-3x-11 \), so the first blank is \(-3\) and the second blank is \(-11\)