Question

1. (3 pts) find the equation of a line that passes through the point (3, -1) and is perpendicular to the equation y = -3x + 2.

Explanation:

Step1: Find slope of given line

The given line is \( y = -3x + 2 \). The slope (\( m_1 \)) of this line is -3 (from \( y = mx + c \) form).

Step2: Find slope of perpendicular line

For two perpendicular lines, the product of their slopes is -1. Let the slope of the required line be \( m_2 \). So, \( m_1 \times m_2 = -1 \). Substituting \( m_1 = -3 \), we get \( -3 \times m_2 = -1 \), so \( m_2 = \frac{1}{3} \).

Step3: Use point - slope form

The point - slope form of a line is \( y - y_1 = m(x - x_1) \), where \( (x_1,y_1)=(3,-1) \) and \( m=\frac{1}{3} \). Substituting these values, we get \( y - (-1)=\frac{1}{3}(x - 3) \).

Step4: Simplify the equation

Simplify \( y + 1=\frac{1}{3}x - 1 \). Subtract 1 from both sides: \( y=\frac{1}{3}x - 2 \).

Answer:

The equation of the line is \( y=\frac{1}{3}x - 2 \)