Question

find the equation of a line perpendicular to y = -x + 6 that passes through the point (1, -3).

Explanation:

Step1: Find the slope of the given line

The given line is \( y = -x + 6 \). The slope-intercept form of a line is \( y = mx + b \), where \( m \) is the slope. So the slope of the given line, \( m_1 \), is \( -1 \).

Step2: Find the slope of the perpendicular line

The slope of a line perpendicular to a line with slope \( m_1 \) is the negative reciprocal of \( m_1 \). Let \( m_2 \) be the slope of the perpendicular line. So \( m_2 = -\frac{1}{m_1} \). Substituting \( m_1 = -1 \), we get \( m_2 = -\frac{1}{-1} = 1 \).

Step3: Use the point-slope form to find the equation of the line

The point-slope form of a line is \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a point on the line and \( m \) is the slope. We know the line passes through \( (1, -3) \) and has a slope \( m_2 = 1 \). Substituting \( x_1 = 1 \), \( y_1 = -3 \), and \( m = 1 \) into the point-slope form:
\( y - (-3) = 1 \times (x - 1) \)
Simplify this equation:
\( y + 3 = x - 1 \)
Subtract 3 from both sides to get the slope-intercept form:
\( y = x - 4 \)

Answer:

The equation of the perpendicular line is \( y = x - 4 \)