Question

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

Explanation:

Step1: Find the slope of the given line

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

Step2: Find the slope of the perpendicular line

If two lines are perpendicular, the product of their slopes is \( -1 \). Let the slope of the perpendicular line be \( m_2 \). Then \( m_1\times m_2=-1 \). Substituting \( m_1 = - 1 \), we get \( -1\times m_2=-1 \), so \( m_2 = 1 \).

Step3: Use point - slope form to find the equation

The point - slope form of a line is \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(1,-3) \) and \( m = 1 \). Substituting these values, we have \( y-(-3)=1\times(x - 1) \), which simplifies to \( y + 3=x - 1 \).

Answer:

\( y + 3=x - 1 \) (or the option corresponding to this equation, looking at the options, it is the fourth option: \( y + 3=x - 1 \))