Question

write the equation of a line that is perpendicular to $y = -x - 6$ and that passes through the point $(-9, -4)$.

Explanation:

Step1: Find the slope of the given line

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

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 \( m_2 = 1 \).

Step2: 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 of the line.

We know that \( m = m_2=1 \) and the point \( (x_1,y_1)=(-9,-4) \). Substituting these values into the point - slope form:

\( y-(-4)=1\times(x - (-9)) \)

Simplify the left - hand side and the right - hand side:

\( y + 4=x + 9 \)

Then, subtract 4 from both sides to get the slope - intercept form:

\( y=x+9 - 4 \)

\( y=x + 5 \)

Answer:

\( y=x + 5 \)