To find the equation of line \( k \) that is perpendicular to line \( j \) and passes through the point \((-1, -3)\), we first need to determine the slope of line \( j \).
The equation of line \( j \) is:
\[ y = \frac{1}{5}x + 7 \]
From this equation, we can see that the slope (\( m_j \)) of line \( j \) is:
\[ m_j = \frac{1}{5} \]
The slope of line \( k \), which is perpendicular to line \( j \), is the negative reciprocal of \( m_j \). Therefore, we calculate the slope of line \( k \) (\( m_k \)) as follows:
\[ m_k = -\frac{1}{m_j} = -\frac{1}{\frac{1}{5}} = -5 \]
Now that we have the slope of line \( k \), we can use the point-slope form of the equation of a line to find its equation. The point-slope form is given by:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) \) is the point the line passes through and \( m \) is the slope. Here, \( (x_1, y_1) = (-1, -3) \) and \( m = -5 \):
\[ y - (-3) = -5(x - (-1)) \]
Simplifying this, we have:
\[ y + 3 = -5(x + 1) \]
Distributing on the right side:
\[ y + 3 = -5x - 5 \]
Next, we isolate \( y \):
\[ y = -5x - 5 - 3 \]
\[ y = -5x - 8 \]
Thus, the equation of line \( k \) in slope-intercept form is:
\[ \boxed{y = -5x - 8} \]
1 answer