Answer :
To find the equation of line \( b \) that is perpendicular to line \( a \) and passes through the point \((-9, 2)\), we need to follow these steps:
1. Determine the Slope of Line \( a \):
The given line \( a \) has the equation \( y = \frac{1}{3}x + c \) (where \( c \) is a constant). The slope \( m \) of line \( a \) is \(\frac{1}{3}\).
2. Find the Slope of Line \( b \):
Two lines are perpendicular if the product of their slopes is \(-1\). Let the slope of line \( b \) be \( m_b \).
We have:
[tex]\[ \left(\frac{1}{3}\right) \cdot m_b = -1 \][/tex]
Solving for \( m_b \):
[tex]\[ m_b = -3 \][/tex]
3. Use the Point-Slope Form:
We now know that the slope of line \( b \) is \(-3\) and it passes through the point \((-9, 2)\). The point-slope form of the equation of a line is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Here, \( (x_1, y_1) = (-9, 2) \) and \( m = -3 \). Substituting these values in, we get:
[tex]\[ y - 2 = -3(x + 9) \][/tex]
Simplifying this equation:
[tex]\[ y - 2 = -3x - 27 \][/tex]
Adding 2 to both sides:
[tex]\[ y = -3x - 25 \][/tex]
Thus, the equation of the line \( b \) that is perpendicular to line \( a \) and passes through the point \((-9, 2)\) is \( y = -3x - 25 \).
Now, checking given options:
- \( \mathbf{A. \ y = \frac{1}{3}x + 5} \)
- \( \mathbf{B. \ y = \frac{1}{3}x - 1} \)
- \( \mathbf{C. \ y = \frac{1}{3}x - \frac{29}{3}} \)
- \( \mathbf{D. \ y = -\frac{1}{3}x + 1} \)
None of these options match [tex]\( y = -3x - 25 \)[/tex]. It seems there might be a mistake in the provided options, or perhaps they meant to ask for something slightly different. However, based on the algebraic work and the derived equation, the correct equation for line [tex]\( b \)[/tex] should be [tex]\( y = -3x - 25 \)[/tex].
1. Determine the Slope of Line \( a \):
The given line \( a \) has the equation \( y = \frac{1}{3}x + c \) (where \( c \) is a constant). The slope \( m \) of line \( a \) is \(\frac{1}{3}\).
2. Find the Slope of Line \( b \):
Two lines are perpendicular if the product of their slopes is \(-1\). Let the slope of line \( b \) be \( m_b \).
We have:
[tex]\[ \left(\frac{1}{3}\right) \cdot m_b = -1 \][/tex]
Solving for \( m_b \):
[tex]\[ m_b = -3 \][/tex]
3. Use the Point-Slope Form:
We now know that the slope of line \( b \) is \(-3\) and it passes through the point \((-9, 2)\). The point-slope form of the equation of a line is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Here, \( (x_1, y_1) = (-9, 2) \) and \( m = -3 \). Substituting these values in, we get:
[tex]\[ y - 2 = -3(x + 9) \][/tex]
Simplifying this equation:
[tex]\[ y - 2 = -3x - 27 \][/tex]
Adding 2 to both sides:
[tex]\[ y = -3x - 25 \][/tex]
Thus, the equation of the line \( b \) that is perpendicular to line \( a \) and passes through the point \((-9, 2)\) is \( y = -3x - 25 \).
Now, checking given options:
- \( \mathbf{A. \ y = \frac{1}{3}x + 5} \)
- \( \mathbf{B. \ y = \frac{1}{3}x - 1} \)
- \( \mathbf{C. \ y = \frac{1}{3}x - \frac{29}{3}} \)
- \( \mathbf{D. \ y = -\frac{1}{3}x + 1} \)
None of these options match [tex]\( y = -3x - 25 \)[/tex]. It seems there might be a mistake in the provided options, or perhaps they meant to ask for something slightly different. However, based on the algebraic work and the derived equation, the correct equation for line [tex]\( b \)[/tex] should be [tex]\( y = -3x - 25 \)[/tex].
