Answer :
To write the equation of the line passing through point \( C(-3, -2) \) and perpendicular to line segment \( AB \) where \( A(2, 9) \) and \( B(8, 4) \), we first need to determine the slope of line \( AB \).
The slope \( m \) of line \( AB \) is calculated using the formula:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Using the coordinates of points \( A \) and \( B \):
[tex]\[ m_{AB} = \frac{4 - 9}{8 - 2} = \frac{-5}{6} \][/tex]
The slope of a line perpendicular to \( AB \) is the negative reciprocal of the slope of \( AB \). Hence, the slope of the perpendicular line is:
[tex]\[ m_{\text{perpendicular}} = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{5}{6}} = \frac{6}{5} \][/tex]
Now, we use the point-slope form of the line equation, which is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Here, \( (x_1, y_1) \) represents point \( C \), and \( m \) is the slope of the perpendicular line.
Substituting the values for point \( C(-3, -2) \) and the slope \( \frac{6}{5} \) into the point-slope form:
[tex]\[ y - (-2) = \frac{6}{5}(x - (-3)) \][/tex]
Simplifying, we have:
[tex]\[ y + 2 = \frac{6}{5}(x + 3) \][/tex]
Expanding and simplifying further to get the equation into slope-intercept form:
[tex]\[ y + 2 = \frac{6}{5}x + \frac{6}{5} \cdot 3 \][/tex]
[tex]\[ y + 2 = \frac{6}{5}x + \frac{18}{5} \][/tex]
Subtracting 2 (or \(\frac{10}{5}\)) from both sides to isolate \( y \):
[tex]\[ y = \frac{6}{5}x + \frac{18}{5} - \frac{10}{5} \][/tex]
[tex]\[ y = \frac{6}{5}x + \frac{8}{5} \][/tex]
Therefore, the equation of the line passing through point \( C \) and perpendicular to \( \overline{ AB } \) is:
[tex]\[ y = 1.2x + 1.6 \][/tex]
To complete the equation in the form \( y = \square x + \square \):
[tex]\[ y = \boxed{1.2} x + \boxed{1.6} \][/tex]
The slope \( m \) of line \( AB \) is calculated using the formula:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Using the coordinates of points \( A \) and \( B \):
[tex]\[ m_{AB} = \frac{4 - 9}{8 - 2} = \frac{-5}{6} \][/tex]
The slope of a line perpendicular to \( AB \) is the negative reciprocal of the slope of \( AB \). Hence, the slope of the perpendicular line is:
[tex]\[ m_{\text{perpendicular}} = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{5}{6}} = \frac{6}{5} \][/tex]
Now, we use the point-slope form of the line equation, which is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Here, \( (x_1, y_1) \) represents point \( C \), and \( m \) is the slope of the perpendicular line.
Substituting the values for point \( C(-3, -2) \) and the slope \( \frac{6}{5} \) into the point-slope form:
[tex]\[ y - (-2) = \frac{6}{5}(x - (-3)) \][/tex]
Simplifying, we have:
[tex]\[ y + 2 = \frac{6}{5}(x + 3) \][/tex]
Expanding and simplifying further to get the equation into slope-intercept form:
[tex]\[ y + 2 = \frac{6}{5}x + \frac{6}{5} \cdot 3 \][/tex]
[tex]\[ y + 2 = \frac{6}{5}x + \frac{18}{5} \][/tex]
Subtracting 2 (or \(\frac{10}{5}\)) from both sides to isolate \( y \):
[tex]\[ y = \frac{6}{5}x + \frac{18}{5} - \frac{10}{5} \][/tex]
[tex]\[ y = \frac{6}{5}x + \frac{8}{5} \][/tex]
Therefore, the equation of the line passing through point \( C \) and perpendicular to \( \overline{ AB } \) is:
[tex]\[ y = 1.2x + 1.6 \][/tex]
To complete the equation in the form \( y = \square x + \square \):
[tex]\[ y = \boxed{1.2} x + \boxed{1.6} \][/tex]
