Answer :
To find the equation of the line passing through point \( C \) and perpendicular to line segment \( \overline{A B} \), follow these steps:
1. Find the slope of line segment \( \overline{A B} \):
The coordinates of point \( A \) are \( (2, 9) \) and those of point \( B \) are \( (8, 4) \).
The slope \( m_{AB} \) is calculated as:
[tex]\[ m_{AB} = \frac{B_y - A_y}{B_x - A_x} = \frac{4 - 9}{8 - 2} = \frac{-5}{6} \][/tex]
2. Find the slope of the line perpendicular to \( \overline{A B} \):
The slope \( m_{\perpendicular} \) of the perpendicular line is the negative reciprocal of \( m_{AB} \):
[tex]\[ m_{\perpendicular} = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{5}{6}} = \frac{6}{5} = 1.2 \][/tex]
3. Use the point-slope form to write the equation of the line passing through \( C(-3, -2) \):
The point-slope form of a line is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where \( (x_1, y_1) \) is point \( C \) and \( m \) is the slope. So:
[tex]\[ y - (-2) = 1.2(x - (-3)) \][/tex]
4. Convert to slope-intercept form \( y = mx + b \):
Simplify and solve for \( y \):
[tex]\[ y + 2 = 1.2(x + 3) \][/tex]
[tex]\[ y + 2 = 1.2x + 3.6 \][/tex]
[tex]\[ y = 1.2x + 3.6 - 2 \][/tex]
[tex]\[ y = 1.2x + 1.6 \][/tex]
So, the complete equation of the line passing through point \( C \) and perpendicular to \( \overline{A B} \) is:
[tex]\[ y = 1.2x + 1.6 \][/tex]
Thus, the equation in the given format is:
[tex]\[ y = \boxed{1.2}x + \boxed{1.6} \][/tex]
1. Find the slope of line segment \( \overline{A B} \):
The coordinates of point \( A \) are \( (2, 9) \) and those of point \( B \) are \( (8, 4) \).
The slope \( m_{AB} \) is calculated as:
[tex]\[ m_{AB} = \frac{B_y - A_y}{B_x - A_x} = \frac{4 - 9}{8 - 2} = \frac{-5}{6} \][/tex]
2. Find the slope of the line perpendicular to \( \overline{A B} \):
The slope \( m_{\perpendicular} \) of the perpendicular line is the negative reciprocal of \( m_{AB} \):
[tex]\[ m_{\perpendicular} = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{5}{6}} = \frac{6}{5} = 1.2 \][/tex]
3. Use the point-slope form to write the equation of the line passing through \( C(-3, -2) \):
The point-slope form of a line is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where \( (x_1, y_1) \) is point \( C \) and \( m \) is the slope. So:
[tex]\[ y - (-2) = 1.2(x - (-3)) \][/tex]
4. Convert to slope-intercept form \( y = mx + b \):
Simplify and solve for \( y \):
[tex]\[ y + 2 = 1.2(x + 3) \][/tex]
[tex]\[ y + 2 = 1.2x + 3.6 \][/tex]
[tex]\[ y = 1.2x + 3.6 - 2 \][/tex]
[tex]\[ y = 1.2x + 1.6 \][/tex]
So, the complete equation of the line passing through point \( C \) and perpendicular to \( \overline{A B} \) is:
[tex]\[ y = 1.2x + 1.6 \][/tex]
Thus, the equation in the given format is:
[tex]\[ y = \boxed{1.2}x + \boxed{1.6} \][/tex]
