Triangle [tex]\( ABC \)[/tex] is defined by the points [tex]\( A(2,9) \)[/tex], [tex]\( B(8,4) \)[/tex], and [tex]\( C(-3,-2) \)[/tex].

Complete the following equation for a line passing through point [tex]\( C \)[/tex] and perpendicular to [tex]\( \overline{AB} \)[/tex]:

[tex]\[ y = \square x + \square \][/tex]



Answer :

To find the equation of the line passing through point [tex]\( C(-3, -2) \)[/tex] and perpendicular to line segment [tex]\(\overline{AB}\)[/tex], we follow these steps:

1. Calculate the slope of line segment [tex]\(\overline{AB}\)[/tex].
2. Determine the slope of the line perpendicular to [tex]\(\overline{AB}\)[/tex] (which will be the negative reciprocal of the slope of [tex]\(\overline{AB}\)[/tex]).
3. Use the point-slope form of the line equation to find the equation of the line passing through point [tex]\( C \)[/tex].

Step-by-step details:

1. Calculate the slope of [tex]\(\overline{AB}\)[/tex]:
- The formula for the slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
- Here, we are given points [tex]\(A(2, 9)\)[/tex] and [tex]\(B(8, 4)\)[/tex]. Therefore,
[tex]\[ \text{slope}_{AB} = \frac{4 - 9}{8 - 2} = \frac{-5}{6} \][/tex]

2. Determine the slope of the perpendicular line:
- The slope of the line perpendicular to [tex]\(\overline{AB}\)[/tex] is the negative reciprocal of the slope of [tex]\(\overline{AB}\)[/tex]:
[tex]\[ \text{perpendicular slope} = -\frac{1}{\text{slope}_{AB}} = -\frac{1}{-\frac{5}{6}} = \frac{6}{5} = 1.2 \][/tex]

3. Find the equation of the line passing through [tex]\( C(-3, -2) \)[/tex]:
- Use the point-slope form of the line equation [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\( m \)[/tex] is the slope and [tex]\((x_1, y_1)\)[/tex] is a point on the line.
- Substitute the slope [tex]\( 1.2 \)[/tex] and point [tex]\( C(-3, -2) \)[/tex]:
[tex]\[ y - (-2) = 1.2(x - (-3)) \][/tex]
Simplifying this:
[tex]\[ y + 2 = 1.2(x + 3) \][/tex]
[tex]\[ y + 2 = 1.2x + 3.6 \][/tex]
Subtract 2 from both sides:
[tex]\[ y = 1.2x + 1.6 \][/tex]

Thus, the equation of the line in slope-intercept form [tex]\( y = mx + b \)[/tex] is:
[tex]\[ y = 1.2x + 1.6 \][/tex]

Therefore, the completed equation is:
[tex]\[ y = \boxed{1.2} x + \boxed{1.6} \][/tex]

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