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Triangle [tex]A B C[/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{A B}[/tex].

[tex]\[ y = \boxed{\ \ } x + \boxed{\ \ } \][/tex]


Sagot :

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

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

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

3. Determine the equation of the line:
- A line equation in point-slope form through a point [tex]\((x_1, y_1)\)[/tex] with a slope [tex]\( m \)[/tex] is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
- Here, the point is [tex]\( C(-3, -2) \)[/tex] and the slope is [tex]\( 1.2 \)[/tex]:
[tex]\[ y - (-2) = 1.2(x - (-3)) \][/tex]
[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]

Therefore, the completed equation for the line passing through point [tex]\( C \)[/tex] and perpendicular to [tex]\( \overline{AB} \)[/tex] is:
[tex]\[ y = 1.2x + 1.6 \][/tex]

So, the correct answer to fill in the boxes is:
[tex]\[ y = \boxed{1.2} x + \boxed{1.6} \][/tex]
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