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Triangle [tex]$A B C$[/tex] is defined by the points [tex]$A(2,9), B(8,4)$[/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 :

GinnyAnswer
To find the equation of the line passing through point \( C \) and perpendicular to line segment \( \overline{AB} \), we start by determining the properties of line \( \overline{AB} \).

Given points \( A(2, 9) \) and \( B(8, 4) \):

1. Calculate the slope of \( \overline{AB} \):
[tex]\[ \text{slope of } \overline{AB} = \frac{B_y - A_y}{B_x - A_x} = \frac{4 - 9}{8 - 2} = \frac{-5}{6} = -\frac{5}{6} \][/tex]

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

3. The equation of a line in slope-intercept form is \( y = mx + b \). We need to find the y-intercept \( b \) of the line that passes through point \( C \) with coordinates \( (-3, -2) \) and slope \( 1.2 \):
[tex]\[ y = 1.2x + b \][/tex]
Substitute \( C(-3, -2) \) into the equation to solve for \( b \):
[tex]\[ -2 = 1.2(-3) + b \][/tex]
[tex]\[ -2 = -3.6 + b \][/tex]
[tex]\[ b = 1.6 \][/tex]

Therefore, the equation of the line passing through point \( C(-3, -2) \) and perpendicular to \( \overline{AB} \) is:
[tex]\[ y = 1.2x + 1.6 \][/tex]

So, completing the equation \( y = \square x + \square \):

[tex]\[ y = 1.2 \;x + 1.6 \][/tex]

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