To find the perpendicular bisector of segment \( HI \) in standard form, you first need to find the midpoint of \( HI \) and then the slope of the perpendicular bisector.
The midpoint \( M \) of \( HI \) can be found using the midpoint formula:
\[ M_x = \frac{H_x + I_x}{2}, M_y = \frac{H_y + I_y}{2} \]
\[ M_x = \frac{-4 + 2}{2}, M_y = \frac{2 + 4}{2} \]
\[ M_x = -1, M_y = 3 \]
So, the midpoint \( M \) is at (-1, 3).
The slope of \( HI \) is:
\[ m_{HI} = \frac{I_y - H_y}{I_x - H_x} \]
\[ m_{HI} = \frac{4 - 2}{2 - (-4)} \]
\[ m_{HI} = \frac{2}{6} \]
\[ m_{HI} = \frac{1}{3} \]
The slope of the perpendicular bisector \( m_{perpendicular} \) is the negative reciprocal of \( m_{HI} \):
\[ m_{perpendicular} = -\frac{1}{m_{HI}} \]
\[ m_{perpendicular} = -\frac{1}{\frac{1}{3}} \]
\[ m_{perpendicular} = -3 \]
Using the point-slope form of the line equation with midpoint \( M \) and \( m_{perpendicular} \), you get:
\[ y - M_y = m_{perpendicular}(x - M_x) \]
\[ y - 3 = -3(x - (-1)) \]
\[ y - 3 = -3(x + 1) \]
\[ y - 3 = -3x - 3 \]
To convert to standard form \( Ax + By = C \), you need to rearrange terms:
\[ 3x + y = 0 \]
This is the equation of the perpendicular bisector of segment \( HI \) in standard form.
Short Answer
Your teacher will grade your responses for questions 7–9 to ensure that you receive proper credit for your answers.
Given ΔGHI with G(4, –3), H(–4, 2), and I(2, 4), find the perpendicular bisector of segment H I in standard form.
(2 points)
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