Answer :
To determine if line segment [tex]\(\overline{GH}\)[/tex] is the perpendicular bisector of line segment [tex]\(\overline{MN}\)[/tex], we need to verify two key conditions:
1. [tex]\(\overline{GH}\)[/tex] is perpendicular to [tex]\(\overline{MN}\)[/tex].
2. [tex]\(\overline{GH}\)[/tex] passes through the midpoint of [tex]\(\overline{MN}\)[/tex].
### Step 1: Midpoint of [tex]\(\overline{MN}\)[/tex]
The midpoint [tex]\(M_MN\)[/tex] of line segment [tex]\(\overline{MN}\)[/tex] with endpoints [tex]\(M (-2,-2)\)[/tex] and [tex]\(N (2,8)\)[/tex] can be calculated using the midpoint formula:
[tex]\[ \text{Midpoint} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \][/tex]
Substituting the coordinates of [tex]\(M\)[/tex] and [tex]\(N\)[/tex]:
[tex]\[ \text{Midpoint} = \left(\frac{-2 + 2}{2}, \frac{-2 + 8}{2}\right) = (0.0, 3.0) \][/tex]
### Step 2: Midpoint of [tex]\(\overline{GH}\)[/tex]
Next, we need to determine the midpoint of line segment [tex]\(\overline{GH}\)[/tex] with endpoints [tex]\(G (-5,5)\)[/tex] and [tex]\(H (5,1)\)[/tex]. Using the midpoint formula:
[tex]\[ \text{Midpoint} = \left(\frac{-5 + 5}{2}, \frac{5 + 1}{2}\right) = (0.0, 3.0) \][/tex]
### Step 3: Slopes of [tex]\(\overline{GH}\)[/tex] and [tex]\(\overline{MN}\)[/tex]
The slope of a line segment with endpoints [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ \text{Slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
#### Slope of [tex]\(\overline{GH}\)[/tex]
For endpoints [tex]\(G (-5,5)\)[/tex] and [tex]\(H (5,1)\)[/tex]:
[tex]\[ \text{Slope of GH} = \frac{1 - 5}{5 - (-5)} = \frac{-4}{10} = -0.4 \][/tex]
#### Slope of [tex]\(\overline{MN}\)[/tex]
For endpoints [tex]\(M (-2,-2)\)[/tex] and [tex]\(N (2,8)\)[/tex]:
[tex]\[ \text{Slope of MN} = \frac{8 - (-2)}{2 - (-2)} = \frac{10}{4} = 2.5 \][/tex]
### Step 4: Perpendicular Slopes
Two lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex]. Alternatively, if the slope of one line is [tex]\(m\)[/tex], the slope of the line perpendicular to it is [tex]\(-\frac{1}{m}\)[/tex].
The negative reciprocal of the slope of [tex]\(\overline{MN}\)[/tex] is:
[tex]\[ \text{Negative reciprocal of slope of MN} = -\frac{1}{2.5} = -0.4 \][/tex]
### Step 5: Check Perpendicularity
Now, we check if the slope of [tex]\(\overline{GH}\)[/tex] is equal to the negative reciprocal of the slope of [tex]\(\overline{MN}\)[/tex]:
[tex]\[ -0.4 = -0.4 \quad \text{(True)} \][/tex]
Thus, [tex]\(\overline{GH}\)[/tex] is perpendicular to [tex]\(\overline{MN}\)[/tex].
### Step 6: Check if [tex]\(\overline{GH}\)[/tex] Bisects [tex]\(\overline{MN}\)[/tex]
We have already determined that the midpoint of [tex]\(\overline{GH}\)[/tex] is [tex]\((0.0, 3.0)\)[/tex]. Since this matches the midpoint of [tex]\(\overline{MN}\)[/tex], [tex]\(\overline{GH}\)[/tex] bisects [tex]\(\overline{MN}\)[/tex].
### Conclusion
Since [tex]\(\overline{GH}\)[/tex] is both perpendicular to and bisects [tex]\(\overline{MN}\)[/tex], we conclude that [tex]\(\overline{GH}\)[/tex] is indeed the perpendicular bisector of [tex]\(\overline{MN}\)[/tex].
Thus, [tex]\(\overline{GH}\)[/tex] is the perpendicular bisector of [tex]\(\overline{MN}\)[/tex].
1. [tex]\(\overline{GH}\)[/tex] is perpendicular to [tex]\(\overline{MN}\)[/tex].
2. [tex]\(\overline{GH}\)[/tex] passes through the midpoint of [tex]\(\overline{MN}\)[/tex].
### Step 1: Midpoint of [tex]\(\overline{MN}\)[/tex]
The midpoint [tex]\(M_MN\)[/tex] of line segment [tex]\(\overline{MN}\)[/tex] with endpoints [tex]\(M (-2,-2)\)[/tex] and [tex]\(N (2,8)\)[/tex] can be calculated using the midpoint formula:
[tex]\[ \text{Midpoint} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \][/tex]
Substituting the coordinates of [tex]\(M\)[/tex] and [tex]\(N\)[/tex]:
[tex]\[ \text{Midpoint} = \left(\frac{-2 + 2}{2}, \frac{-2 + 8}{2}\right) = (0.0, 3.0) \][/tex]
### Step 2: Midpoint of [tex]\(\overline{GH}\)[/tex]
Next, we need to determine the midpoint of line segment [tex]\(\overline{GH}\)[/tex] with endpoints [tex]\(G (-5,5)\)[/tex] and [tex]\(H (5,1)\)[/tex]. Using the midpoint formula:
[tex]\[ \text{Midpoint} = \left(\frac{-5 + 5}{2}, \frac{5 + 1}{2}\right) = (0.0, 3.0) \][/tex]
### Step 3: Slopes of [tex]\(\overline{GH}\)[/tex] and [tex]\(\overline{MN}\)[/tex]
The slope of a line segment with endpoints [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ \text{Slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
#### Slope of [tex]\(\overline{GH}\)[/tex]
For endpoints [tex]\(G (-5,5)\)[/tex] and [tex]\(H (5,1)\)[/tex]:
[tex]\[ \text{Slope of GH} = \frac{1 - 5}{5 - (-5)} = \frac{-4}{10} = -0.4 \][/tex]
#### Slope of [tex]\(\overline{MN}\)[/tex]
For endpoints [tex]\(M (-2,-2)\)[/tex] and [tex]\(N (2,8)\)[/tex]:
[tex]\[ \text{Slope of MN} = \frac{8 - (-2)}{2 - (-2)} = \frac{10}{4} = 2.5 \][/tex]
### Step 4: Perpendicular Slopes
Two lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex]. Alternatively, if the slope of one line is [tex]\(m\)[/tex], the slope of the line perpendicular to it is [tex]\(-\frac{1}{m}\)[/tex].
The negative reciprocal of the slope of [tex]\(\overline{MN}\)[/tex] is:
[tex]\[ \text{Negative reciprocal of slope of MN} = -\frac{1}{2.5} = -0.4 \][/tex]
### Step 5: Check Perpendicularity
Now, we check if the slope of [tex]\(\overline{GH}\)[/tex] is equal to the negative reciprocal of the slope of [tex]\(\overline{MN}\)[/tex]:
[tex]\[ -0.4 = -0.4 \quad \text{(True)} \][/tex]
Thus, [tex]\(\overline{GH}\)[/tex] is perpendicular to [tex]\(\overline{MN}\)[/tex].
### Step 6: Check if [tex]\(\overline{GH}\)[/tex] Bisects [tex]\(\overline{MN}\)[/tex]
We have already determined that the midpoint of [tex]\(\overline{GH}\)[/tex] is [tex]\((0.0, 3.0)\)[/tex]. Since this matches the midpoint of [tex]\(\overline{MN}\)[/tex], [tex]\(\overline{GH}\)[/tex] bisects [tex]\(\overline{MN}\)[/tex].
### Conclusion
Since [tex]\(\overline{GH}\)[/tex] is both perpendicular to and bisects [tex]\(\overline{MN}\)[/tex], we conclude that [tex]\(\overline{GH}\)[/tex] is indeed the perpendicular bisector of [tex]\(\overline{MN}\)[/tex].
Thus, [tex]\(\overline{GH}\)[/tex] is the perpendicular bisector of [tex]\(\overline{MN}\)[/tex].
