It seems that your request contains problems related to mathematics, specifically geometry and vectors in a coordinate plane. Let's go through each problem step by step.
1. Find the Equation of the Line Segment AB
Given points A(-7, 6) and B(-1, 2), we can find the equation of the line segment between these two points.
Step 1: Find the slope ( m ) of the line.
[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 6}{-1 - (-7)} = \frac{-4}{6} = -\frac{2}{3}
]
Step 2: Use point-slope form to find the equation of the line.
The point-slope form is given by ( y - y_1 = m(x - x_1) ).
Using point A(-7, 6):
[
y - 6 = -\frac{2}{3}(x + 7)
]
Step 3: Rewrite the equation in slope-intercept form:
[
y - 6 = -\frac{2}{3}x - \frac{14}{3}
]
[
y = -\frac{2}{3}x + \frac{18}{3} - \frac{14}{3}
]
[
y = -\frac{2}{3}x + \frac{4}{3}
]
2. Find the Vector AB
To find the vector ( \overrightarrow{AB} ), subtract the coordinates of point A from point B.
[
\overrightarrow{AB} = B - A = (-1 - (-7), 2 - 6) = (6, -4)
]
3. Find the Length of the Segment MN
Given points M(6, -5) and N(3, -9), the distance (length) between two points can be found using the distance formula:
[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
]
[
d = \sqrt{(3 - 6)^2 + (-9 + 5)^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5
]
4. Find the Midpoint of MN
The midpoint ( M ) can be found using the midpoint formula:
[
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
]
[
M = \left( \frac{6 + 3}{2}, \frac{-5 - 9}{2} \right) = \left( \frac{9}{2}, \frac{-14}{2} \right) = \left( 4.5, -7 \right)
]
5. Find the center of the triangle ABC
To find the centroid ( C ) of triangle ABC with vertices A(-2, 2), B(5, -6), and C(6, 4), use:
[
C = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)
]
[
C = \left( \frac{-2 + 5 + 6}{3}, \frac{2 - 6 + 4}{3} \right) = \left( \frac{9}{3}, \frac{0}{3} \right) = (3, 0)
]
If you need further explanations or more problems solved, let me know!
