Let's solve each problem step by step and draw the required figures.

Problem 1: Midpoints of the Sides of a Triangle

Vertices of the triangle: (1, 7), (6, -1), and (0, 3).

To find the midpoints of the sides of the triangle, we'll use the midpoint formula:

$\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Calculations:

1. Midpoint of side between (1, 7) and (6, -1):

   $\left( \frac{1 + 6}{2}, \frac{7 + (-1)}{2} \right) = \left( \frac{7}{2}, \frac{6}{2} \right) = (3.5, 3)$

2. Midpoint of side between (6, -1) and (0, 3):

   $\left( \frac{6 + 0}{2}, \frac{-1 + 3}{2} \right) = \left( \frac{6}{2}, \frac{2}{2} \right) = (3, 1)$

3. Midpoint of side between (0, 3) and (1, 7):

   $\left( \frac{0 + 1}{2}, \frac{3 + 7}{2} \right) = \left( \frac{1}{2}, \frac{10}{2} \right) = (0.5, 5)$

Midpoints: (3.5, 3), (3, 1), and (0.5, 5).

Problem 2: Extended Line Segment

Segment endpoints: (-1, 4) and (2, 2).

The length of the line segment is extended three times its own length.

1. Length of the segment: Using the distance formula:

   $\text{Length} = \sqrt{(2 - (-1))^2 + (2 - 4)^2} = \sqrt{(2 + 1)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}$

2. Direction vector from (-1, 4) to (2, 2):

   $(2 - (-1), 2 - 4) = (3, -2)$

3. Extending the segment: Scale the direction vector by 3:

   $(3 \times 3, 3 \times -2) = (9, -6)$

4. Finding the terminal point:

   $(2, 2) + (9, -6) = (11, -4)$

Terminal Point: (11, -4).

Problem 3: Dividing the Segment into Two Segments

Endpoints: (2, -4) and (2, 3).

Division ratio: One segment is $\frac{3}{4}$ of the other.

1. Total length of the segment:

   $\text{Length} = |3 - (-4)| = 7$

2. Finding the ratio: Let $x$ be the length of one segment and $\frac{3}{4}x$ be the other.

   $x + \frac{3}{4}x = 7 \implies \frac{7}{4}x = 7 \implies x = 4$

   The two segments are 4 and 3 units long.

3. Point of division:

   From (2, -4) to (2, 3), the point is 4 units above (2, -4):

   $(-4 + 4) = 0 \implies (2, 0)$

Point of Division: (2, 0).

Problem 4: Finding the Other End of a Diameter of a Circle

Center of circle: (4, -2).

One endpoint of the diameter: (7, 2).

1. Midpoint (center) formula:

   $\text{Center} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

2. Setting up equations:

   $4 = \frac{7 + x}{2}, \quad -2 = \frac{2 + y}{2}$

   Solving:

   $8 = 7 + x \implies x = 1, \quad -4 = 2 + y \implies y = -6$

Other Endpoint: (1, -6).

Problem 5: Height of the Light Bulb

Person's height: 6 ft.

Ratio: $\frac{4}{10}$ of the distance from the pole to the tip of the shadow.

Using similar triangles:

1. Let $x$ be the height of the pole, and $d$ be the distance of the shadow.

2. From the ratio:

   $6 = \frac{4}{10} \times x$

   Solving for $x$:

   $6 \times \frac{10}{4} = x \implies x = 15$

Height of the Light Bulb: 15 ft.

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Would you like a detailed explanation or any drawings for these solutions? Here are some related questions:

1. How do you find the centroid of a triangle given its vertices?
2. What is the formula for finding the length of a segment given two points?
3. How do you calculate the slope of a line between two points?
4. How can you determine if three points are collinear?
5. What is the equation of a circle given its center and radius?

Tip: When solving geometric problems, drawing figures can help visualize and simplify the problem!