https://file.aiquickdraw.com/mathbot/file/1728881282278y5l61dza.jpg

Let's break down and solve the questions from the image.

### a) **Midpoint and distance between the points (-2, 5) and (3, 7):**

1. **Midpoint Formula**: 
\[
\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]
Given points: \((-2, 5)\) and \((3, 7)\)
\[
\text{Midpoint} = \left( \frac{-2 + 3}{2}, \frac{5 + 7}{2} \right) = \left( \frac{1}{2}, 6 \right)
\]

2. **Distance Formula**:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
\[
d = \sqrt{(3 - (-2))^2 + (7 - 5)^2} = \sqrt{(3 + 2)^2 + (2)^2} = \sqrt{25 + 4} = \sqrt{29}
\]
So, the distance is \(\sqrt{29}\).

---

### b) **Equation of the line through (2, 8) and (-1, 4):**

1. **Slope Formula**:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]
\[
m = \frac{4 - 8}{-1 - 2} = \frac{-4}{-3} = \frac{4}{3}
\]

2. **Point-Slope Form**:
The equation of the line is:
\[
y - y_1 = m(x - x_1)
\]
Substitute \(m = \frac{4}{3}\) and the point \((2, 8)\):
\[
y - 8 = \frac{4}{3}(x - 2)
\]
Simplifying to slope-intercept form:
\[
y - 8 = \frac{4}{3}x - \frac{8}{3}
\]
\[
y = \frac{4}{3}x + 8 - \frac{8}{3} = \frac{4}{3}x + \frac{24}{3} - \frac{8}{3}
\]
\[
y = \frac{4}{3}x + \frac{16}{3}
\]

---

### c) **Radius and center of the circle given by the equation \((x - 1)^2 + (y + 3)^2 = 36\):**

1. **Standard form of a circle**: 
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Where \((h, k)\) is the center, and \(r\) is the radius.

- From the equation \((x - 1)^2 + (y + 3)^2 = 36\), we can see that:
  - The center is \((1, -3)\)
  - The radius is \(\sqrt{36} = 6\)

---

### d) **Standard equation for a circle with a diameter having endpoints \((5, -2)\) and \((-3, 4)\):**

1. **Midpoint of the diameter (center of the circle)**:
\[
\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{5 + (-3)}{2}, \frac{-2 + 4}{2} \right) = \left( \frac{2}{2}, \frac{2}{2} \right) = (1, 1)
\]
So, the center of the circle is \((1, 1)\).

2. **Radius (half the distance between the endpoints)**:
We already have the formula for distance:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(-3 - 5)^2 + (4 + 2)^2} = \sqrt{(-8)^2 + (6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10
\]
The radius is half of the diameter:
\[
r = \frac{10}{2} = 5
\]

3. **Standard equation**:
The standard equation of the circle is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Substituting \(h = 1\), \(k = 1\), and \(r = 5\):
\[
(x - 1)^2 + (y - 1)^2 = 25
\]

---

Would you like further details on any of these parts? Here are five related questions:

1. How would you find the equation of a circle given different types of information?
2. Can you generalize the distance formula for three-dimensional space?
3. How can you derive the equation of a line in parametric form?
4. What is the significance of the slope in different geometrical interpretations?
5. How can midpoint and distance formulas be extended to more complex geometric shapes?

**Tip:** Always check if equations are in their simplest form when solving geometry problems.

Determine:
 a) The midpoint and distance between the points (-2,5) and (3,7)
 b) The equation of the line through (2,8) and (-1,4)
 c) The radius and center of the circle (x-1)^2 + (y + 3)^2 = 36
 d) The standard equation for a circle having a diameter with endpoints (x1, y1) = (5, -2) and (x2, y2) = (-3, 4)

This problem covers key concepts of coordinate geometry, including finding the midpoint and distance between two points, determining the equation of a line through two points, and finding the standard form of a circle's equation. Explore how to calculate the radius and center of a circle, as well as the equation of a circle with given endpoints.

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