Math Problem Statement

Find the distance between the points C(-1, 8) and D(5, 9). Find the coordinates of the midpoint M of the segment joining H(2, -1) and K(3, 8). The coordinates of the midpoint of segment AC are (2, -3). The coordinates of A are (0, 5). Find the coordinates of C.

Solution

Let's analyze the image and solve the three math problems based on the provided formulas.

17. Find the distance between the points C(1,8)C(-1, 8) and D(5,9)D(5, 9). Round your answer to the nearest tenth.

The distance formula is given as: d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Where:

  • x1=1x_1 = -1, y1=8y_1 = 8 (coordinates of CC),
  • x2=5x_2 = 5, y2=9y_2 = 9 (coordinates of DD).

Substituting the values:

d=(5(1))2+(98)2=(5+1)2+(1)2=62+12=36+1=37d = \sqrt{(5 - (-1))^2 + (9 - 8)^2} = \sqrt{(5 + 1)^2 + (1)^2} = \sqrt{6^2 + 1^2} = \sqrt{36 + 1} = \sqrt{37}

Now, calculating 37\sqrt{37} to the nearest tenth: d6.1d \approx 6.1

Thus, the distance between points CC and DD is approximately 6.1 units.


18. Find the coordinates of the midpoint MM of the segment joining H(2,1)H(2, -1) and K(3,8)K(3, 8).

The midpoint formula is: M=(x1+x22,y1+y22)M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)

Where:

  • x1=2x_1 = 2, y1=1y_1 = -1 (coordinates of HH),
  • x2=3x_2 = 3, y2=8y_2 = 8 (coordinates of KK).

Substituting the values:

M=(2+32,1+82)=(52,72)=(2.5,3.5)M = \left( \frac{2 + 3}{2}, \frac{-1 + 8}{2} \right) = \left( \frac{5}{2}, \frac{7}{2} \right) = (2.5, 3.5)

Thus, the midpoint MM is (2.5,3.5)(2.5, 3.5).


19. The coordinates of the midpoint of segment ACAC are (2,3)(2, -3). The coordinates of AA are (0,5)(0, 5). Find the coordinates of CC.

The midpoint formula is: M=(x1+x22,y1+y22)M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) Let the coordinates of CC be (x2,y2)(x_2, y_2), and we know:

  • M=(2,3)M = (2, -3),
  • A=(0,5)A = (0, 5).

Using the midpoint formula, we can set up two equations: 2=0+x22(for the x-coordinates)2 = \frac{0 + x_2}{2} \quad \text{(for the x-coordinates)} 3=5+y22(for the y-coordinates)-3 = \frac{5 + y_2}{2} \quad \text{(for the y-coordinates)}

Solve each equation:

  1. 2=0+x222 = \frac{0 + x_2}{2} x2=4x_2 = 4

  2. 3=5+y22-3 = \frac{5 + y_2}{2} 6=5+y2y2=65=11-6 = 5 + y_2 \quad \Rightarrow \quad y_2 = -6 - 5 = -11

Thus, the coordinates of point CC are (4,11)(4, -11).


Let me know if you'd like more details or if you have any further questions.

Here are five related questions to deepen your understanding:

  1. How would the distance formula change in 3D space?
  2. Can you apply the midpoint formula to find the center of a triangle?
  3. What are the properties of line segments that share the same midpoint?
  4. How would you calculate the slope of a line given two points?
  5. How would the distance between two points change if the coordinate system was non-Cartesian?

Tip: Always visualize the points on a graph to better understand their spatial relationships.

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Distance Formula
Midpoint Formula

Formulas

Distance formula: d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Midpoint formula: M = ((x1 + x2)/2, (y1 + y2)/2)

Theorems

Pythagorean Theorem (underlying concept for the distance formula)

Suitable Grade Level

Grades 8-10