`A B C D` is a square `A-=(1,2),B-=(3,-4)dot` If line `C D` passes through `(3,8),` then the midpoint of `C D` is (a) (2, 6) (b) `(6,2)` (c) `(2,5)` (d) `((28)/5,1/5)`

Options

A. (2,6)
B. (6,2)
C. (2,5)
D. (28/5,1/5)

Explanation

To solve the problem step by step, we will find the midpoint of line segment CD, given that AB is a square and the coordinates of points A and B are provided.

### Step 1: Identify the coordinates of points A and B
- A = (1, 2)
- B = (3, -4)

### Step 2: Calculate the slope of line AB
The slope (m) of line AB can be calculated using the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting the coordinates of A and B:
\[ m_{AB} = \frac{-4 - 2}{3 - 1} = \frac{-6}{2} = -3 \]

### Step 3: Determine the slope of line CD
Since AB is parallel to CD, the slope of line CD will be the same as that of AB:
\[ m_{CD} = -3 \]

### Step 4: Write the equation of line CD
Line CD passes through the point (3, 8). We can use the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting the slope and the point (3, 8):
\[ y - 8 = -3(x - 3) \]
Expanding this:
\[ y - 8 = -3x + 9 \]
\[ y = -3x + 17 \]

### Step 5: Find the coordinates of point C
Since ABCD is a square, we need to find the coordinates of point C. The slope of line BC (which is perpendicular to line AB) can be calculated as:
\[ m_{BC} = -\frac{1}{m_{AB}} = \frac{1}{3} \]

### Step 6: Write the equation of line BC
Using point B (3, -4) and slope \( \frac{1}{3} \):
\[ y - (-4) = \frac{1}{3}(x - 3) \]
Expanding this:
\[ y + 4 = \frac{1}{3}x - 1 \]
\[ y = \frac{1}{3}x - 5 \]

### Step 7: Find the intersection of lines CD and BC
To find the coordinates of point C, we need to solve the equations of lines CD and BC:
1. \( y = -3x + 17 \)
2. \( y = \frac{1}{3}x - 5 \)

Setting these equations equal to each other:
\[ -3x + 17 = \frac{1}{3}x - 5 \]
Multiplying through by 3 to eliminate the fraction:
\[ -9x + 51 = x - 15 \]
Combining like terms:
\[ -10x = -66 \]
\[ x = \frac{66}{10} = \frac{33}{5} \]

### Step 8: Substitute x back to find y
Substituting \( x = \frac{33}{5} \) into the equation of line CD:
\[ y = -3\left(\frac{33}{5}\right) + 17 \]
\[ y = -\frac{99}{5} + \frac{85}{5} = -\frac{14}{5} \]

Thus, the coordinates of point C are:
\[ C = \left(\frac{33}{5}, -\frac{14}{5}\right) \]

### Step 9: Find the midpoint of CD
The midpoint M of line segment CD can be calculated as:
\[ M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
Substituting the coordinates of points C and D (D is not given, but we can assume D has coordinates that maintain the square):
Assuming D is symmetric to C across the line AB, we can find the coordinates of D as well.

### Final Step: Calculate the midpoint
After determining the coordinates of D, we can substitute them into the midpoint formula to find the midpoint of line segment CD.