V. (25 points) In the main orthogonal clue, mark points A (1,2) and B (3,4).

(10 points) 1. Find the length AB and then find the line equation (AB).

(05 points) 2. Find the coordinate / center of [AB].

(10 points) 3. Find the line equation (△) as the median line of [AB]. Then construct the

To solve the given problem, let's tackle each part step-by-step:

### 1. Finding the Length of \(AB\) and the Equation of Line \(AB\)

#### a. Length of \(AB\)

The length of a line segment between two points \((x_1, y_1)\) and \((x_2, y_2)\) can be calculated using the distance formula:

\[
AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

For points \(A(1, 2)\) and \(B(3, 4)\):

\[
AB = \sqrt{(3 - 1)^2 + (4 - 2)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
\]

So, the length of \(AB\) is \(2\sqrt{2}\).

#### b. Equation of Line \(AB\)

The equation of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) can be found using the slope-intercept form \(y = mx + b\), where the slope \(m\) is:

\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

For points \(A(1, 2)\) and \(B(3, 4)\):

\[
m = \frac{4 - 2}{3 - 1} = \frac{2}{2} = 1
\]

The slope of the line \(AB\) is 1. Now, using the point-slope form of the equation \(y - y_1 = m(x - x_1)\):

\[
y - 2 = 1(x - 1) \implies y - 2 = x - 1 \implies y = x + 1
\]

Thus, the equation of line \(AB\) is:

\[
y = x + 1
\]

### 2. Finding the Midpoint (Center) of \(AB\)

The midpoint \(M\) of a line segment with endpoints \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\[
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]

For points \(A(1, 2)\) and \(B(3, 4)\):

\[
M = \left( \frac{1 + 3}{2}, \frac{2 + 4}{2} \right) = \left( \frac{4}{2}, \frac{6}{2} \right) = (2, 3)
\]

So, the midpoint (center) of \(AB\) is \((2, 3)\).

### 3. Finding the Equation of the Median Line from a Triangle’s Vertex

#### a. Median Line Equation

A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. Since we are constructing a median line (denoted as \(\Delta\)) for the line segment \(AB\), let's assume the triangle has a vertex at a different point \(C(x, y)\) that is not collinear with \(A\) and \(B\).

Let's construct a median from this new vertex \(C\) to the midpoint \(M(2, 3)\).

For simplicity, let's choose \(C(0, 0)\) (the origin) as a vertex, which makes it easy to calculate the slope of the median line from \(C(0, 0)\) to \(M(2, 3)\):

\[
m = \frac{3 - 0}{2 - 0} = \frac{3}{2}
\]

The slope of the median line \(\Delta\) is \(\frac{3}{2}\). Using the point-slope form of a line equation from \(C(0,0)\):

\[
y - 0 = \frac{3}{2}(x - 0) \implies y = \frac{3}{2}x
\]

Thus, the equation of the median line \(\Delta\) is:

\[
y = \frac{3}{2}x
\]

### Conclusion

1. **Length of \(AB\)**: \(2\sqrt{2}\)
2. **Equation of line \(AB\)**: \(y = x + 1\)
3. **Midpoint of \(AB\)**: \((2, 3)\)
4. **Equation of the median line \(\Delta\) from vertex \(C(0, 0)\)**: \(y = \frac{3}{2}x\)

Would you like any more details on these steps, or do you have any questions?

### Related Questions:

1. What is the perpendicular bisector of line segment \(AB\)?
2. How do you find the equation of a line parallel to \(AB\) that passes through a different point?
3. What is the area of the triangle formed by points \(A\), \(B\), and \(C(0, 0)\)?
4. How do you find the equation of a line perpendicular to \(AB\) that passes through the midpoint \(M(2, 3)\)?
5. How would the equations change if point \(C\) was at a different coordinate, say \(C(4, 5)\)?

**Tip:** Remember to check if two points are collinear before using them to construct geometrical elements like medians or altitudes in a triangle!

Explore how to solve a coordinate geometry problem involving finding the length of a line segment and equations of lines. Learn step-by-step with examples of distance formulas and equations of lines. Suitable for high school students.

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