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!