The problem provides the coordinates of three points $A$, $B$, and $C$ in an orthonormal coordinate system. It asks several questions, including finding the coordinates of vector $AB$, calculating the length of $AB$, finding the midpoint of $AB$, verifying the equation of line $AB$, checking if point $C$ lies on line $AB$, and finding the equation of a line perpendicular to $AB$ passing through $C$, and also the equation of a line parallel to $AB$ and passing through $E(2,5)$.

1. Problem Description

The problem provides the coordinates of three points $A$, $B$, and $C$ in an orthonormal coordinate system. It asks several questions, including finding the coordinates of vector $AB$, calculating the length of $AB$, finding the midpoint of $AB$, verifying the equation of line $AB$, checking if point $C$ lies on line $AB$, and finding the equation of a line perpendicular to $AB$ passing through $C$, and also the equation of a line parallel to $AB$ and passing through $E(2,5)$.

2. Solution Steps

2. Coordinates of Vector AB:

The coordinates of vector $AB$ are given by $(x_B - x_A, y_B - y_A)$.

Given $A(1,3)$ and $B(2,5)$, we have:

$AB = (2-1, 5-3) = (1, 2)$

3. Calculate AB:

The length of $AB$ is given by the distance formula:

$AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$

$AB = \sqrt{(2-1)^2 + (5-3)^2} = \sqrt{1^2 + 2^2} = \sqrt{1+4} = \sqrt{5}$

4. Coordinates of Midpoint of AB:

The midpoint $M$ of segment $AB$ has coordinates $(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2})$.

$M = (\frac{1+2}{2}, \frac{3+5}{2}) = (\frac{3}{2}, \frac{8}{2}) = (\frac{3}{2}, 4)$

5. Show that the equation of line AB is y = 2x + 1:

We have the points $A(1,3)$ and $B(2,5)$. The slope $m$ is $\frac{y_B - y_A}{x_B - x_A} = \frac{5-3}{2-1} = \frac{2}{1} = 2$.

Using the point-slope form, $y - y_A = m(x - x_A)$, we have $y - 3 = 2(x - 1)$.

$y - 3 = 2x - 2$, so $y = 2x + 1$.

6. Does point C belong to line AB?

We have $C(-2,1)$. Substitute $x=-2$ into the equation $y = 2x + 1$:

$y = 2(-2) + 1 = -4 + 1 = -3$.

Since the $y$-coordinate of $C$ is 1, and the $y$-coordinate on the line $AB$ when $x=-2$ is -3, $C$ does not belong to the line $AB$.

7. Equation of the line (D) perpendicular to (AB) passing through C:

The slope of line $AB$ is

2. The slope of a line perpendicular to $AB$ is $-\frac{1}{2}$.

Using the point-slope form with $C(-2,1)$ and slope $-\frac{1}{2}$:

$y - 1 = -\frac{1}{2}(x - (-2)) = -\frac{1}{2}(x + 2)$

$y - 1 = -\frac{1}{2}x - 1$

$y = -\frac{1}{2}x$

8. Equation of the line (A) parallel to (AB) passing through E(2,5):

Since the line is parallel to $AB$, its slope is the same as $AB$, which is

2. Using the point-slope form with $E(2,5)$ and slope 2:

$y - 5 = 2(x - 2)$

$y - 5 = 2x - 4$

$y = 2x + 1$

3. Final Answer

2. $AB = (1, 2)$

3. $AB = \sqrt{5}$

4. $M = (\frac{3}{2}, 4)$

5. $y = 2x + 1$

6. No, point C does not belong to line AB.

7. $y = -\frac{1}{2}x$

8. $y = 2x + 1$