The problem is to perform various geometric calculations and determinations based on the given points $A(1,3)$, $B(2,5)$, and $C(-2,1)$ in an orthonormal coordinate system. The specific tasks include placing the points, finding the coordinates of vector $\vec{AB}$, calculating the distance $AB$, finding the coordinates of the midpoint $I$ of segment $[AB]$, determining the equation of line $(AB)$, checking if $C$ lies on $(AB)$, finding the equation of the line perpendicular to $(AB)$ through $C$, and finding the equation of the line parallel to $(AB)$ through $E(2,5)$.

1. Problem Description

The problem is to perform various geometric calculations and determinations based on the given points $A(1,3)$, $B(2,5)$, and $C(-2,1)$ in an orthonormal coordinate system. The specific tasks include placing the points, finding the coordinates of vector $\vec{AB}$, calculating the distance $AB$, finding the coordinates of the midpoint $I$ of segment $[AB]$, determining the equation of line $(AB)$, checking if $C$ lies on $(AB)$, finding the equation of the line perpendicular to $(AB)$ through $C$, and finding the equation of the line parallel to $(AB)$ through $E(2,5)$.

2. Solution Steps

(2) Determine the coordinates of the vector $\vec{AB}$.

$\vec{AB} = (x_B - x_A, y_B - y_A) = (2 - 1, 5 - 3) = (1, 2)$

(3) Calculate $AB$.

The distance formula is $AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$.

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

(4) Determine the coordinates of the midpoint $I$ of segment $[AB]$.

The midpoint formula is $I = (\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2})$.

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

(5) Show that the reduced equation of line $(AB)$ is $y = 2x + 1$.

The slope of line $(AB)$ is $m = \frac{y_B - y_A}{x_B - x_A} = \frac{5 - 3}{2 - 1} = \frac{2}{1} = 2$.

The equation of the line $(AB)$ is $y - y_A = m(x - x_A)$.

$y - 3 = 2(x - 1)$

$y - 3 = 2x - 2$

$y = 2x - 2 + 3$

$y = 2x + 1$

(6) Does point $C$ belong to line $(AB)$?

The equation of line $(AB)$ is $y = 2x + 1$.

For point $C(-2, 1)$, substitute the coordinates into the equation:

$1 = 2(-2) + 1$

$1 = -4 + 1$

$1 = -3$

This is false, so point $C$ does not belong to line $(AB)$.

(7) Determine the reduced equation of the line $(D)$ perpendicular to $(AB)$ and passing through point $C$.

The slope of line $(AB)$ is $2$. The slope of a line perpendicular to $(AB)$ is $m' = -\frac{1}{2}$.

The equation of line $(D)$ is $y - y_C = m'(x - x_C)$.

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

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

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

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

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

(8) Determine the reduced equation of the line $(\Delta)$ parallel to $(AB)$ and passing through point $E(2, 5)$.

The slope of line $(AB)$ is $2$. Since $(\Delta)$ is parallel to $(AB)$, its slope is also $2$.

The equation of line $(\Delta)$ is $y - y_E = m(x - x_E)$.

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

$y - 5 = 2x - 4$

$y = 2x - 4 + 5$

$y = 2x + 1$

3. Final Answer

(2) $\vec{AB} = (1, 2)$

(3) $AB = \sqrt{5}$

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

(5) $y = 2x + 1$

(6) No, $C$ does not belong to $(AB)$.

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

(8) $y = 2x + 1$