The problem consists of three exercises: Exercise 1: Given points $A(1, 3)$, $B(2, 5)$, and $C(-2, 1)$, we need to: 1. Plot these points in a coordinate system.

1. Problem Description

The problem consists of three exercises:

Exercise 1: Given points $A(1, 3)$, $B(2, 5)$, and $C(-2, 1)$, we need to:

1. Plot these points in a coordinate system.

2. Find the coordinates of vector $\vec{AB}$.

3. Calculate the length $AB$.

4. Find the coordinates of the midpoint $I$ of segment $[AB]$.

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

6. Determine if point $C$ belongs to the line $(AB)$.

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

8. Find the equation of the line $(\Delta)$ parallel to $(AB)$ and passing through point $E(2, 5)$, which is coincidentally the same as point B.

Exercise 2: Given that $f$ is a linear function with $f(7) = 14$, we need to:

1. Determine the expression of $f(x)$.

2. Find the number $x$ such that $f(x) = 21$.

3. Sketch the graph of function $f$.

Exercise 3: Given that $g$ is an affine function with $g(0) = 3$ and $g(2) = 7$, we need to:

1. Determine the expression of $g(x)$.

2. Calculate $g(1)$ and $g(3)$.

3. Find the number $x$ such that $g(x) = 21$.

4. Sketch the graph of function $g$.

2. Solution Steps

Exercise 1:

1. Plotting the points is a graphical task.

2. The coordinates of vector $\vec{AB}$ are given by $(x_B - x_A, y_B - y_A) = (2 - 1, 5 - 3) = (1, 2)$.

3. The distance $AB$ is given by $\sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} = \sqrt{(2 - 1)^2 + (5 - 3)^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$.

4. The midpoint $I$ of segment $[AB]$ has coordinates $\left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right) = \left(\frac{1 + 2}{2}, \frac{3 + 5}{2}\right) = \left(\frac{3}{2}, \frac{8}{2}\right) = (1.5, 4)$.

5. Let the equation of line $(AB)$ be $y = mx + b$. The slope $m = \frac{y_B - y_A}{x_B - x_A} = \frac{5 - 3}{2 - 1} = \frac{2}{1} = 2$. Thus, $y = 2x + b$. Since point $A(1, 3)$ is on the line, we have $3 = 2(1) + b$, so $b = 3 - 2 = 1$. Therefore, the equation of line $(AB)$ is $y = 2x + 1$.

6. For point $C(-2, 1)$ to belong to line $(AB)$, it must satisfy the equation $y = 2x + 1$. Substituting $x = -2$, we get $y = 2(-2) + 1 = -4 + 1 = -3$. Since the y-coordinate of point C is 1 and not -3, point $C$ does not belong to line $(AB)$.

7. The line $(D)$ is perpendicular to $(AB)$, so its slope is $m_D = -\frac{1}{m_{AB}} = -\frac{1}{2}$. Since line $(D)$ passes through point $C(-2, 1)$, its equation is $y - 1 = -\frac{1}{2}(x - (-2))$, which simplifies to $y - 1 = -\frac{1}{2}x - 1$. Therefore, $y = -\frac{1}{2}x$.

8. The line $(\Delta)$ is parallel to $(AB)$, so its slope is $m_\Delta = m_{AB} = 2$. Since line $(\Delta)$ passes through point $E(2, 5)$, its equation is $y - 5 = 2(x - 2)$, which simplifies to $y - 5 = 2x - 4$. Therefore, $y = 2x + 1$.

Exercise 2:

1. Since $f$ is a linear function, $f(x) = kx$ for some constant $k$. We are given $f(7) = 14$, so $7k = 14$, which means $k = \frac{14}{7} = 2$. Therefore, $f(x) = 2x$.

2. We want to find $x$ such that $f(x) = 21$, so $2x = 21$, which means $x = \frac{21}{2} = 10.5$.

3. Sketching the graph of $f(x) = 2x$ is a graphical task.

Exercise 3:

1. Since $g$ is an affine function, $g(x) = ax + b$ for some constants $a$ and $b$. We are given $g(0) = 3$ and $g(2) = 7$.

$g(0) = a(0) + b = 3$, so $b = 3$.

$g(2) = a(2) + 3 = 7$, so $2a = 4$, which means $a = 2$. Therefore, $g(x) = 2x + 3$.

2. $g(1) = 2(1) + 3 = 2 + 3 = 5$.

$g(3) = 2(3) + 3 = 6 + 3 = 9$.

3. We want to find $x$ such that $g(x) = 21$, so $2x + 3 = 21$, which means $2x = 18$, so $x = \frac{18}{2} = 9$.

4. Sketching the graph of $g(x) = 2x + 3$ is a graphical task.

3. Final Answer

Exercise 1:

1. Plotting the points is a graphical task.

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

3. $AB = \sqrt{5}$

4. $I = (1.5, 4)$

5. $y = 2x + 1$

6. Point $C$ does not belong to line $(AB)$.

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

8. $y = 2x + 1$

Exercise 2:

1. $f(x) = 2x$

2. $x = 10.5$

3. Sketching the graph is a graphical task.

Exercise 3:

1. $g(x) = 2x + 3$

2. $g(1) = 5$, $g(3) = 9$

3. $x = 9$

4. Sketching the graph is a graphical task.