Soit les points A(1, 3), B(-2, 1), C(5, -3). 1) Donner les composantes des vecteurs AB, AC et BC. 2) Montrer que le triangle ABC est rectangle en A et déterminer le centre et le rayon du cercle (C) ci

$$\vec{BC} = (x_C - x_B, y_C - y_B) = (5 - (-2), -3 - 1) = (5 + 2, -4) = (7, -4)$$

The components of the vectors are:

$$\boxed{\vec{AB} = (-3, -2)}$$ $$\boxed{\vec{AC} = (4, -6)}$$ $$\boxed{\vec{BC} = (7, -4)}$$

2) Montrer que le triangle ABC est rectangle en A et déterminer le centre et le rayon du cercle (C) circonscrit au triangle ABC.

To show that the triangle is right-angled at A, we calculate the dot product of vectors $\vec{AB}$ and $\vec{AC}$. If the dot product is zero, the vectors are orthogonal, and the angle at A is $90^\circ$.

$$\vec{AB} \cdot \vec{AC} = (-3)(4) + (-2)(-6)$$ $$\vec{AB} \cdot \vec{AC} = -12 + 12$$ $$\vec{AB} \cdot \vec{AC} = 0$$

Since the dot product is 0, the vectors $\vec{AB}$ and $\vec{AC}$ are orthogonal. Therefore, triangle ABC is right-angled at A.

For a right-angled triangle, the hypotenuse is the diameter of the circumscribed circle. The center of the circumscribed circle is the midpoint of the hypotenuse. In this case, BC is the hypotenuse.
Let $M$ be the center of the circle.

$$M = \left(\frac{x_B + x_C}{2}, \frac{y_B + y_C}{2}\right)$$ $$M = \left(\frac{-2 + 5}{2}, \frac{1 + (-3)}{2}\right)$$ $$M = \left(\frac{3}{2}, \frac{-2}{2}\right)$$ $$M = \left(\frac{3}{2}, -1\right)$$

The center of the circumscribed circle is $\boxed{M\left(\frac{3}{2}, -1\right)}$.

The radius $R$ is half the length of the hypotenuse BC.
First, calculate the length of $\vec{BC}$:

$$|\vec{BC}| = \sqrt{(x_C - x_B)^2 + (y_C - y_B)^2} = \sqrt{7^2 + (-4)^2}$$ $$|\vec{BC}| = \sqrt{49 + 16} = \sqrt{65}$$

Now, calculate the radius:

$$R = \frac{1}{2} |\vec{BC}| = \frac{\sqrt{65}}{2}$$

The radius of the circumscribed circle is $\boxed{R = \frac{\sqrt{65}}{2}}$.

3) Trouver les coordonnées du point S tel que $\vec{AS} = \frac{3}{4}\vec{AB} + \frac{4}{3}\vec{BC}$.

We found $\vec{AB} = (-3, -2)$.

$$\frac{3}{4}\vec{AB} = \frac{3}{4}(-3, -2) = \left(-\frac{9}{4}, -\frac{6}{4}\right) = \left(-\frac{9}{4}, -\frac{3}{2}\right)$$

We found $\vec{BC} = (7, -4)$.

$$\frac{4}{3}\vec{BC} = \frac{4}{3}(7, -4) = \left(\frac{28}{3}, -\frac{16}{3}\right)$$

$$\vec{AS} = \left(-\frac{9}{4} + \frac{28}{3}, -\frac{3}{2} - \frac{16}{3}\right)$$

For the x-component:

$$-\frac{9}{4} + \frac{28}{3} = \frac{-9 \times 3}{12} + \frac{28 \times 4}{12} = \frac{-27 + 112}{12} = \frac{85}{12}$$

For the y-component:

$$-\frac{3}{2} - \frac{16}{3} = \frac{-3 \times 3}{6} - \frac{16 \times 2}{6} = \frac{-9 - 32}{6} = -\frac{41}{6}$$

So, $\vec{AS} = \left(\frac{85}{12}, -\frac{41}{6}\right)$.

Let $S(x_S, y_S)$. We know $A(1, 3)$.

$$\vec{AS} = (x_S - x_A, y_S - y_A) = (x_S - 1, y_S - 3)$$

Equating the components of $\vec{AS}$:

$$x_S - 1 = \frac{85}{12}$$ $$x_S = 1 + \frac{85}{12} = \frac{12}{12} + \frac{85}{12} = \frac{97}{12}$$ $$y_S - 3 = -\frac{41}{6}$$ $$y_S = 3 - \frac{41}{6} = \frac{18}{6} - \frac{41}{6} = -\frac{23}{6}$$

The coordinates of point S are $\boxed{S\left(\frac{97}{12}, -\frac{23}{6}\right)}$.