Riješite zadatke.

Tražimo simetralu dužine $\overline{AB}$$\overline{AB}$ s točkama $A(-1, 3)$$A(-1, 3)$ i $B(9, -5)$$B(9, -5)$.
1. Polovište $P$$P$: $(\frac{-1+9}{2}, \frac{3-5}{2}) = (4, -1)$$(\frac{-1+9}{2}, \frac{3-5}{2}) = (4, -1)$.
2. Koeficijent smjera $AB$$AB$: $k_{AB} = \frac{-5-3}{9-(-1)} = \frac{-8}{10} = -\frac{4}{5}$$k_{AB} = \frac{-5-3}{9-(-1)} = \frac{-8}{10} = -\frac{4}{5}$.
3. Koeficijent smjera simetrale (okomica): $k_s = -\frac{1}{k_{AB}} = \frac{5}{4}$$k_s = -\frac{1}{k_{AB}} = \frac{5}{4}$.
4. Jednadžba pravca kroz $P(4, -1)$$P(4, -1)$:
$y - (-1) = \frac{5}{4}(x - 4) \implies y = \frac{5}{4}x - 5 - 1 \implies y = \frac{5}{4}x - 6$$y - (-1) = \frac{5}{4}(x - 4) \implies y = \frac{5}{4}x - 5 - 1 \implies y = \frac{5}{4}x - 6$.
Implicitno: $5x - 4y - 24 = 0$$5x - 4y - 24 = 0$.

Dokazujemo da je trokut $ABC$$ABC$ pravokutan koristeći vektor $\overrightarrow{AC}$$\overrightarrow{AC}$.
Zadani su vektori $\overrightarrow{BC}=4\vec{i}+2\vec{j}$$\overrightarrow{BC}=4\vec{i}+2\vec{j}$ i $\overrightarrow{AB}=-2\vec{i}-6\vec{j}$$\overrightarrow{AB}=-2\vec{i}-6\vec{j}$.
Vektor $\overrightarrow{AC}$$\overrightarrow{AC}$ jednak je zbroju vektora $\overrightarrow{AB}$$\overrightarrow{AB}$ i $\overrightarrow{BC}$$\overrightarrow{BC}$:
$\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = (-2+4)\vec{i} + (-6+2)\vec{j} = 2\vec{i} - 4\vec{j}$$\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = (-2+4)\vec{i} + (-6+2)\vec{j} = 2\vec{i} - 4\vec{j}$.
Provjeravamo duljine stranica (kvadrate duljina):
$|AB|^2 = (-2)^2 + (-6)^2 = 4 + 36 = 40$$|AB|^2 = (-2)^2 + (-6)^2 = 4 + 36 = 40$
$|BC|^2 = 4^2 + 2^2 = 16 + 4 = 20$$|BC|^2 = 4^2 + 2^2 = 16 + 4 = 20$
$|AC|^2 = 2^2 + (-4)^2 = 4 + 16 = 20$$|AC|^2 = 2^2 + (-4)^2 = 4 + 16 = 20$
Vrijedi $|AC|^2 + |BC|^2 = 20 + 20 = 40 = |AB|^2$$|AC|^2 + |BC|^2 = 20 + 20 = 40 = |AB|^2$.
Prema obratu Pitagorinog poučka, trokut je pravokutan s pravim kutom u vrhu $C$$C$.
Odgovor: Trokut je pravokutan.