Olá Marianne.
1°) Vamos calcular a equação de cada reta (diagonal), para isso, temos que descobrir os coeficientes e jogar na equação fundamental.
[tex]\underline{diagonal \ AC \rightarrow A(1,1) \ \ C(6,5)} \\\\ m = \frac{\Delta y}{\Delta x} = \frac{y_{f}-y_{i}}{x_{f}-x_{i}} = \frac{5-1}{6-1} = \boxed{\frac{4}{5}} \\\\ y-y_{0} = m ( x-x_{0}) \\\\ y - 1 = \frac{4}{5} (x-1) \\\\ y - 1 = \frac{4}{5}x - \frac{4}{5} \\\\ \frac{4}{5}x - y + 1 - \frac{4}{5} = 0 \\\\ \frac{4}{5}x - y + \frac{1}{5} = 0 \ \ \ \ \ \times 5 \\\\ \boxed{4x - 5y + 1 = 0}[/tex]
[tex]\underline{diagonal \ BD \rightarrow B(5,2) \ \ D(2,4)} \\\\ m = \frac{\Delta y}{\Delta x} = \frac{y_{f}-y_{i}}{x_{f}-x_{i}} = \frac{4-2}{2-5} = \boxed{-\frac{2}{3}} \\\\ y-y_{0} = m (x-x_{0}) \\\\ y - 2 = -\frac{2}{3}(x-5) \\\\ y-2 = -\frac{2}{3}x + \frac{10}{3} \\\\ \frac{2}{3}x + y - 2 - \frac{10}{3} = 0 \\\\ \frac{2}{3}x + y - \frac{16}{3} = 0 \ \ \ \ \times 3 \\\\ \boxed{2x + 3y - 16 = 0}[/tex]
Agora para descobrir o ponto de intersecção, basta fazer um sisteminha:
4x - 5y + 1 = 0 => 4x - 5y = -1
2x + 3y - 16 = 0 => 2x + 3y = 16
[tex]\left \{ {{4x-5y = -1 \ \ \ \ \times 3} \atop {2x+3y = 16 \ \ \ \ \times 5}} \right. \\\\ \left \{ {{12x-15y = -3} \atop {10x+15y = 80}} \right. \\\\ 22x = 77 \\\\ x = \frac{77}{22} \\\\ \boxed{x = 3,5} \\\\\\ \rightarrow 2 (3,5) + 3y = 16 \\ 7 + 3y = 16 \\ 3y = 9 \\\\ \boxed{y = 3}[/tex]
[tex]\therefore \boxed{P(3,5; 3)}[/tex]
