Bonjour,
Question 1 :
[tex]\vec{AM} = (x - 8 \ ; y - 15) \\\\ \\det (\vec{AM} ; \vec{u}) = (x-8) \times (-5) - (y +15) \times 6 = 0 \\ \Leftrightarrow-5x + 40 -6y + 90 = 0 \\ \Leftrightarrow-5x -6y +130 = 0[/tex]
Question 2 :
[tex]\vec{BM} = (x + 9 \ ; y + 7) \\\\ \\det (\vec{BM} ; \vec{v}) = (x+9) \times (-1) - (y +7) \times (-4)= 0 \\ \Leftrightarrow-x -9 + 4y + 28 = 0 \\ \Leftrightarrow-x + 4y + 19 = 0[/tex]
Question 3 :
-5x - 6y + 130 = 0
-6y = 5x - 130
[tex]y = \frac{5}{-6} x - \frac{130}{-6} \\y = - \frac{5}{6} x + \frac{130}{6}[/tex]
-x + 4y + 19 = 0
4y = x - 19
[tex]y = \frac{1}{4}x - \frac{19}{4}[/tex]
Question 4 :
[tex]- \frac{5}{6} x + \frac{130}{6} = \frac{1}{4}x - \frac{19}{4}\\ \\ - \frac{5}{6} x - \frac{1}{4}x = - \frac{19}{4} - \frac{130}{6}\\ \\ -\frac{13}{12} x = - \frac{317}{12} \\ \\ x = - \frac{317}{12} \times (-\frac{12}{13}) \\ \\ x = \frac{317}{13}[/tex]
[tex]y = \frac{1}{4} \times \frac{317}{13} - \frac{19}{4} = \frac{35}{26}[/tex]
Les coordonnées du point d'intersection sont [tex](\frac{317}{13} ; \frac{35}{26} )[/tex]
