Respuesta :

gmany

[tex]k:\ y=m_1x+b_1\\l:\ y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}[/tex]

[tex] 115)\ -x-y=3\ \ \ |+x\\-y=x+3\ \ \ |\cdot(_1)\\y=-x-3\to m_1=-1\to\boxed{m_2=-\dfrac{1}{-1}=1}\\\\116)\ 0=-4x-y+5\ \ \ |+y\\y=-4x+5\to m_1=-4\to\boxed{m_2=-\dfrac{1}{-4}=\dfrac{1}{4}}\\\\117)\ 6-2y=x\ \ \ \ |-6\\-2y=x-6\ \ \ \ |:(-2)\\y=-\dfrac{1}{2}x+3\to m_1=-\dfrac{1}{2}\to\boxed{m_2=-\dfrac{1}{-\frac{1}{2}}=2}\\\\118)\ 5y=25+4x\ \ \ \ |:5\\y=5+\dfrac{4}{5}x\to m_1=\dfrac{4}{5}\to\boxed{m_2=-\dfrac{1}{\frac{4}{5}}=-\dfrac{5}{4}} [/tex]

[tex] 119)\ -15x-9y=9\ \ \ \ |+15x\\-9y=15x+9\ \ \ \ |:(-9)\\y=-\dfrac{15}{9}x-1\\y=-\dfrac{5}{3}x-1\to m_1=-\dfrac{5}{3}\to\boxed{m_2=-\dfrac{1}{-\frac{5}{3}}=\dfrac{3}{5}} [/tex]