Question

85 баллов даю ток помогите срочно

Answer 1

$2)\; \; l_1:\; \; 5x+y-1=0\; ,\quad l_2:\; kx-y-6=0\; ,\; \; l_1\perp l_2\\\\l_1\perp l_2\; \; \Leftrightarrow \; \; \vec{n}_1\perp \vec{n}_2\; \; \Leftrightarrow \; \; \vec{n}_1\cdot \vec{n}_2=0\\\\\vec{n}_1=(5,1)\; ,\; \; \vec{n}_2=(k,-1)\\\\\vec{n}_1\cdot \vec{n}_2=5k-1=0\; \; \to \; \; k=\frac{1}{5}$

$1)\; \; A(-3,2)\; ,\; \; B(3,4)\; ,\; \; C(9,-4)\\\\1.1)\; \; |\overline {AB|}=\sqrt{(3+3)^2+(4-2)^2}=\sqrt{6^2+2^2}=\sqrt{40}=2\sqrt{10}\\\\1.2)\; \; AB:\; \frac{x+3}{6} = \frac{y-2}{2} \; \; \Rightarrow \; \; 2(x+3)=6(y-2)\\\\2x+6=6y-12\; \; \to \; \; 6y=2x+18\; ,\; y=\frac{1}{3}x+3$

$BC=(9-3,-4-4)=(6,-8)\; ;\\\\ \frac{x-3}{6} = \frac{y-4}{-8} \; \; \; \Rightarrow \; \; \; \frac{x-3}{3}=\frac{y-4}{-4} \; \; \Rightarrow \; \; -4(x-3)=3(y-4)\\\\-4x+12=3y-12\; \; \to \; \; 3y=-4x+24=0\; ,\; y=-\frac{4}{3}x+8$

$k_{AB}=\frac{1}{3}\; ,\; \; k_{BC}=-\frac{4}{3}\\\\tgB=\Big | \frac{k_1-k_2}{1+k_1k_2} \Big |=\Big | \frac{\frac{1}{3}+\frac{4}{3}}{1-\frac{1}{3}\cdot \frac{4}{3}} \Big |= \frac{5\cdot 9}{3(9-4)} = \frac{15}{5} =3$

$1.4)\; \; x_{D}= \frac{x_{B}+x_{C}}{2}= \frac{3+9}{2}=6\; ;\; \; y_{D}= \frac{4-4}{2} =0\\\\AD:\; \; \frac{x+3}{6+3}=\frac{y-2}{0-2} \; \; ,\; \; \; \frac{x+3}{9}= \frac{y-2}{-2} \\\\1.5)\; \; CE\perp AB\; \; \Rightarrow \; \; \vec{s}_{AB}=\vec{n}_{CE}=(6,-2)\\\\CE:\; \; 6(x-9)-2(y+4)=0\; |:2\\\\3(x-9)-(y+4)=0\\\\CE:\; \; 3x-y-21=0$

$1.6)\; \; BM\parallel AC\; \; \Rightarrow \; \; \vec{s}_{AC}=\vec{s}_{BM}=(9+3,-4-2)=(12,-6)\\\\BM:\; \; \frac{x-3}{12} = \frac{y-4}{-6} \; \; \to \; \; \frac{x-3}{2} =\frac{y-4}{-1}$