Question

алгебра прошу помогите ​

Answer 1

Ответ:

$l_1:\ 4x-6y+16=0\\\\l_2:\ 5x+9y+12=0\ \ ,\ \ \ M(\ 2\ ;-3\ )\ ,\ \ M\notin l_1\, ,\, l_2\\\\\\1)\ \ \ l_1\parallel l_3\ \ \Rightarrow \ \ \vec{n}_1=\vec{n}_3\ \ ,\ \ \vec{n}_1=(\ 4\ ;-6\ )\\\\l:\ 4(x-2)-6(y+3)=0\ \ ,\ \ \ \underline {\ 4x-6y-26=0\ }\\\\\\2)\ \ l_2\perp l_4\ \ \Rightarrow \ \ \vec{n}_2=\vec{s}_4\ \ ,\ \ \vec{n}_2=(\ 5\ ;\ 9\ )\\\\l:\ \dfrac{x-2}{5}=\dfrac{y+3}{9}\ \ ,\ \ \ 9(x-2)=5(y+3)\ \ ,\ \ \ \underline {\ 9x-5y-33=0\ }$

$3)\ \ coa\varphi =\dfrac{\vec{n}_1\cdot \vec{n}_2}{|\vec{n}_1|\cdot |\vec{n}_2|}=\dfrac{4\cdot 5-8\cdot 9}{\sqrt{4^2+6^2}\cdot \sqrt{5^2+9^2}}=\dfrac{-52}{\sqrt{52}\cdot \sqrt{106}}=-\dfrac{26}{\sqrt{1378}}\\\\\varphi =arccos\dfrac{26}{\sqrt{1378}} - ostruj\ ygol\\\\\\4)\ \ d\, (M;l_1)=\dfrac{|Ax_{0}+By_{0}+D|}{\sqrt{A^2+B^2}}=\dfrac{|4\cdot 2-6\cdot (-3)+16|}{\sqrt{4^2+6^2}}=\dfrac{42}{\sqrt{52}}=\dfrac{21}{\sqrt{13}}$