Ребят, помогите срочно! Координаты: A(13;10) B(3;5) C(15;-4). Created by moetyan. matematika-ru

Ребят, помогите срочно! Координаты: A(13;10) B(3;5) C(15;-4)...

moetyan @moetyan

August 2022 1 22 Report

Ребят, помогите срочно!
Координаты:
A(13;10)
B(3;5)
C(15;-4)

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NNNLLL54

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$A(13,10)\; ,\; \; B(3,5)\; ,\; \; C(15,-4)\\\\1)\; \; \overline {AB}=(3-13,5-10)=(-10,-5)\; \; ,\; \; \vec{s}_{AB}=(5,1)\\\\|\overline {AB}|=\sqrt{10^2+5^2}=\sqrt{125}=5\sqrt5\\\\\overline {AC}=(15-13,-4-10)=(2,-14)\; \; ,\; \; \vec{s}_{AC}=(1,-7)\\\\|\overline {AC}|=\sqrt{2^2+14^2}=\sqrt{200}=10\sqrt2\\\\\overline {BC}=(15-3,-4-5)=(12,-9)\; \; ,\; \; \vec{s}_{BC}=(4,-3)\\\\|\overline {BC}|=\sqrt{12^2+9^2}=\sqrt{225}=15\\\\2)\; \; AB:\; \; \frac{x-3}{-10}=\frac{y-5}{-5}\; \; \Rightarrow \; \; \frac{x-3}{2}=\frac{y-5}{1}\\\\\vec{s}=(2,1)\; \; \Rightarrow \; \; \vec{n}=(-1,2)\; \; \; \Big (\; \vec{s}\cdot \vec{n}=0\; \Big )$

$x-3=2(y-5)\; \; \to \; \; y=\frac{1}{2}x+\frac{7}{2}\; \; \Rightarrow \; \; k= \frac{1}{2}\\\\AC:\; \; \frac{x-15}{2}=\frac{y+4}{-14}\; \; \to \; \; \frac{x-15}{1}=\frac{y+4}{-7}\\\\\vec{s}=(1,-7)\; \; \to \; \; \vec{n}=(7,1)\\\\-7(x-15)=y+4\; \; \to \; \; y=-7x+101\; \; \Rightarrow \; \; k=-7\\\\BC:\; \; \frac{x-3}{12}=\frac{y-5}{-9}\; \; \to \; \; \frac{x-3}{4}=\frac{y-5}{-3}\\\\\vec{s}=(4,-3)\; \; \to \; \; \vec{n}=(3,4)\\\\-3(x-3)=4(y-5)\; \; \to \; \; y=-\frac{3}{4}x+\frac{29}{4}\; \; \Rightarrow \; \; k=-\frac{3}{4}$

$3)\; \; cosC=\frac{(\overline {BC}\cdot \overline {AC})}{|\overline {BC}|\cdot |\overline {AC}|}=\frac{2\cdot 4+14\cdot 3}{15\cdot 10\sqrt2}=\frac{50}{150\sqrt2}=\frac{1}{3\sqrt2}\\\\\angle C=arccos\frac{1}{3\sqrt2}$

$4)\; \; AL:\; \; \vec{n}=\vec{s}_{BC}=(4,-3)\; ,\; \; 4(x-13)-3(y-10)=0\; ,\\\\4x-52-3y+30=0\\\\AL:\; \; 4x-3y-22=0\; ;\\\\5)\; \; BK:\; \; x_{K}=\frac{x_{A}+x_{C}}{2}=\frac{13+15}{2}=14\\\\y_{K}=\frac{y_{A}+y_{C}}{2}=\frac{10-4}{2}=3\; ,\qquad K(14,3)$

$\overline {BK}=(14-3,3-5)=(11,-2)\\\\BK:\; \; \frac{x-3}{11}=\frac{y-5}{-2}\; ,\; \; -2(x-3)=11(y-5)\; ,\; \; \underline {2x+11y=61}\\\\6)\; \; \vec{n}_{AL}=\vec{s}_{BC}=(4,-3)\; \; \to \; \; AL:\; \; 4(x-13)-3(y-10)=0\; ,\\\\AL:\; \; \underline {4x-3y=22}\\\\\left \{ {{2x+11y=61\, |\cdot (-2)} \atop {4x-3y=22\; \; \; \; }} \right. \oplus \left \{ {{2x=61-11y} \atop {-25y=-100}} \right.\; \left \{ {{x=8,5} \atop {y=4}} \right. \; \; P(8,5\, ;\, 4)$

$7)\; \; \vec{s}_{AC}=(1,-7)\; ,\; AC\parallel\; l\; ,\; \; P(8,5\, ;\, 4)\in l\\\\l:\; \; \frac{x-8,5}{1}=\frac{y-4}{-7}\\\\8)\; \; x_{A}=\frac{x_{T}+x_{C}}{2}\; ,\; \; x_{T}=2x_{A}-x_{C}=2\cdot 13-15=11\\\\y_{T}=2y_{A}-y_{С}=2\cdot 10+4=24\\\\T(11,24)$

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