\(M\left(x;y\right)\left(d\right):3x-y=0\Leftrightarrow y=3x\)

\(\Rightarrow M\left(x;3x\right)\)

\(d\left(M;\left(\Delta\right)\right)=\dfrac{\left|4x+3.3x+2\right|}{\sqrt{4^2+3^2}}=3\)

\(\Leftrightarrow\dfrac{\left|13x+2\right|}{5}=3\)

\(\Leftrightarrow\left|13x+2\right|=15\)

\(\Leftrightarrow\left[{}\begin{matrix}13x+2=15\\13x+2=-15\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{17}{13}\end{matrix}\right.\)

\(x=1\Rightarrow y=3.1=3\)

\(x=-\dfrac{17}{13}\Rightarrow y=3.\left(-\dfrac{17}{13}\right)=-\dfrac{51}{13}\)

\(\Rightarrow\left[{}\begin{matrix}x+y=1+3=4\\x+y=-\dfrac{17}{13}-\dfrac{51}{13}=-\dfrac{68}{13}\end{matrix}\right.\)

a) \(R=IA=\sqrt{\left(2-1\right)^2+\left(2+4\right)^2}=\sqrt{37}\)

Phương trình đường tròn \(\left(C\right)\) có tâm I(1; -4) và bán kính \(R=\sqrt{37}\) là:

\(\left(C\right):\left(x-1\right)^2+\left(y+4\right)^2=37\)

b) Giả sử \(M\left(0;2\right)\in\left(C\right)\Leftrightarrow\left(0-1\right)^2+\left(2+4\right)^2=37\left(đúng\right)\Rightarrow M\left(0;2\right)\)

Vectơ chỉ phương của tiếp tuyến \(\) tại M là \(\overrightarrow{IM}\). Ta có:

\(\overrightarrow{IM}=\left(-1;6\right)\)

Phương trình tiếp tuyến \(\left(d\right)\) tại \(M\left(0;2\right)\) có dạng:

\(\left(d\right):-\left(x-0\right)+6\left(y-2\right)=0\)

\(\Rightarrow\left(d\right):x-6y+12=0\)

\(\dfrac{W_d}{W'_d}=\dfrac{\dfrac{1}{2}m\cdot v^2}{\dfrac{1}{2}m\cdot\left(2v\right)^2}=\dfrac{v^2}{4v^2}=\dfrac{1}{4}\\ \Rightarrow W'_d=4W_d\)

a: Tọa độ trọng tâm của ΔABC là:

\(\left\{{}\begin{matrix}x=\dfrac{-1+1+2}{3}=\dfrac{2}{3}\\y=\dfrac{2+1+3}{3}=\dfrac{6}{3}=2\\z=\dfrac{3+0+\left(-4\right)}{3}=-\dfrac{1}{3}\end{matrix}\right.\)

b:

A(-1;2;3); B(1;1;0); C(2;3;-4); H(a;b;c)

\(\overrightarrow{AH}=\left(a+1;b-2;c-3\right);\overrightarrow{BC}=\left(2-1;3-1;-4-0\right)=\left(1;2;-4\right)\)

\(\overrightarrow{BH}=\left(1-x;1-y;0-c\right)=\left(1-x;1-y;-c\right)\)

\(\overrightarrow{AC}=\left(2+1;3-2;-4-3\right)=\left(3;1;-7\right)\)

\(\overrightarrow{CH}=\left(2-a;3-b;-4-c\right)\)

\(\overrightarrow{AB}=\left(1+1;1-2;0-3\right)=\left(2;-1;-3\right)\)

H(a;b;c) là trực tâm của ΔABC

=>\(\left\{{}\begin{matrix}\overrightarrow{AH}\cdot\overrightarrow{BC}=0\\\overrightarrow{BH}\cdot\overrightarrow{AC}=0\\\overrightarrow{CH}\cdot\overrightarrow{AB}=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}1\left(a+1\right)+2\left(b-2\right)+\left(-4\right)\left(c-3\right)=0\\3\left(1-a\right)+1\left(1-b\right)+\left(-7\right)\cdot\left(-c\right)=0\\2\left(2-a\right)+\left(-1\right)\left(3-b\right)+\left(-3\right)\left(-4-c\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a+1+2b-4-4c+12=0\\3-3a+1-b+7c=0\\4-2a+b-3+12+3c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+2b-4c=-12+4-1=-12+3=-9\\-3a-b+7c=-4\\-2a+b+3c=-13\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a+2b-4c=-9\\-6a-2b+14c=-8\\-4a+2b+6c=-26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5a+10c=-17\\5a-10c=17\\a+2b-4c=-9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}0a=0\\5a-10c=17\\2b=-9-a+4c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\in R\\10c=5a-17\\2b=-a+4c-9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a\in R\\c=\dfrac{5a-17}{10}\\b=\dfrac{-a+4c-9}{2}=\dfrac{-a+4\cdot\dfrac{5a-17}{10}-9}{2}=\dfrac{-a+\dfrac{20a-34}{10}-9}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a\in R\\c=\dfrac{5a-17}{10}\\b=\dfrac{-10a+20a-34-90}{20}=\dfrac{10a-124}{20}=\dfrac{5a-62}{10}\end{matrix}\right.\)

\(T=2a+b:c=2a+\dfrac{5a-62}{10}:\dfrac{5a-17}{10}=2a+\dfrac{5a-62}{5a-17}=\dfrac{2a\left(5a-17\right)+5a-62}{5a-17}\)

=>\(T=\dfrac{10a^2-34a+5a-62}{5a-17}=\dfrac{10a^2-29a-62}{5a-17}\)