Đáp án:
Giải thích các bước giải:
$$\eqalign{
& 1)\,\,\overrightarrow {AB} = \left( {5;6} \right);\,\,\overrightarrow {AC} = \left( {4; - 2} \right) \cr
& \Rightarrow \overrightarrow {AB} ,\,\,\overrightarrow {AC} \,\,khong\,\,cung\,\,phuong \cr
& \Rightarrow A,\,\,B,\,\,C\,\,khong\,\,\,thang\,\,hang \cr
& \Rightarrow Ton\,\,tai\,\,tam\,\,giac\,\,ABC. \cr
& 2)\,\,Pt\,\,AC:\,\,{{x + 3} \over {1 + 3}} = {{y + 1} \over { - 3 + 1}} \cr
& \Leftrightarrow {{x + 3} \over 4} = {{y + 1} \over { - 2}} \Leftrightarrow x + 3 = - 2\left( {y + 1} \right) \cr
& \Leftrightarrow x + 2y + 5 = 0. \cr
& Goi\,\,H\,\,la\,\,truc\,\,tam\,\,\Delta ABC. \cr
& Pt\,\,dg\,\,cao\,\,ung\,\,voi\,\,canh\,\,BC \cr
& \left\{ \matrix{
di\,\,qua\,\,A\left( { - 3; - 1} \right) \hfill \cr
\overrightarrow n = \overrightarrow {BC} = \left( { - 1; - 8} \right)//\left( {1;8} \right) \hfill \cr} \right. \cr
& \Rightarrow {d_1}:\,\,1\left( {x + 3} \right) + 8\left( {y + 1} \right) = 0 \Leftrightarrow x + 8y + 11 = 0 \cr
& Pt\,\,dg\,\,cao\,\,ung\,\,voi\,\,canh\,\,AC \cr
& \left\{ \matrix{
di\,\,qua\,\,B\left( {2;5} \right) \hfill \cr
\overrightarrow n = \overrightarrow {AC} = \left( {4; - 2} \right)//\left( {2; - 1} \right) \hfill \cr} \right. \cr
& \Rightarrow {d_2}:\,\,2\left( {x - 2} \right) - \left( {y - 5} \right) = 0 \Leftrightarrow 2x - y + 1 = 0 \cr
& \Rightarrow H = {d_1} \cap {d_2} = \left( { - {{19} \over {17}}; - {{21} \over {17}}} \right) \cr
& 3)\,\,Goi\,\,BH \bot AC \cr
& BH = d\left( {B;AC} \right) = {{\left| {2 + 2.5 + 5} \right|} \over {\sqrt {{1^2} + {2^2}} }} = {{17} \over {\sqrt 5 }} \cr
& AC = \sqrt {{4^2} + {2^2}} = 2\sqrt 5 \cr
& \Rightarrow {S_{\Delta ABC}} = {1 \over 2}BH.AC = {1 \over 2}.{{17} \over {\sqrt 5 }}.2\sqrt 5 = 17 \cr} $$
$$\eqalign{
& Goi\,\,I\left( {a;b} \right)\,\,la\,\,tam\,\,dg\,\,tron\,\,ngoai\,\,tiep\,\,\Delta ABC \cr
& \Rightarrow IA = IB = IC \cr
& \Rightarrow \left\{ \matrix{
I{A^2} = I{B^2} \hfill \cr
I{A^2} = I{C^2} \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
{\left( {x + 3} \right)^2} + {\left( {y + 1} \right)^2} = {\left( {x - 2} \right)^2} + {\left( {y - 5} \right)^2} \hfill \cr
{\left( {x + 3} \right)^2} + {\left( {y + 1} \right)^2} = {\left( {x - 1} \right)^2} + {\left( {y + 3} \right)^2} \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
6x + 9 + 2y + 1 = - 4x + 4 - 10y + 25 \hfill \cr
6x + 9 + 2y + 1 = - 2x + 1 + 6y + 9 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
10x + 12y = 19 \hfill \cr
8x - 4y = 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
x = {19 \over {34}} \hfill \cr
y = {{617} \over {34}} \hfill \cr} \right. \Rightarrow I\left( {{19 \over {34}};{{17} \over {34}}} \right) \cr} $$
R=IA, bạn tính bán kính sau đó viết pt đường tròn nhé, ra số to quá
