\(A\left(1;-1\right);B\left(2;-3\right);C\left(3;4\right)\)
\(AB=\sqrt{\left(2-1\right)^2+\left(-3+1\right)^2}=\sqrt{5}\)
\(AC=\sqrt{\left(3-1\right)^2+\left(4+1\right)^2}=\sqrt{29}\)
\(BC=\sqrt{\left(3-2\right)^2+\left(4+3\right)^2}=5\sqrt{2}\)
c:
Gọi D(x;y) là chân đường phân giác kẻ từ A xuống BC
B(2;-3); D(x;y); C(3;4)
\(\overrightarrow{BD}=\left(x-2;y+3\right);\overrightarrow{BC}=\left(1;7\right)\)
Xét ΔABC có AD là phân giác
nên \(\dfrac{DB}{DC}=\dfrac{AB}{AC}=\dfrac{\sqrt{5}}{\sqrt{29}}\)
=>\(\dfrac{DB}{BC}=\dfrac{\sqrt{5}}{\sqrt{5}+\sqrt{29}}\)
=>\(\overrightarrow{BD}=\dfrac{\sqrt{5}}{\sqrt{5}+\sqrt{29}}\cdot\overrightarrow{BC}\)
=>\(\left\{{}\begin{matrix}x-2=1\cdot\dfrac{\sqrt{5}}{\sqrt{5}+\sqrt{29}}=\dfrac{\sqrt{5}}{\sqrt{5}+\sqrt{29}}\\y+3=\dfrac{7\sqrt{5}}{\sqrt{5}+\sqrt{29}}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{3\sqrt{5}+2\sqrt{29}}{\sqrt{5}+\sqrt{29}}\\y=\dfrac{4\sqrt{5}-3\sqrt{29}}{\sqrt{5}+\sqrt{29}}\end{matrix}\right.\)
A(1;-1); \(D\left(\dfrac{3\sqrt{5}+2\sqrt{29}}{\sqrt{5}+\sqrt{29}};\dfrac{4\sqrt{5}-3\sqrt{29}}{\sqrt{5}+\sqrt{29}}\right)\)
\(\overrightarrow{AD}=\left(\dfrac{3\sqrt{5}+2\sqrt{29}}{\sqrt{5}+\sqrt{29}}-1;\dfrac{4\sqrt{5}-3\sqrt{29}}{\sqrt{5}+\sqrt{29}}+1\right)\)
=>\(\overrightarrow{AD}=\left(\dfrac{2\sqrt{5}+\sqrt{29}}{\sqrt{5}+\sqrt{29}};\dfrac{5\sqrt{5}-2\sqrt{29}}{\sqrt{5}+\sqrt{29}}\right)\)
=>Vecto pháp tuyến là \(\left(\dfrac{-5\sqrt{5}+2\sqrt{29}}{\sqrt{5}+\sqrt{29}};\dfrac{2\sqrt{5}+\sqrt{29}}{\sqrt[]{5}+\sqrt{29}}\right)\)
Phương trình đường thẳng AD là:
\(\dfrac{-5\sqrt{5}+2\sqrt{29}}{\sqrt{5}+\sqrt{29}}\left(x-1\right)+\dfrac{2\sqrt{5}+\sqrt{29}}{\sqrt{5}+\sqrt{29}}\left(y+1\right)=0\)
=>\(\dfrac{-5\sqrt{5}+2\sqrt{29}}{\sqrt{5}+\sqrt{29}}x+\dfrac{2\sqrt{5}+\sqrt{29}}{\sqrt{5}+\sqrt{29}}y+\dfrac{5\sqrt{5}-2\sqrt{29}+2\sqrt{5}+\sqrt{29}}{\sqrt{5}+\sqrt{29}}=0\)
=>\(\dfrac{-5\sqrt{5}+2\sqrt{29}}{\sqrt{5}+\sqrt{29}}x+\dfrac{2\sqrt{5}+\sqrt{29}}{\sqrt{5}+\sqrt{29}}y+\dfrac{7\sqrt{5}-\sqrt{29}}{\sqrt{5}+\sqrt{29}}=0\)
