Bài 6: M(x;y); A(1;3); B(4;2)

\(\overrightarrow{MA}=\left(1-x;3-y\right);\overrightarrow{MB}=\left(4-x;2-y\right)\)

=>\(MA^2=\left(1-x\right)^2+\left(3-y\right)^2;MB^2=\left(4-x\right)^2+\left(2-y\right)^2\)

ΔMAB cân tại M

=>MA=MB

=>\(MA^2=MB^2\)

=>\(\left(1-x\right)^2+\left(3-y\right)^2=\left(4-x\right)^2+\left(2-y\right)^2\)

=>\(x^2-2x+1+y^2-6y+9=x^2-8x+16+y^2-4y+4\)

=>-2x-6y+10=-8x-4y+20

=>-2x-6y+8x+4y=20-10

=>6x-2y=10

=>3x-y=5

=>y=3x-5

ΔMAB vuông tại M

=>\(\overrightarrow{MA}\cdot\overrightarrow{MB}=\overrightarrow{0}\)

=>(1-x)(4-x)+(3-y)(2-y)=0

=>(x-1)(x-4)+(y-3)(y-2)=0

=>(x-1)(x-4)+(3x-5-3)(3x-5-2)=0

=>(x-1)(x-4)+(3x-8)(3x-7)=0

=>\(x^2-5x+4+9x^2-45x+56=0\)

=>\(10x^2-50x+60=0\)

=>\(x^2-5x+6=0\)

=>(x-2)(x-3)=0

=>x=2 hoặc x=3

TH1: x=2

=>y=3x-5=3*2-5=6-5=1

=>M(2;1)

TH2: x=3

=>y=3x-5=3*3-5=9-5=4

=>M(3;4)

Bài 5:

a: A(1;3); B(4;2); C(1;0)

\(AB=\sqrt{\left(4-1\right)^2+\left(2-3\right)^2}=\sqrt{3^2+\left(-1\right)^2}=\sqrt{10}\)

\(AC=\sqrt{\left(1-1\right)^2+\left(0-3\right)^2}=3\)

\(BC=\sqrt{\left(1-4\right)^2+\left(0-2\right)^2}=\sqrt{\left(-3\right)^2+\left(-2\right)^2}=\sqrt{13}\)

Chu vi của tam giác ABC là:

AB+AC+BC

\(=3+\sqrt{10}+\sqrt{13}\)

b: A(1;3); B(4;2); C(1;0); H(x;y)

H là trực tâm của ΔABC

=>AH⊥BC và BH⊥AC

\(\overrightarrow{AH}=\left(x-1;y-3\right);\overrightarrow{BC}=\left(1-4;0-2\right)=\left(-3;-2\right)\)

\(\overrightarrow{BH}=\left(x-4;y-2\right);\overrightarrow{AC}=\left(1-1;0-3\right)=\left(0;-3\right)\)

BH⊥AC
=>\(\overrightarrow{BH}\cdot\overrightarrow{AC}=0\)

=>0(x-4)+(-3)(y-2)=0

=>-3(y-2)=0

=>y-2=0

=>y=2

AH⊥BC

=>\(\overrightarrow{AH}\cdot\overrightarrow{BC}=0\)

=>-3(x-1)+(-2)(y-3)=0

=>3(x-1)+2(y-3)=0

=>3(x-1)+2(2-3)=0

=>3(x-1)+2*(-1)=0

=>3x-3-2=0

=>3x=5

=>x=5/3

=>H(5/3;2)