Let P(x₁ , y₁) and Q(x₂ , y₂) are two points which divide AB in three equal parts.
By section formula
\(P\left( { x }_{ 1 }{ y }_{ 1 } \right) =\left( \frac { 1\times \left( -4 \right) +2\times \left( 2 \right) }{ 1+2 } ,\frac { 1\times \left( -6 \right) +2\times \left( -3 \right) }{ 1+2 } \right) \)
\(=\left( \frac { -4+4 }{ 3 } ,\frac { -6+(-6) }{ 3 } \right) \)
= (0, -4)
\(Q\left( { x }_{ 2 }{ y }_{ 2 } \right) =\left( \frac { 2\times \left( -4 \right) +1\times \left( 2 \right) }{ 2+1 } ,\frac { 2\times \left( -6 \right) +1\times \left( -3 \right) }{ 2+1 } \right) \)
\(=\left( \frac { -8+2 }{ 3 } ,\frac { -12+(-3) }{ 3 } \right) \)
= (- 2, - 5)

a = 1 and b = -3, a² + b² = 10

Let points be A(1,5), B(2,3) and C(-2,-11)
\(AB=\sqrt { \left( 2-1 \right) ^{ 2 }+\left( 3-5 \right) ^{ 2 } } =\sqrt { 1+4 } =\sqrt { 5 } units\)
\(BC=\sqrt { \left( -2-2 \right) ^{ 2 }+\left( -11-3 \right) ^{ 2 } } =\sqrt { 16+196 } =\sqrt { 212 } =2\sqrt { 53 } units.\)
\(AC=\sqrt { \left( -2-1 \right) ^{ 2 }+\left( -11-5 \right) ^{ 2 } } =\sqrt { 9+256 } =\sqrt { 265 } =\sqrt { 5\times 53 } units \)
AB+BC≠ AC
Hence, the given points are not collinear. 

Let A(-5,7) and B(-1,3)
\(AB=\sqrt{(-1+5)^2+(3-7)^2}\)=\(\sqrt{16+16}\)=\(\sqrt{32}\)
=\(\sqrt{16X2}=4\sqrt2 units\)

\(\therefore \quad \frac { 1 }{ 2 } \left[ { { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) } \right] =0\)
\(\Rightarrow \quad \frac{1}{2}[0(2-y)+1(y-0)+x(0-2)]=0\)
\(\Rightarrow \quad \frac { 1 }{ 2 }[y-2x]=0\)
\(\Rightarrow \quad 2x-y=0\)

Area of triangle
\(=\frac { 1 }{ 2 } \left[ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right] \)
\(=\frac { 1 }{ 2 } [ 0(0-5)+6(5-0)+0(0+0)]\)
\(=\frac { 1 }{ 2 }[6\times5]\) = 15 sq. units

Given A(3,0), B(6,4) and C(-1,3)
\(\therefore\)  AB² = (3-6)² + (0-4)² = 9 + 16 = 25
BC² = (6+1)² + (4-3)² = 49 + 1 = 50
CA² = (-1-3)² + (3-0)² = 16 + 9 = 25
AB² = CA² \(\Rightarrow\) AB = CA
\(\therefore\)  Triangle is isosceles
Also, 25 + 50 = 50
\(\Rightarrow\)  AB² + CA² = BC²
Since pythagoras theorem is verified, therefore triangle is a right angled triangle.

Points are (-4, -7) and (0, -7)
Distance = \(=\sqrt { { \left( 0+4 \right) }^{ 2 }+{ \left( -7+7 \right) }^{ 2 } } =\sqrt { { 4 }^{ 2 }+0 } =\sqrt { 16 } \) = 4 units

Here, \(A\rightarrow \left( 0,5 \right) \) and \(B\rightarrow \left( -3,1 \right) \)
\(AB=\sqrt { { { { \left( { { x }_{ 2 }-{ x }_{ 1 } } \right) ^{ 2 }+\left( { { y }_{ 2 }-{ y }_{ 1 } } \right) ^{ 2 } } } } } \)
\(AB=\sqrt { { { { \left( { -3-0 } \right) ^{ 2 }+\left( { 1-5 } \right) ^{ 2 } } } } } \)
\(AB=\sqrt { { { { 9+16 } } } } =\sqrt { 25 } =\) 5 units

Here, coordinates of \(A\equiv ({ x }_{ 1 },{ y }_{ 2 })\) = (-6, 10),
Coordinates of B \(\equiv \) (x₂, y₂) = (-4, 6) and 
Coordinates of C \(\equiv \) (x₃, y₃) = (3, -8).
We know that,
Area of triangle = \(\frac { 1 }{ 2 } \left| { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) \right| \)
\(\therefore\) Area of \(\Delta ABC\) = \(\frac { 1 }{ 2 } \left| -6\{ 6-(-8)\} +(-4)(-8-10)+3(10-6) \right| \)
\(=\frac { 1 }{ 2 } \left| -6(14)+(-4)(-18)+2(4) \right| \)
\(=\frac { 1 }{ 2 } \left| -84+72+12 \right| =0\)
Since, area of \(\Delta ABC\)  is zero. So, points A, B and C are collinear.
Now, \(AB=\sqrt { { (-4+6) }^{ 2 }+{ (6-10) }^{ 2 } } \)
[ using distance formula ]
\(=\sqrt { { 2 }^{ 2 }+{ (-4) }^{ 2 } } =\sqrt { 4+16 } =\sqrt { 20 } \)
\(=2\sqrt { 5 } units\)
\(AC=\sqrt { ({ 3+6) }^{ 2 }+(-8-10)^{ 2 } } \)
\(=\sqrt { { 9 }^{ 2 }+({ -18 })^{ 2 } } =\sqrt { 81+324 } \)
\(=\sqrt { 405 } =\sqrt { 81\times 5 } =9\sqrt { 5 } units\)
\(\therefore \quad AB=2\sqrt { 5 } \times \frac { 9 }{ 9 } =\frac { 2 }{ 9 } AC\)      
Hence proved.

Let A = (x₁,y₁) = (-4, 0), B = (x₂,y₂) = (4, 0) and C = (x₃, y₃) = (0, 3)
Now, AB = \(\sqrt { { [4-(-4)] }^{ 2 }+{ (0-0) }^{ 2 } } \) [ using distance formula ]
\(=\sqrt { { (4+4) }^{ 2 } } =\sqrt { { 8 }^{ 2 } } =8\quad units\)
\(BC=\sqrt { (0-4{ ) }^{ 2 }+{ (3-0) }^{ 2 } } =\sqrt { { (-4) }^{ 2 }+{ (3) }^{ 2 } } \)
\(=\sqrt { 16+9 } =\sqrt { 25 } =5\quad units\)
and AC \(=\sqrt { [0-(-4){ ] }^{ 2 }+(3-0{ ) }^{ 2 } } \)
\(=\sqrt { 16+9 } =\sqrt { 25 } =5\quad units\)
\(\because BC = AC\)
So, \(\triangle \) ABC is an isosceles triangle.

scalene triangle

A(4, -8), B(3, 6) and C(5, -4) are vertices of AABC and D is the mid-point of BC

Coordinates of D are [Given] 
\(\left( \frac { 3+5 }{ 2 } ,\frac { 6+(-4) }{ 2 } \right) =\left( \frac { 8 }{ 2 } ,\frac { 8 }{ 2 } \right) =\left( 4,1 \right) \)
\(\frac { AP }{ PD } =2\quad \) [Given]
\(\Rightarrow \ AP:PD\ =\ 2:1\)
⇒ Coordinates of P are \(\left( \frac { 2\times 4+1\times 3 }{ 2+1 } ,\frac { 2\times 1+1\times 6 }{ 2+1 } \right) =\left( \frac { 8+3 }{ 3 } ,\frac { 2+6 }{ 3 } \right) =\left( \frac { 11 }{ 3 } ,\frac { 8 }{ 3 } \right) \)

\(AB=\sqrt { ({ 1+2) }^{ 2 }+({ 2-3) }^{ 2 } } =\sqrt { 9+1 } =\sqrt { 10 } \)
\(BC=\sqrt { ({ 7-1) }^{ 2 }+({ 0-2) }^{ 2 } } =\sqrt { 36+4 } =\sqrt { 40 } =2\sqrt { 10 } \)
\(AC=\sqrt { ({ 7+2) }^{ 2 }+({ 0-3) }^{ 2 } } =\sqrt { 81+9 } =\sqrt { 90 } =3\sqrt { 10 } \)
Since  \(AB+BC=\sqrt { 10 } +2\sqrt { 10 } =(1+2)\sqrt { 10 } =3\sqrt { 10 } =AC\)
Hence, the points A, B and C are collinear.

Let the vertices of \(\triangle ABC\) be A(1, -1), B(0, 4) and C(-5,3)
Using distance formula, 
\(AB=\sqrt { { \left( 1-0 \right) }^{ 2 }+{ \left( -1-4 \right) }^{ 2 } } \)
\(=\sqrt { 1+{ 5 }^{ 2 } } =\sqrt { 26 } \)
\(BC=\sqrt { { \left( -5-0 \right) }^{ 2 }+{ \left( 3-4 \right) }^{ 2 } } \)
\(=\sqrt { 25+1 } =\sqrt { 26 } \)
\(AC=\sqrt { { \left( -5-1 \right) }^{ 2 }+{ \left( 3+1 \right) }^{ 2 } } \)
\(=\sqrt { 36+16 } =\sqrt { 52 } =2\sqrt { 13 } \)
\(\Rightarrow \quad AB=BC\neq AC\)
\(\Rightarrow \quad \triangle ABC\) is isosceles.
Now, using mid-section formula, the co-ordinates of mid-point of BC are
\(x=\frac { -5+0 }{ 2 } =-\frac { 5 }{ 2 } \)
\(y=\frac { 3+4 }{ 2 } =\frac { 7 }{ 2 } \)
\(\Rightarrow \quad D\left( x,y \right) =\left( -\frac { 5 }{ 2 } ,\frac { 7 }{ 2 } \right) \)
Length of median AD
\(=\sqrt { { \left( \frac { -5 }{ 2 } -1 \right) }^{ 2 }+{ \left( \frac { 7 }{ 2 } +1 \right) }^{ 2 } } \)
\(=\sqrt { { \left( \frac { -7 }{ 2 } \right) }^{ 2 }+{ \left( \frac { 9 }{ 2 } \right) }^{ 2 } } \)
\(=\sqrt { \frac { 130 }{ 4 } } =\sqrt { \frac { 130 }{ 2 } } \)
Length of median AD is \(\frac { \sqrt { 130 } }{ 2 } \) units.

Diagonals of a parallelogram bisect each other.
Mid-points of AC and BD are same
\(\Rightarrow \quad \left( 3,\frac { -1+y }{ 2 } \right) =\left( 3,2 \right) \)
\(\frac { -1+y }{ 2 } =2\quad \Rightarrow \quad y=5\)

Let  P(9a-2, -b) divides AB internally in the ratio 3: 1.
\(\therefore \) By section formula, we get 9a-2 = \(\frac { 3(8a)+1(3a+1) }{ 3+1 } \) and -b = \(\frac { 3(5)+1(-3) }{ 3+1 } \) [\(\because \) for internally ratio, coordinates are \(\left( \frac { { m }_{ 1 }{ x }_{ 2 }+{ m }_{ 2 }{ x }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } ,\frac { { m }_{ 1 }{ y }_{ 2 }+{ m }_{ 2 }{ y }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) \)]
\(\Rightarrow 9a-2=\frac { 24a+3a+1 }{ 4 } \) and \(-b=\frac { 15-3 }{ 4 } \)
\(\Rightarrow 9a-2=\frac { 27a+1 }{ 4 }\)  and \(-b=\frac { 12 }{ 4 } \)
\(\Rightarrow \) 36a-8 = 27a+1 and b = -3
\(\Rightarrow \)  9a-9 =0  and b=-3
\(\Rightarrow \)  a=1   and  b=-3
Now,\({ a }^{ 2 }+{ b }^{ 2 }={ (1) }^{ 2 }+{ (-3) }^{ 2 }\) = 1+9 = 10 units.