Let the vertices of a triangle be A(1, −6), B(−5, 2) and C(x₃, y₃)
Given the centroid of a triangle as (−2, 1) we get,
\(\frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } \) = -2
\(\frac { 1-5+{ x }_{ 3 } }{ 3 } \) = -2
−4 + x₃ = −6
x₃ = −2
\(\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \) = 1
\(\frac { 6+2+{ y }_{ 3 } }{ 3 } \) = 1
−4 + y₃ = 3
y₃ = 7
Therefore, third vertex is (−2, 7).

The centroid G(x, y) of a triangle whose vertices are (x₁, y₁), (x₂ , y₂ ) and (x₃ , y₃) is given by

G(x,y)=G\(\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 }3 }{ } \right) \)
We have (x₁, y₁) = (6, −1); (x₂ , y₂ ) = (8, 3); (x₃ , y₃) = (10, −5)
The centroid of the triangle
G(x, y) =G\(\left( \frac { 6+8+10 }{ 3 } ,\frac { -1+3-5 }{ 3 } \right) \)
= G \((\frac{24}{3},\frac{-3}{3})\)= G(8, -1)

Let P, Q, R be the points on the line segment joining A(−11, 4) and B(9, 8) such that AP = PQ = QR = RB .
Here Q is the mid-point of AB, P is the mid-point of AQ and R is the mid-point of QB.
Q is the mid-point of AB =\(\left( \frac { -11+9 }{ 2 } ,\frac { 4+8 }{ 2 } \right) =\left( \frac { -2 }{ 2 } ,\frac { 12 }{ 2 } \right) \)= (-1, 6)
P is the mid-point of AQ =\(\left( \frac { -11-1 }{ 2 } ,\frac { 4+6 }{ 2 } \right) =\left( \frac { -12 }{ 2 } ,\frac { 10 }{ 2 } \right) \) = (-6, 5)
R is the mid-point of QB =\(\left( \frac { -1+9 }{ 2 } ,\frac { 6+8 }{ 2 } \right) =\left( \frac { 8 }{ 2 } ,\frac { 14 }{ 2 } \right) \) = (4, 7)
Hence the points which divides AB into four equal parts are P(–6, 5), Q(–1, 6) and R(4, 7).

Let A(x, 3), B(6, y), C(8, 2) and D(9, 4) be the vertices of the parallelogram ABCD. By definition, diagonals AC and BD bisect each other.

Mid-point of AC = Mid-point of BD
\(\left( \frac { x+8 }{ 2 } ,\frac { 3+2 }{ 2 } \right) =\left( \frac { 6+9 }{ 2 } ,\frac { y+4 }{ 2 } \right) \)
equating the coordinates on both sides, we get
\(\frac { x+8 }{ 2 } =\frac { 15 }{ 2 } \)
x + 8 = 15
x = 7
\(\frac { 5 }{ 2 } =\frac { y+4 }{ 2 } \)
5 = y + 4
y = 1
Hence, x = 7 and y = 1.

Let the coordinates of A be (x₁, y₁) and the given point is B(5,−6). Since the centre is the mid-point of the diameter AB, we have
\(\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } \) = 3
x₁ + 5 = 6
x₁ = 6 − 5
x₁ = 1
\(\frac { { y}_{ 1 }+{ y }_{ 2 } }{ 2 } \) = -4
y₁ − 6 = −8
y₁ = −8 + 6
y₁ = −2

Therefore, the coordinates of A is (1, −2) .

Let A(−3, 5) and B(x₂, y₂ ) be the given two points.
Given P(−2, 3) divides AB internally in the ratio 1:6.
By section formula, P\(\left( \frac { m{ x }_{ 2 }+n{ x }_{ 1 } }{ m+n } ,\frac { m{ y }_{ 2 }+n{ y }_{ 1 } }{ m+n } \right) \) = P(-2, 3)
P\(\left( \frac { 1({ x }_{ 2 })+6(-33) }{ 1+6 } ,\frac { 1({ y }_{ 2 })+6(5) }{ 1+6 } \right) \) = P(-2, 3)
Equating the coordinates
\(\frac { { x }_{ 2 }-18 }{ 7 } \) = -2
x₂ −18 = −14
x₂ = 4
\(\frac { { y }_{ 2 }-30 }{ 7 } \) = 3
y₂ + 30 = 21
y₂ = −9
Therefore, the coordinate of B is (4, −9)

Let the vertices of the ABC be A(x₁, y₁), B(x₂, y₂ ) and C(x₃, y₃) and the given mid-points of the sides AB, BC and CA are (5, 1), (3, −5) and (−5, −1) respectively. Therefore
\(\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } =5 \Rightarrow\)x₁ + x₂ = 10 ...(1)
\(\frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } =3 \Rightarrow\)x₂ + x₃ = 6 ...(2)
\(\frac { { x }_{ 3 }+{ x }_{ 1 } }{ 2 } =-5 \Rightarrow\)x₃ + x₁ = –10 ...(3)
Adding (1), (2) and (3)
2x₁ + 2x₂ + 2x₃ = 6
x₁ + x₂ + x₃ = 3 ...(4)
(4) − (2) \(\Rightarrow\) x₁ = 3 − 6 = −3 
(4) − (3) \(\Rightarrow\) x₂ = 3 +10 = 13
(4) − (1) \(\Rightarrow\) x₃ = 3 −10 = −7
\(\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } =1\)\(\Rightarrow\)y₁ + y₂ = 2 …(5)
\(\frac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } =-5\)\(\Rightarrow\)y₂ + y₃ = –10 …(6)
\(\frac { { y }_{ 3 }+{ y }_{ 1 } }{ 2 } =1\)\(\Rightarrow\)y₃ + y₁ = –2 …(7)
Adding (5), (6) and (7),
2y₁ + 2y₂ + 2y₃ = −10
y₁ + y₂ + y₃ = −5 ...(8)
(8) − (6) \(\Rightarrow\) y₁ = −5 +10 = 5
(8) − (7) \(\Rightarrow\) y₂ = −5 + 2 = −3
(8) − (5) \(\Rightarrow\) y₃ = −5 − 2 = −7
Therefore the vertices of the triangles are A(−3, 5), B(13, −3) and C(−7, −7).

"The centroid of the triangle obtained by joining the mid points of the sides of a triangle is the same as the centroid of the original triangle."
\(\therefore\) The mid points of the sides of the triangle are given as
  x₁y_{1 ,} x₂y, x₃y₃
\(\left( \frac { 3 }{ 2 } ,5 \right) \left( 7,\frac { -9 }{ 2 } \right) \left( \frac { 13 }{ 2 } ,\frac { -13 }{ 2 } \right) \)
\(\therefore\)Centroid \(=\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) \)
\(=\left( \frac { \frac { 3 }{ 2 } +\frac { 7 }{ 1 } +\frac { 13 }{ 2 } }{ 3 } ,\frac { 5+\left( \frac { -9 }{ 2 } \right) +\frac { (-13) }{ 2 } }{ 3 } \right) =\left( \frac { \frac { 3+14+13 }{ 2 } }{ 3 } ,\frac { \frac { 10-9-13 }{ 2 } }{ 3 } \right) \)
\(=\left( \frac { \frac { 30 }{ 2 } }{ 3 } ,\frac { \frac { -12 }{ 2 } }{ 3 } \right) =\left( \frac { 15 }{ 3 } ,\frac { -6 }{ 3 } \right) =(5,-2)\)

Centroid G (x,y) = (4, -2)
two vertices (x₁, y₁) = (3, -2)
(x₂, y₂) = (5, 2), (x₃, y₃) =?

\(\frac { 8+{ x }_{ 3 } }{ 3 } =4\ \ \frac { { y }_{ 3 } }{ 3 } =-2\)
8 + x₃ = 12  y₃ = -6
x₃ = 4
∴ The third vertex (x₃, y₃) = (4, -6)

x₁ y₁ x₂ y₂
A(-2, -3) B(2, 1)
m : n = 3 : 1
The point P divides AB in the ratio 3 : 1
\(P(x,y)=\left( \frac { { mx }_{ 2 }+{ nx }_{ 1 } }{ m+n } ,\frac { { my }_{ 2 }+{ ny }_{ 1 } }{ m+n } \right) \)
\(=\left( \frac { 3\times 2+1\times -2 }{ 3+1 } ,\frac { 3\times 1+1\times -3) }{ 3+1 } \right) \)
\(=\left( \frac { 6-2 }{ 4 } ,\frac { 3-3 }{ 4 } \right) =\left( \frac { 4 }{ 4 } ,\frac { 0 }{ 4 } \right) =(1,0)\)
P is at 25% distance from B on its left and C is at 25% distance from B on its right
\(\therefore\) B is the mid point of PC
Mid point of  \(\bar { PC } =\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \)
\((2,1)=\left( \frac { 1+{ x }_{ 2 } }{ 2 } ,\frac { 0+{ y }_{ 2 } }{ 2 } \right) \)
\(\frac { 1+{ x }_{ 2 } }{ 2 } =2\quad \frac { { y }_{ 2 } }{ 2 } =1\)
1 + x₂ = 4 y₂ = 2
⇒ x₂ = 3, y₂ = 2
∴ C(x₂, y₂) = (3, 2) is the solutions.

\(\bar { AB } =\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (-1-6) }^{ 2 }+{ (-4-3) }^{ 2 } } \)
\(=\sqrt { { (-7) }^{ 2 }+{ (-7) }^{ 2 } } =\sqrt { 49+49 } =\sqrt { (49)2 } \)
\(\frac { 1 }{ 2 } AB=\frac { 7\sqrt { 2 } }{ 2 } =\frac { 7\sqrt { 2 } }{ \sqrt { 2 } \times \sqrt { 2 } } =\frac { 7 }{ \sqrt { 2 } } \)
\(AB=\left( \frac { 6+(-1) }{ 2 } ,\frac { 3+(-4) }{ 2 } \right) =\left( \frac { 5 }{ 2 } ,\frac { -1 }{ 2 } \right) \)
Mid point of \(AB=\left( \frac { 6+(-1) }{ 2 } ,\frac { 3+(-4) }{ 2 } \right) =\left( \frac { 5 }{ 2 } ,\frac { -1 }{ 2 } \right) \)
SinceC is at the distance \(\frac{1}{2}\) AB from B
Midpoint of MC = B
\(\left( \frac { \frac { 5 }{ 2 } +{ x }_{ 3 } }{ 2 } ,\frac { \frac { -1 }{ 2 } +{ y }_{ 3 } }{ 2 } \right) =(-1,-4)\)
\(\left( \frac { \frac { 5 }{ 2 } +{ x }_{ 3 } }{ 2 } \right) =-1\)
\(\frac { 5 }{ 2 } +{ x }_{ 3 }=-2\)
\({ x }_{ 3 }=-2-\frac { 5 }{ 2 } \)
\(=\frac { -4-5 }{ 2 } \)
\(=\frac { -9 }{ 2 } \)
\(\therefore C({ x }_{ 3 },{ y }_{ 3 })=\left( \frac { -9 }{ 2 } ,\frac { -15 }{ 2 } \right) \)
\(\left( \frac { -\frac { 1 }{ 2 } +{ y }_{ 3 } }{ 2 } \right) =-4\)
|\(-\frac { 1 }{ 2 } +{ y }_{ 3 }=-8\)
\({ y }_{ 3 }=-8+\frac { 1 }{ 2 } \)
\(=\frac { -16+1 }{ 2 } \)
\(=\frac { -15 }{ 2 } \)
Similarly by Mid point of DM = A(6, 3)
\(\left( \frac { { x }_{ 4 }+\frac { 5 }{ 2 } }{ 2 } ,\frac { { y }_{ 4 }+\left( \frac { -1 }{ 2 } \right) }{ 2 } \right) =(6,3)\)
\(\frac { { x }_{ 4 }+\frac { 5 }{ 2 } }{ 2 } =6\)
\({ x }_{ 4 }+\frac { 5 }{ 2 } =12\)
\({ x }_{ 4 }=12-\frac { 5 }{ 2 } =\frac { 24-5 }{ 2 } =\frac { 19 }{ 2 } \)
\(\frac { { y }_{ 4 }+\left( \frac { -1 }{ 2 } \right) }{ 2 } =3\)
\({ y }_{ 4 }+\left( \frac { -1 }{ 2 } \right) =6\)
\({ y }_{ 4 }=6+\frac { 1 }{ 2 } =\frac { 12+102 }{ 2 } =\frac { 13 }{ 2 } \)
\(\therefore\) The other end D (x₄, y₄)\(=\left( \frac { 19 }{ 2 } ,\frac { 13 }{ 2 } \right) \)

x₁ y₁   x₂ y₂  
A(-3, 5) B(4, -9) P(2, -5)
\(=P(x,y)\left( \frac { { mx }_{ 2 }+{ nx }_{ 1 } }{ m+n } ,\frac { { my }_{ 2 }+{ ny }_{ 1 } }{ m+n } \right) \)
\(P(2,-5)=\left( \frac { m\times 4+n(-5) }{ m+n } ,\frac { m\times (-9)+n(5) }{ m+n } \right) \)
\(\left( \frac { 4m-3n }{ m+n } \right) =2\)
4m -3n = 2m+ 2n
4m -2m = 2n + 3n
2m = 5n
\(\frac { m }{ n } =\frac { 5 }{ 2 } \)
∴ The ratio m : n = 5 : 2

Mid point of Diagonal AC:
\(=\left( \frac { -a+a }{ 2 } ,\frac { 0+0 }{ 2 } \right) =\left( \frac { 0 }{ 2 } ,\frac { 0 }{ 2 } \right) =(0,0)\)
Mid point Diagonal of BD:
\(=\left( \frac { 0+0 }{ 2 } ,\frac { b+(-b) }{ 2 } \right) =\left( \frac { 0 }{ 2 } ,\frac { 0 }{ 2 } \right) =(0,0)\)
Mid point of AC = Mid point of BD
∴ Diagonals bisect each other

In a parallelogram diagonals bisect each other and diagonals are not equal.
Mid point of DE \(=\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \)
(-3, 6)\(=\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \)
\(\frac { x_{ 1 }+{ x }_{ 2 } }{ 2 } =-3\ \ \ \ \ \frac { y_{ 1 }+y_{ 2 } }{ 2 } =6\)
x₁ + x₂ = -6........(1)   
y₁ + y₂ = 12.......(2)
Mid point of EF\(\left( \frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } ,\frac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } \right) \)
(0, 7)=\(\left( \frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } ,\frac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } \right) \)
\(\frac { x_{ 2 }+{ x }_{ 3 } }{ 2 } =0\ \ \ \frac { y_{ 2 }+{ y }_{ 3 } }{ 2 } =7\)
x₂+x₃= 0..............(3)
Mid point of FD=\(\left( \frac { { x }_{ 3 }+{ x }_{ 1 } }{ 2 } ,\frac { { y }_{ 3 }+{ y }_{ 1 } }{ 2 } \right) =(-3,-2)\)
x₃ + x₁ = -6.......(5)     
y₃ + y₁ = -4..........(6)

∴ D (x₁, y₁) = (-6, -3)
Mid points of the diagonals are equal in parallelogram
∴ We have to prove this
Mid point of AC \(=\left( \frac { (-3)+(-3) }{ 2 } ,\frac { 6+(-2) }{ 2 } \right) =\left( \frac { -6 }{ 2 } ,\frac { 4 }{ 2 } \right) =(-3,2)\)
Mid Point of BD \(=\left( \frac { -6+0 }{ 2 } ,\frac { -3+7 }{ 2 } \right) =\left( \frac { -6 }{ 2 } ,\frac { 4 }{ 2 } \right) =(-3,2)\)
∴ Mid point of AC = Mid point of BD
∴ ABCD is a parallelogram

Mid point
\(M(x,y)=\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \)
Mid point AB(2, 4)=\(\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \)
\(\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } =2\Rightarrow { x }_{ 1 }+{ x }_{ 2 }=4\ \ \ ...(1)\)
\(\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } =4\Rightarrow { y }_{ 1 }+{ y }_{ 2 }\ \ \ ...(2)\)
Mid point of BC (-2, 3) =\(\left( \frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } ,\frac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } \right) \)
\(\left( \frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } \right) =-2\Rightarrow { x }_{ 2 }+{ x }_{ 3 }=-4 \quad ...(3)\)
\(\left( \frac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } \right) =3\Rightarrow { y }_{ 2 }+y_{ 3 }=6\ \quad ...(4)\)
Mid point of AC (5, 2) =\(\left( \frac { { x }_{ 1 }+{ x }_{ 3 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 3 } }{ 2 } \right) \)
\(\left( \frac { x_{ 1 }+{ x }_{ 3 } }{ 2 } \right) =3\Rightarrow x_{ 1 }+x_{ 3 }=10 \quad ...(5)\)
\(\frac { { y }_{ 1 }+{ y }_{ 3 } }{ 2 } =3\Rightarrow { y }_{ 1 }+y_{ 3 }=4 \quad ...(6)\)


Substitue x₁ = 9 in (5)
         9 +X₃  = 10
              X₃  = 1
Substitue X₁ = 1 in(3)
x₂ + 1 = -4 ⇒ x₂ = -5

substitute  y₁ = 3 in (6)
3 + y₃ = 4
y₃ = 1
substitute y₃ = 1 in (4)
1 + y₂ = 6
y₃ = 5
ஃ The vertices of the triangle A(x₁ y₁) = (9, 3)
B (x₂ y₂) = (-5, 5)
 C(x₃y₃) = (1, 1)

\(M(x,y)=\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \)
\((-4,2)=\left( \frac { -3+{ x }_{ 2 } }{ 2 } ,\frac { 7+{ y }_{ 2 } }{ 2 } \right) \left( { x }_{ 1 }{ y }_{ 1 } \right) \)
\(\frac { -3+{ x }_{ 2 } }{ 2 } =-4\quad \frac { 7+{ y }_{ 2 } }{ 2 } =2\)
-3 + x₂ = -8
7 +y₂ = 4
x₂ = -8 +3
y₂ = 4 -7 = -3
x₂ = -5
The other end is (-5, -3)