\(G\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) =\left( \frac { 2+5+9 }{ 3 } ,\frac { -5+11+9 }{ 9 } \right) =\left( \frac { 16 }{ 3 } ,5 \right) \)

\(120=\frac { 8(50)+4(y) }{ 12 } \)
1440 = 400 + 4y
4y = 1040
\(y=\frac { 1040 }{ 4 } =260\ km\)

\((9,-3)=\left( \frac { m(12+n(7) }{ m+n } ,\frac { m(1)+n(-5) }{ m+n } \right) \)
\(\frac {13m+37n}{ m+n } =9\)
13m + 7n = 9m + 9n
4m = 2n
\(\frac { m }{ n } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \)
\(\frac{m-5n}{m+nn}=-3\)
m -5n = -3m -3n
m + 3m= 5n-3n
4m=2n
\(\frac { m }{ n } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \)

"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 =\(\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) \)
\(\therefore G(x,y)=\left( \frac { 3+(-2)+5 }{ 3 } ,\frac { 4+(-1)+3 }{ 3 } \right) \)
\(=\left( \frac { 8-2 }{ 3 } ,\frac { 7-1 }{ 3 } \right) =\left( \frac { 6 }{ 3 } ,\frac { 6 }{ 3 } \right) =(2,2)\)
In a parallelogram diagonals bisect each other
∴ Mid point of DG = Mid point of BC
\(\left( \frac { x+2 }{ 2 } ,\frac { y+2 }{ 2 } \right) =\left( \frac { -2+5 }{ 2 } ,\frac { -1+3 }{ 2 } \right) \)
\(\frac { x+2 }{ 2 } =\frac { 3 }{ 2 } \)
x+2 = 3
x = 3 - 2 = 1
\(\frac { y+2 }{ 2 } =\frac { 2 }{ 2 } \)
y = 2 - 2 = 0
y + 2 = 2
\(\therefore\) The co-ordinates of the vertex D (x, y) = (1, 0)

Vertices of a triangle
(x₁ ,y₁) = (1,2)
(x₂,y₂) = (h,-3)
(x₃, y₃) = (-4, k)
Centroid G (x,y) = (5, -1)
\(G=\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) =(5,-1)\)
\(\left( \frac { 1+h+(-4) }{ 3 } ,\frac { 2+(-3)+k }{ 3 } \right) =(5,-1)\)
\(\Rightarrow \frac { -3+h }{ 3 } =5\)
-3 + h = 15
h = 18
\(\frac { 2-3+k }{ 3 } =-1\)
\(\therefore \sqrt { { (h+k) }^{ 2 }+{ (h+3k) }^{ 2 } } \)
\(=\sqrt { (18+(-2))^{ 2 }+{ (18+3(-2)) }^{ 2 } } \)
\(=\sqrt { { 16 }^{ 2 }+{ 12 }^{ 2 } } =\sqrt { 256+144 } =\sqrt { 400 } =20\)

Mid point of AC

Mid point of BC
\(=\left( \frac { 2a+2b }{ 2 } ,\frac { 0+2c }{ 2 } \right) \)

Distance between two points \(=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } \)
\(\therefore \bar { AB } =\sqrt { { (2a-0) }^{ 2 }+({ 0-0) }^{ 2 } } =\sqrt { { 4a }^{ 2 } } =2a\)

\(=\left( \frac { 2+(-3) }{ 2 } ,\frac { 2+(-2) }{ 2 } \right) \)
\(=\left( \frac { 0 }{ 2 } ,\frac { 0 }{ 2 } \right) =(0,0)\)
Distance between two points
\(d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } \)
\(\bar { OA } =\sqrt { { (-3-0) }^{ 2 }+{ (2-0) }^{ 2 } } \)
\(=\sqrt { { (-3) }^{ 2 }+{ 2 }^{ 2 } } =\sqrt { 9+4 } =\sqrt { 13 } \)
\(\bar { OB } =\sqrt { { (-3-0) }^{ 2 }+{ (2-0 })^{ 2 } } =\sqrt { 9+4 } =\sqrt { 13 } \)
\(\bar { OC } =\sqrt { { (-3-0) }^{ 2 }+{ (-2-0) }^{ 2 } } =\sqrt { 9+4 } =\sqrt { 13 } \)
\(\therefore \overline { OA } =\overline{ OB } =\overline { OC } \) Hence Proved

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)

\(M(x,y)=\left( \frac { 3+p }{ 2 } ,\frac { 4+7 }{ 2 } \right) \)
\(\frac { 3+p }{ 2 } =x\quad \frac { 4+7 }{ 2 } =y\)
\(y=\frac { 11 }{ 2 } \)
substitute the values of (x, y) in 2x + 2y +1 = 0

3 + p + 11 + 1 = 0
P + 15 = 0
p = -15