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).

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 POQ be the right angled triangle and O be placed at the origin. Let OQ = a units and OP be b units. Let us name the coordinates of P as (0,b) and Q as (a,0).
By mid-point formula, if M is the mid-point of the hypotenuse PQ [PM=MQ], then M is

\(\left( \frac { a+0 }{ 2 } ,\frac { b+0 }{ 2 } \right) =\left( \frac { a }{ 2 } ,\frac { b }{ 2 } \right) \)
We now use the distance formula and find that
OM=\(\sqrt { { \left( \frac { a }{ 2 } -0 \right) }^{ 2 }{ +\left( \frac { b }{ 2 } -0 \right) }^{ 2 } } =\sqrt { \frac { { a }^{ 2 } }{ 4 } +\frac { { b }^{ 2 } }{ 4 } } \) which is the same value as
QM=\(\sqrt { { \left( a-\frac { a }{ 2 } \right) }^{ 2 }{ +\left( 0-\frac { b }{ 2 } \right) }^{ 2 } } =\sqrt { \frac { { a }^{ 2 } }{ 4 } +\frac { { b }^{ 2 } }{ 4 } } \) and similarly PM=\(\sqrt { \frac { { a }^{ 2 } }{ 4 } +\frac { { b }^{ 2 } }{ 4 } } \)
This shows OM = QM = PM, which we desired to prove.

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 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 =\(\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) \)
\(\therefore G(x,y)=\left( \frac { 3+(-2)+5 }{ 3 } ,\frac { 4+(-1)+3 }{ 3 } \right) \)
\(=\left( \frac { 8-2 }{ 3 } ,\frac { 7-1 }{ 3 } \right) =\left( \frac { 6 }{ 3 } ,\frac { 6 }{ 3 } \right) =(2,2)\)
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)

Dividing point \((x,y)=\left( \frac { { mx }_{ 2 }+{ nx }_{ 1 } }{ m+n } ,\frac { { my }_{ 2 }+{ ny }_{ 1 } }{ m+n } \right) \)
\((6,360)=\left( \frac { m\times 12+n\times 2 }{ m+n } ,\frac { m\alpha +n\times 180 }{ m+n } \right) \)
⇒\(\frac { 12m+2n }{ m+n } =6\)
12m + 2n = 6m + 6n
12m - 6m = 6n -2n
6m=4n

m=\(\frac{2}{3}\)n  ...(1)
\(\Rightarrow \frac { m\alpha +180n }{ m+n } =360\)   ...(2)
\(\Rightarrow \frac { \frac { 2 }{ 3 } n\alpha +180n }{ \frac { 2 }{ 3 } n+n } =360\)

\(\Rightarrow \frac { 2 }{ 3 } \alpha =240+360-180\)
\(\frac { 2 }{ 3 } \alpha =600-180\)
\(\frac { 2 }{ 3 } \alpha =420\)

\(\alpha\)=630
\(\therefore\)The car will reach 630km at 12midnight

\(=\left( \frac { 2+(-3) }{ 2 } ,\frac { 2+(-2) }{ 2 } \right) \)
\(=\left( \frac { 0 }{ 2 } ,\frac { 0 }{ 2 } \right) =(0,0)\)
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