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 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 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 A(3,5), B(8,−10) be the given points and let the point P(x, y) divides the line segment AB internally in the ratio 3:2.
By section formula,
P(x, y) =P\(\left( \frac { m{ x }_{ 2 }+n{ x }_{ 1 } }{ m+n } ,\frac { m{ y }_{ 2 }+n{ y }_{ 1 } }{ m+n } \right) \)
Here x₁ = 3, y₁ = 5, x₂ = 8, y₂ = −10 and m = 3, n = 2
Therefore P(x, y) = P\(\left( \frac { 3(8)+2(3) }{ 3+2 } ,\frac { 3(-10)+2(5) }{ 3+2 } \right) \) = P(6, -4)

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)

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
\(\frac { 2-3+k }{ 3 } =-1\)
\(\therefore \sqrt { { (h+k) }^{ 2 }+{ (h+3k) }^{ 2 } } \)
\(=\sqrt { (18+(-2))^{ 2 }+{ (18+3(-2)) }^{ 2 } } \)
\(=\sqrt { { 16 }^{ 2 }+{ 12 }^{ 2 } } =\sqrt { 256+144 } =\sqrt { 400 } =20\)

D (x, y) is the Mid point BC
\(\therefore \ D(x,y)=\left( \frac { 1+5 }{ 2 } ,\frac { -1+1 }{ 2 } \right) \)
\(=\left( \frac { 6 }{ 2 } ,\frac { 0 }{ 2 } \right) \)
= (3, 0)
AD is the median through A
Here A x₂ y₂ D x₃ y₃
(-1,3)     (3,0)
\( \therefore\) Length of AD \(=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (3-(-1)) }^{ 2 }+{ (0-3) }^{ 2 } } \)
\(=\sqrt { { (3+1) }^{ 2 }+{ (-3) }^{ 2 } } =\sqrt { 16+9 } =\sqrt { 25 } =5\) units 

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)

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

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
Here (-5,11)
x₂ y₂
(4,-7)
m  n
7 : 2
\(=P(x,y)\left( \frac { { mx }_{ 2 }+{ nx }_{ 1 } }{ m+n } ,\frac { { my }_{ 2 }+{ ny }_{ 1 } }{ m+n } \right) \)
\(=\left( \frac { 7\times 4+2\times -5 }{ 7+2 } ,\frac { 7\times -7+2\times 11) }{ 7+2 } \right) \)
\(=\left( \frac { 28-10 }{ 9 } ,\frac { -49+22 }{ 9 } \right) \)
\(=\left( \frac { 18 }{ 9 } ,\frac { -27 }{ 9 } \right) =(2,-3)\)

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

Mid point of AC

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

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

In squares the diagonals are equal and bisect each other
∴ Mid point of BD = Mid point of AC
\(\left( \frac { -1+x }{ 2 } ,\frac { -2+y }{ 2 } \right) =\left( \frac { -5+5 }{ 2 } ,\frac { 4+2 }{ 2 } \right) \)
\(\frac { -1+x }{ 2 } =\frac { 0 }{ 2 } \quad \frac { -2+y }{ 2 } =\frac { 6 }{ 2 } \)
-1 + x = -2 + y = 6
x =1  y = 8
∴ The vertex D(x, y) = (1, 8)

Mid point = \(\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \)
Mid point of AB=\(\left( \frac { 8+10 }{ 2 } ,\frac { 6+10 }{ 2 } \right) \)
\(=\left( \frac { 18 }{ 2 } ,\frac { 6 }{ 2 } \right) =(9,3)\)
Mid point M(x,y) = Mid point of OD
\(=\left( \frac { 0+9 }{ 2 } ,\frac { 0+3 }{ 2 } \right) =\left( \frac { 9 }{ 2 } ,\frac { 3 }{ 2 } \right) =(4.5,1.5) \)

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)