Let P(4, 3), A(9, 3), B(7, –1) and C(–1, 3)
If P is the centre of the circle which passes through the points A, B, and C, then P is equidistant from A, B and C (i.e.) PA = PB = PC
By distance formula,
\(d=\sqrt{(x_2 -x_1)^2 +(y_2 -y_1)^2}\)
\(AP=PA =\sqrt{(4-9)^2+(3-3)^2}=\sqrt{(-5)^2+0}=\sqrt{25}=5\)
\(BP=PB =\sqrt{(4-7)^2+(3+1)^2}=\sqrt{(3)^2+(4)^2}=\sqrt{9+16}=\sqrt{25}=5\)
\(CP=CB =\sqrt{(4+1)^2+(3-3)^2}=\sqrt{(5)^2+0}=\sqrt{25}=5\)
PA = PB = PC, Radius = 5
Therefore P is the centre of the circle, passing through A, B and C

Distance =\(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+(y_{ 2 }-y_{ 1 })^{ 2 } } \)
AB = \(\sqrt { (7-3)^{ 2 }+(6+2)^{ 2 } } =\sqrt { (4)^{ 2 }+(8)^{ 2 } } \)
= \(\sqrt { 16+64 } =\sqrt { 80 } \)
BC = \(\sqrt { (-1-7)^{ 2 }+(2-6)^{ 2 } } =\sqrt { (-8)^{ 2 }+(-4)^{ 2 } } \)
= \(\sqrt { 64+16 } =\sqrt { 80 } \)
CD = \(\sqrt { (-5+1)^{ 2 }+(-6-2)^{ 2 } } =\sqrt { (-4)^{ 2 }+(-8)^{ 2 } } \)
= \(\sqrt { 16+64 } =\sqrt { 80 } \)
AD =\(\sqrt { (-5-3)^{ 2 }+(-6-2)^{ 2 } } =\sqrt { (-8)^{ 2 }+(-4)^{ 2 } } \)
= \(\sqrt { 64+16 } =\sqrt { 80 } \)
AB = BC = CD = AD = \(\sqrt { 80 } \)
All the four sides are equal
∴  ABCD is a rhombus

Distance = \(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+(y_{ 2 }-y_{ 1 })^{ 2 } } \)
AB = \(\sqrt { (2-1)^{ 2 }+(1-1)^{ 2 } } \)
= \(\sqrt { (1)^{ 2 }+(0)^{ 2 } } =\sqrt { 1 } =1\)
BC = \(\sqrt { (2-1)^{ 2 }+(2-1)^{ 2 } } \)
= \(\sqrt { (0)^{ 2 }+(1)^{ 2 } } =\sqrt { 1 } =1\)
CD =\(\sqrt { (1-2)^{ 2 }+(2-2)^{ 2 } } \)
= \(\sqrt { (-1)^{ 2 }+(0)^{ 2 } } =\sqrt { 1 } =1\)
AD =\(\sqrt { (1-1)^{ 2 }+(2-1)^{ 2 } } \)
= \(\sqrt { (0)^{ 2 }+(1)^{ 2 } } =\sqrt { 1 } =1\)
AB = BC = CD = AD =1
All the four sides are equal.
ABCD is a rhombus

(-1,1), (1, 3) and (3, a)
Distance =\(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+({ y }_{ 2 }-y_{ 1 })^{ 2 } } \)
AB =\(\sqrt { (1+1)^{ 2 }+(3-1)^{ 2 } } \)
= \(\sqrt { (2)^{ 2 }+(2)^{ 2 } } =\sqrt { 4+4 } =\sqrt { 8 } \)
BC =\(\sqrt { (3-1)^{ 2 }+(a-3)^{ 2 } } \)
= \(\sqrt { (2)^{ 2 }+(a-3)^{ 2 } } =\sqrt { 4+(a-3)^{ 2 } } \)
But given AB = BC ⇒\(\sqrt { 8 } =\sqrt { 4+(a-3)^{ 2 } } \)
a-3 = 2 (or) a - 3 = -2
a = 2+3 (or) a = 3-2
a = 5 (or) a = 1
∴ 4+(a - 3)² = 8
(a - 3)² = 8 - 4
(a - 3)² = 4
a - 3 = \(\sqrt { 4 } =\pm 2\)
∴ The value of a = 5 or a = 1

(i) The point P (0, -4), Q(0, -1), R(0, 3), S(0, 6) lies on the y-axis.
(ii) Distance = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
(a) PR=\(\sqrt{(0-0)^2+(3+4)^2}\)
\(=\sqrt{0+7^2}=\sqrt{49}=7 \ uint\)
(b) QS =\(\sqrt{(0-0)^2+(6+1)^2}\)
\(=\sqrt{0+7^2}=\sqrt{49}=7 \ uint\)

Measure BD and DC.
BD = 3.3 cm and DC = 3.3 cm
\(\therefore\) BD=DC=3.3 cm
Measure CE and EA.
CE =4 cm andEA= 4 cm
\(\therefore\) CE=EA=4cm
Distance =\(\sqrt{(x_2+x_1)^2+(y_2-y_1)^2}\)
CG =\(\sqrt{(-3+3)^2+(3-7)^2}\)
\(=\sqrt{0+(-4)^2}=\sqrt{16}=4 \ uints\)
GF =\(\sqrt{(-3+3)^2+(3-1)^2}\)  (The point F is (-3,1) )
\(=\sqrt{0+2^2}=\sqrt{4}=2 \ uints\)
AG =\(\sqrt{(-3-1)^2+(3-0)^2}\)
\(=\sqrt{(-4)^2+(3)^2}=\sqrt{16+9}=\sqrt{25}=5 \ units\)
GD =\(\sqrt{(-3+5)^2+(3-4.5)^2}\)
\(=\sqrt{2^2+(-1.5)^2}=\sqrt{4+2.25}=\sqrt{6.25}=2.5 \ uints\)
BG =\(\sqrt{(-7+3)^2+(2-3)^2}\)
\(=\sqrt{(-4)^2+(-1)^2}=\sqrt{16+1}\)
\(=\sqrt{17}=4.12 \ units\) (Approximately BG = 4)
GE =\(\sqrt{(-3+1)^2+(3-3.5)^2}\)
\(=\sqrt{2^2+(0.5)^2}=\sqrt{4+0.25}\)
\(=\sqrt{4.25}=2.06 \ units\) (Approximately GE = 2)
AG : GD = 5 : 2.5
= 1 : 0.5
=2: 1
BG: GE =4: 2
= 2: 1
CG: GF =4: 2
= 2: 1
AG : GD = BG : GE = CG = GF = 2 : 1

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).

Let A(−2, −1) and B(4, 8) are the given points.

Let P(a,b) and Q(c,d) be the points of trisection of AB, so that AP = PQ = QB .
By the formula proved above,
P is the point.
\(\left( \frac { { x }_{ 2 }+{ 2x }_{ 1 } }{ 3 } ,\frac { { y }_{ 2 }+{ 2y }_{ 1 } }{ 3 } \right) =\left( \frac { 4+2(-2) }{ 3 } ,\frac { 8+2(-1) }{ 3 } \right) \) = (0, 2)
Q is the point
\(\left( \frac { { 2x }_{ 2 }+{ x }_{ 1 } }{ 3 } ,\frac { {2 y }_{ 2 }+{ y }_{ 1 } }{ 3 } \right) =\left( \frac { 2(4)-2 }{ 3 } ,\frac { 2(8)-1 }{ 3 } \right) \) = (2, 5)

Given points are A(–3, 6) and B(1, –2), P(–2, 4) divide AB internally in the ratio m : n.
By section formula,
\(P(x, y)=P\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)\)
= P(−2, 4) .......(1)
Here x₁ = −3, y₁ = 6, x₂ = 1, y₂ = −2
\((1)\Rightarrow \left( \frac { m(1)+n(-3) }{ m+n } ,\frac { m(-2)+n(6) }{ m+n } \right) \)
Equating x-coordinates, we get
\(\frac{m-3 n}{m+n}=-2\) or m− 3n = −2m− 2n
3m = n
\(\frac{m}{n}=\frac{1}{3}\)
m : n = 1: 3
Hence P divides AB internally in the ratio 1 : 3.

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)