Since O is mid point of BC and coordinates of C are (0.-3)
Coordinate of B are (0,3)
Now, AO will be the perpendicular bisector of BC. Therefore, A will lie on  x-axis. Let coordinates of A are (x, 0). 
Now AB=BC
\(\Rightarrow \)  \(\sqrt { (x-0)^{ 2 }+(0-3)^{ 2 } } =6\)
 \(\sqrt { { x }^{ 2 }+9 } =6\)
x² +9 =36
\(\Rightarrow \) x² =27 x=\(\pm \) 3 \(\sqrt 3\)
 Coordinates of A are or (3\(\sqrt 3\),0) or (-3\(\sqrt 3\) ,0)
When A is ,(3\(\sqrt 3\) ,0)then D will be (-3\(\sqrt 3\) ,0) so that BACD is a rhombus

Let P(2, - 2). Q (-2. l) and R(5, 2) are vertices of the right-angled triangle. 
PQ = \(\sqrt { (-2-2)^{ 2 }+[1=(-2)]^{ 2 } } \)
= \(\sqrt { 16+9 } \)  =\(\sqrt { 25 } \)=5 units
QR=  \(\sqrt { [5-(-2)]^{ 2 }+(2-1)^{ 2 } } \)
   \(\sqrt { { 7 }^{ 2 }+1^{ 2 } } =\sqrt { 49+1 } =\sqrt { 50 } \)  units
PR=   \(\sqrt { (2-5)^{ 2 }+(-2-2)^{ 2 } } \)
=   \(\sqrt { { 3 }^{ 2 }+4^{ 2 } } =\sqrt { 9+6 } \)        
= \(\sqrt { 25 } \) =5 units
PQ²+PR²=52+52 =50 =QR²
\(\therefore \triangle \) PQR is right angled traingle
Hypotenuse  =\(\sqrt { 50 } \) units  =5 \(\sqrt { 2 } \)units
Area of \(\triangle \) PQR 
=\(\frac { 1 }{ 2 } \) |x₁(y₂-y₃)+x₂(y₃-y₂)+x₃(y₁-y₂)|
= \(\frac { 1 }{ 2 } \)|2(1-2)+(-2){2-(-2)}+5(-2-1)|
= \(\frac { 1 }{ 2 } \)2(-1)+(-2)(4)+5(-3)|
= \(\frac { 1 }{ 2 } \) |-25|=\(\frac { 25 }{ 2 } \)  sq.units

Let points P and Q trisect the line joining the points.

∴ AP=PQ=QB
P divides AB in the ration 1:2 and Q divides AB in the ration 2:1
P (coordinate x) = \(\frac { 1\times \left( -2 \right) +2\times 4 }{ 1+2 } =\frac { 6 }{ 3 } \)=2; P (coordinate y) = \(\frac { 1\times \left( -3 \right) +2\times \left( -1 \right) }{ 1+2 } =-\frac { 5 }{ 3 } \)
The coordinates of P are \(\left( 2,\frac { 5 }{ 3 } \right) \)
Q(x-coordinate) =  \(\frac { 2\times \left( -2 \right) +1\times \left( 4 \right) }{ 2+1 } =-\frac { -4+4 }{ 3 } =0\) Q(y-coordinates) = \(\frac { 2\times \left( -3 \right) +1\times \left( -1 \right) }{ 2+1 } =-\frac { -7 }{ 3 } \)
The coordinates of Q are \(\left( 0,-\frac { 7 }{ 3 } \right) \).

Let A(2, 3) and B(4, 1) be the given points.
Here, x₁ = 2, y₁ = 3 and x₂ = 4, y₂ = 1
Now, AB = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
[by distance formula]
\(\begin{aligned} & =\sqrt{(4-2)^2+(1-3)^2} \end{aligned}\)
\(\begin{aligned} & =\sqrt{(2)^2+(-2)^2}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2} \text { units } \end{aligned}\)

\(\therefore \quad \frac { 1 }{ 2 } \left[ { { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) } \right] =0\)
\(\Rightarrow \quad \frac{1}{2}[0(2-y)+1(y-0)+x(0-2)]=0\)
\(\Rightarrow \quad \frac { 1 }{ 2 }[y-2x]=0\)
\(\Rightarrow \quad 2x-y=0\)

Here, x₁ = \(10\ \cos { 30° } \), y₁=0 and x₂=0, y₂=\(10\).\( \cos { 60° } \)
Distance between the points = \(\sqrt { { \left( 0-10\quad \cos { 30° } \right) }^{ 2 }+{ \left( 10\quad \cos { 60° } -0 \right) }^{ 2 } } \)  [\(\because \) distance=\(\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }-{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } } \)]
\(=\sqrt { { \left( 10 \right) }^{ 2 }{ \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }+{ \left( 10 \right) }^{ 2 }{ \left( \frac { 1 }{ 2 } \right) }^{ 2 } } =\sqrt { \frac { 300 }{ 4 } +\frac { 100 }{ 4 } } \quad \left[ \because \quad \cos { 30° } =\frac { \sqrt { 3 } }{ 2 } \quad and\quad \cos { 60° } =\frac { 1 }{ 2 } \right] \)
\(=\sqrt { \frac { 400 }{ 4 } } =\sqrt { 100 } =10\) units

Here, (x₁, y₁)=(3, -7), (x₂, y₂) = (-8, 6) and (x₃, y₃) = (5, 10)
\(\therefore \) Coordinates of the centroid of a triangle are \(\left( \frac { 3-8+5 }{ 3 } ,\frac { -7+6+10 }{ 3 } \right) \) i.e. (0, 3).
\(\therefore\) centroid of triangle\(=\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) \)

It is clear from the figure, perpendicular distance of A from the Y-axis is 5 units.

Here, |PQ|=10
⇒ PQ²=100
⇒ (x-2)²+(5+3)²=100
⇒ x²-4x+4+64=100
⇒ x²-4x-3²=0
⇒ (x-8)(x+4)=0
⇒ x=8 or -4
Hence, the values of x are -4 and 8

Coordinates of the centroid are :
\(\left( \frac { 8+8+2 }{ 3 } ,\frac { 7-2-2 }{ 3 } \right) i.e.,(6,1)\)

\(AB=\sqrt { \left( 0-0 \right) ^{ 2 }+\left( 6-0 \right) ^{ 2 } } =6\quad units\\ BC=\sqrt { \left( 8-0 \right) ^{ 2 }+\left( 0-0 \right) ^{ 2 } } =8\quad units\\ AC=\sqrt { \left( 0-8 \right) ^{ 2 }+\left( 6-0 \right) ^{ 2 } } =10\quad units\)
∴ Perimeter of ΔABC=6+8+10=24 units

∵ P(-1,3), Q(2,p) and C(5,-1) are colinear
∴ x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)=0
⇒ -1{p-(-1)}+2(-1-3)+5(3-p)=0 ⇒ -1(p+1)+2(-4)+5(3-p)=0
⇒ -p-1-8+15-5p =0 ⇒ -6p+6=0 ⇒ -6p=-6 ⇒ p=1

\(5\sqrt { 2 }\ \ units\)

\(\frac { 3\times x+4\times 2 }{ 3+4 } =-1\ \Rightarrow \ \frac { 3x+8 }{ 7 } =-1\)
3x+8=-7 ⇒ 3x=-15
x=-5
∴ Coordinates of B are(-5,-2).
\(\frac { 3\times y+4\times 2y5 }{ 3+4 } =2\ \Rightarrow \ \frac { 3y+20 }{ 7 } =2\)
3y+20=14 ⇒ 3y=14-20
3y=-6 ⇒ y=-2.

(-2, -3)

m = 6

5 units

Let coordinates of the required point be R(x, y) this means R divides the join of P(4, -3) and Q(8, 5) in the ratio 3:1 internally.

Using the formula for internal division.
\(\left( \frac { { kx }_{ 2 }+{ x }_{ 1 } }{ k+1 } ,\frac { { ky }_{ 2 }+{ y }_{ 1 } }{ k+1 } \right) \)
\(x=\frac { 3(8)+1(4) }{ 3+1 } =\frac { 24+4 }{ 4 } =\frac { 28 }{ 4 } =7\) and \( y=\frac { 3(5)+1(-3) }{ 3+1 } =\frac { 15-3 }{ 4 } =\frac { 12 }{ 4 } =3\)
Thus, the coordinates of R (7, 3) divides PQ in the ratio 3:1.

\(AB=\sqrt { (14-{ 2) }^{ 2 }+(10+2)^{ 2 } } =12\sqrt { 2 } units\)
\(BC=\sqrt { (11-{ 14) }^{ 2 }+(13-10)^{ 2 } } =3\sqrt { 2 } units\)
\(CD=\sqrt { (-1-{ 11) }^{ 2 }+(1-13)^{ 2 } } =12\sqrt { 2 } units\)
\(AD=\sqrt { (-1-{ 2) }^{ 2 }+(1+2)^{ 2 } } =3\sqrt { 2 } units\)
⇒ AB=CD and BC = AD
∴ ABCD is a || gm
Now, \(AC=\sqrt { (11-2)^{ 2 }+({ 13+2) }^{ 2 } } =\sqrt { 306 } \)
⇒ AC²=306 units, AB² = 288 units, BC² =18 units
AB²+BC²=306 units
⇒ AC²=AB²+BC²⇒∠ABC = 90⁰
⇒ ABCD is a rectangle.

True because distance between centre (origin) and Q(6,8) is greater than its radius. i.e., 5

Let A(-2,-3), B(1,6) is divides by P in ratio 1:2.

⇒ \(P\left( \frac { 1\times 1+2(-2) }{ 1+2 } ,\frac { 6\times 1+2(-3) }{ 1+2 } \right) \)
⇒ P\((\frac{1-4}{6},\frac{6-6}{3})\)
⇒ P(-1,0)

Since point P(1,0) is equidistant from Q(3,s) and R(s, -3).
∴ |PQ| = |PR|
⇒ \(\sqrt { ({ 3-1) }^{ 2 }+({ s-0) }^{ 2 } } =\quad \sqrt { ({ s-1) }^{ 2 }+({ -3-0) }^{ 2 } } \)
Squaring both sides, we have
⇒ 4+s²=s²+1-2s+9
⇒ 2s=6
⇒ s=3
Hence, the value of s is 3. 

P divides the joining of A and B in ratio 1:4.
The coordinates of P are \((\frac {6+4}{1+4}, \frac{7+8}{1+4})\) i.e., P(2,3).
Q divides the joining of A and B in the rario 2:3.
The coordinates of Q are \((\frac{12+3}{2+3}, \frac{14+6}{1+4})\) i.e., Q(3,4).
R divides the joining of A and B in the ratio 3:2.
The coordinates of R are \((\frac{18+2}{3+2},\frac{21+4}{3+2})\), i.e., R(4,5).

Yes

AB = \(\sqrt { { (3-0 })^{ 2 }+(\sqrt { 3 } -0)^{ 2 } } \) = \(\sqrt{9+3}\) =\(\sqrt {12}\) = 2\(\sqrt{3}\)
∴ Side of equilateral △ = 2\(\sqrt{3}\) units
Area of equilateral △ = \(\frac{\sqrt 3} {4}\) (Sides)²
=\(\frac {\sqrt {3}} {4} \times (2\sqrt {3})^{2} = 3{\sqrt{3}}\) sq.units