Let the coordinates of B and C be (a,b) and (x,y) respectively
Then \(\left( \frac { 1+a }{ 2 } ,\frac { -4+b }{ 2 } \right) \)  =(2.-1)

\(\Rightarrow \) \(\frac { 1+a }{ 2 } =2\quad and\quad \frac { -4+b }{ 2 } =-1\)
\(\Rightarrow \) 1+a=4 and -4+b=-2
\(\Rightarrow \) a=3 and b=2
Also  \(\left( \frac { 1+x }{ 2 } ,\frac { -4+y }{ 2 } \right) \)=(0,-1)
\(\Rightarrow \) \(\frac { 1+x }{ 2 } =0\quad and\quad \frac { -4+y }{ 2 } =-1\)
\(\Rightarrow \)  1+x=0 and -4+y=-2
\(\Rightarrow \)   x=-1 and y=2
Thus the coordinates of the vertices of ABC are A(1,-4),B(3,2) and C(-1,2)
Area of ABC =\(\frac { 1 }{ 2 } \) |1(2-2)+3(2+4)+(-1)(-4-2)|
= \(\frac { 1 }{ 2 } \)|0+18+6|
= \(\frac { 1 }{ 2 } \)(24)=12 sq.units 

(0,0)

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 A(-5,7) and B(-1,3)
\(AB=\sqrt{(-1+5)^2+(3-7)^2}\)=\(\sqrt{16+16}\)=\(\sqrt{32}\)
=\(\sqrt{16X2}=4\sqrt2 units\)

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\(=\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) \)

True.
Here, P(0, -7) lies the y-axis
\(\left| PA \right| =\sqrt { { \left( -1-0 \right) }^{ 2 }+{ \left( 0+7 \right) }^{ 2 } } \) 
 \(=\sqrt { 1+49 } =\sqrt { 50 } \) units
\(\left| PB \right| =\sqrt { { \left( -7-0 \right) }^{ 2 }+{ \left( -6+7 \right) }^{ 2 } } \) 
\(=\sqrt { 49+1 } =\sqrt { 50 } \) units
P(0, -7) lies on the y-axis and its distance from both A(-1, 0) and B(7, -6) is \(\sqrt { 50 } \) units.

Here, P (0, y), A (1, 5) and B (4, 6) AP=BP
⇒ \(\sqrt { \left( 0-1 \right) ^{ 2 }+\left( y-5 \right) ^{ 2 } } =\sqrt { \left( 0-4 \right) ^{ 2 }+\left( y-6 \right) ^{ 2 } } \)
⇒ 1+y²+25-10y=16+y²+36-12y
⇒ 2y=26
⇒ y=13
Thus, required point is (0,13).

Let (x,y) be the coordinates of third vertex of the triangle.
\(\therefore \frac { x-3+0 }{ 3 } =0\) and \(\frac { y+1-2 }{ 3 } =0\) 
\(\Rightarrow \ x=3\ \) and x=1
Hence, the coordinates of the third vertex is (3,1).

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

\({-5-1\over2}={a\over2}\)
-6=a
a=-6

(11,2)

Area of triangle
=\(\frac{1}{2}\)|x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)
⇒\(\frac{1}{2}\)|x(1-2)+1(2-0)+0(0-1)|=4
⇒ |-x+2|=8 ⇒ -x+2=土8
⇒ -x=8-2 or -x=-8-2
⇒ x=-6 or x=10

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

No

The given points are (1,0) and (4, p)
\(\mathrm{d}=\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}\)
\( \Rightarrow 5=\sqrt{(4-1)^2+\mathrm{p}^2}\)
\( \Rightarrow 25=(4-1)^2+\mathrm{p}^2\)
\(\Rightarrow 25-9=\mathrm{p}^2 \)
\( \Rightarrow \mathrm{p}^2=16\)
\( \Rightarrow \mathrm{p}=+4,-4\)
p=±4

No

A(-6,7),B(-1,-5)
Let x₁=-6,y₁=7;x₂=-1,y₂=-5
Distance, \(AB=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } \)
\(=\sqrt { { 5 }^{ 2 }+\left( -12 \right) ^{ 2 } } =\sqrt { 25+144 } =\sqrt { 169 } =13\)
2AB=2x13=26

Distance between \((-{8\over5},2) \ and \ ({2\over5},2)\)
=\(\sqrt{(-{8\over5}-{2\over5})^2+(2-2)^2}=\sqrt4=2\)

Given vertices of a parallelogram are A (6, 1), B(8, 2), C(9, 4) and D(p, 3).
Here, we have to find the value of p i.e. vertex D (p, 3).
We know that diagonals of a parallelogram bisect each other.
\(\therefore \) Coordinates of mid-point of diagonal AC = Coordinates of mid-point of diagonal BD
\(\Rightarrow \left( \frac { 6+9 }{ 2 } ,\frac { 1+4 }{ 2 } \right) =\left( \frac { 8+p }{ 2 } ,\frac { 2+3 }{ 2 } \right)\)
\(\Rightarrow \left( \frac { 15 }{ 2 } ,\frac { 5 }{ 2 } \right) =\left( \frac { 8+p }{ 2 } ,\frac { 5 }{ 2 } \right) \)
On equating x-coordinate from both sides, we get
 \(\frac { 15 }{ 2 } =\frac { 8+p }{ 2 } \)
\(\Rightarrow 15=8+p\quad \)
\(\Rightarrow p=15-8\quad \)
\(\Rightarrow p=7\quad \)
Hence, the required value of p is 7.

Given, coordinates of point M = (5,-3) and coordinates of point N = (-3,-3)

(ii) Length of MN = Perpendicular distance of M from negative Y-axis + Perpendicular distance of N from negative Y-axis
= Abscissa of M +Abscissa of N (without sign)
= 5 + 3 = 8 units
(iii) Here, points A, 11 and C divide MN into 4 equal parts of length\(\frac { 8 }{ 4 } \),  i.e. 2 units.
 \(\therefore \) Coordinates of A = (3, - 3)
Coordinates of B = (1,-3)
and coordinates of C = (-1, - 3)

Let the coordinates of the fourth vertex D be  D(x, y). We know that diagonals of a parallelogram  bisect each other.
   
Therefore mid point of AC=midpoint of BD
\(\Rightarrow \) \(\left( \frac { { x }_{ 1 }+{ x }_{ 3 } }{ 2 } ,\frac { { y }_{ 1 }+y_{ 3 } }{ 2 } \right) =\left( \frac { { x }+{ x }_{ 2 } }{ 2 } ,\frac { { y }+y_{ 2 } }{ 2 } \right) \)
\(\Rightarrow \)  \(\frac { { x }_{ 1 }+{ x }_{ 3 } }{ 2 } =\frac { { x }+{ x }_{ 2 } }{ 2 } \)
and  \(\frac { y_{ 1 }+y_{ 3 } }{ 2 } =\frac { { y }+{ y }_{ 2 } }{ 2 } \)
\(\Rightarrow \) x₁+x₃=x+x₂ and y₁+y₃=y+y₂
\(\Rightarrow \)  x=x₁+x₃ +x₂ and y=y₁+y₃-y₂
Hence,the coordinates of the fourth vertex in terms of x₁,x₂,x₃,y₁,y₂ and y₃ are
(x₁+x₃-x₂,y₁+y₃-y₂)

Let the coordinates of the centre C of circle passing through the points P(0,0, Q(-2,1) and R(-3,2) and C(x,y). 
∴ |PC| = |QC| = |RC| = radius  --- (i)
|PC| = \(\sqrt { { (x-0) }^{ 2 }+({ y-0) }^{ 2 } } =\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \)
\(|QC|\ =\ \sqrt { { (x+2) }^{ 2 }+({ y-1) }^{ 2 } } \)
\(=\ \sqrt { { { x }^{ 2 }+4+4x+{ y }^{ 2 }+1-2y } } \)
\(=\sqrt { { { x }^{ 2 }+4x-2y+{ y }^{ 2 }+5 } } \)
\(|RC|\ =\ \sqrt { { (x+3) }^{ 2 }+({ y-2) }^{ 2 } } \)
\(=\sqrt { { { x }^{ 2 }+9+6x+{ y }^{ 2 }+4-4y } } \)
\(=\sqrt { { { x }^{ 2 }+6x-4y+{ y }^{ 2 }+ }13 } \)
From (i), we get
PC²=QC²=RC²
⇒ x²+y²=x²+4x-2y+y²+5
=x²+6x-4y+y²+13
⇒ 4x-2y=-5     ---(ii)
and 6x-4y=-13  --- (iii)
Multiplying (ii) by 2 and subtracting from (iii), we have
-2x=-3 ⇒ \(x=\frac{3}{2}\)
From (ii), we have
\(4\times \frac{3}{2}-2y\)=-5  ⇒ 6-2y=-5}
⇒ \(y=\frac{11}{2}\)
Hence,the coordinates of the centre of circle of circle are C\((\frac{3}{2},\frac{11}{2})\) and  radius = \(\sqrt { \frac { 9 }{ 4 } +\frac { 121 }{ 4 } } =\sqrt { \frac { 130 }{ 4 } } =\frac { \sqrt { 130 } }{ 2 } \)units.

Let the vertices of △ABC be A(x₁,y₁),  B(x₂,y₂) and C(x₃, y₃). Let D(3, 1), F.(5, 6) and F(- 3, 2) be the mid-points of BC, CA and AB respectively. 

then \(\frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } =3\ \Rightarrow \ { x }_{ 2 }+{ x }_{ 3 }=6\) ---- (i)
\(\frac { y_{ 2 }+{ y }_{ 3 } }{ 2 } =3\ \Rightarrow \ { y }_{ 2 }+{ y }_{ 3 }=2\) --- (ii)
\(\frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } =5\ \Rightarrow \ { x }_{ 3 }+{ x }_{ 1 }=10\) --- (iii) and \(\frac { y_{ 3 }+{ y }_{ 1 } }{ 2 } =6\ \Rightarrow \ { y }_{ 1 }+{ y }_{ 3 }=12\) --- (iv)
\(\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } =-3\ \Rightarrow \ { x }_{ 1 }+{ x }_{ 2 }=-6\) --- (v) and \(\frac { y_{ 1 }+{ y }_{ 2 } }{ 2 } =2\ \Rightarrow \ { y }_{ 1 }+{ y }_{ 2 }=4\) --- (vi)
Adding (i), (iii) and (v), we get
2(x₁+x₂+x₃) = 10 ⇒ x₁+x₂+x₃=5 --- (vii)
Subtrcting (i), (iii), (v) separtely from (vii), we get x₁=-1 , x₂=-5, x₃=11 
Adding (ii), (iv) and (vi), we get
2(y₁+y₂+y₃) = 18 ⇒ y₁+y₂+y₃=9  --- (viii)
Subtracting (ii), (iv) and (vi) separetely from (viii), we get y₁=7, y₂=-3, y₃=5.
Hence, the vertices of the triangle ABC are A(-1,7), B(-5,-3) and C(11,5).