\(\frac { 3 }{ 2 } \)  sq unit

Given vertices of \(\Delta\)ABC are A(1, -1), B(-4, 6) and C (-3, -5), respectively.
Now, \(AB=\sqrt { { (-4-1) }^{ 2 }+{ (6+1) }^{ 2 } } \)   [by distance formula]
\(=\sqrt { { (-5) }^{ 2 }+{ (7) }^{ 2 } } =\sqrt { 25+49 } \)
\(=\sqrt { 74 } =8.6\quad units\)
\(BC=\sqrt { (-3+{ 4) }^{ 2 }+{ (-5-6) }^{ 2 } } \)
\(=\sqrt { { (1) }^{ 2 }+{ (-11) }^{ 2 } } =\sqrt { 1+121 } \)
\(=\sqrt { 122 } =11.01\quad units\)
\(\\ and\quad CA=\sqrt { { (-3-1) }^{ 2 }+{ (-5+1) }^{ 2 } } \)
\(=\sqrt { { (-4) }^{ 2 }+{ (-4) }^{ 2 } } =\sqrt { 16+16 } \)
\(=\sqrt { 32 } =5.7\quad units\)
Here, we can see that AB \(\neq \) BC \(\neq \) CA
and BC² \(\neq \) AB² \(\neq \) + CA²
So, given triangle is a scalene triangle.
Now, area of  \(\Delta\)ABC
\(=\frac { 1 }{ 2 } \left| { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) \right| \)
\(=\frac { 1 }{ 2 } \left| 1(6+5)+(-4)(-5+1)+(-3)(-1-6) \right| \)
\(=\frac { 1 }{ 2 } \left| 11-4(-4)-3(-7) \right| =\frac { 1 }{ 2 } \left| 11+16+21 \right| \)
\(=\frac { 1 }{ 2 } \left| 48 \right| =\frac { 1 }{ 2 } \times 48=24\quad sq\quad units\)

(i) Let A(2,-2) = A(x₁,y₁), B(4,8) = B(x₂,y₂) and mid - point of AB be C(x,y).
\(\therefore \ \ x=\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } \)
\(and\ y=\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \)
\(\Rightarrow \ \ \ x=\frac { 2+4 }{ 2 } \ and\ y=\frac { -2+8 }{ 2 } \)
\(\Rightarrow \ x=\frac { 6 }{ 2 } \ and\ y=\frac { 6 }{ 2 } \)
\(\therefore\) x = 3  and y = 3
Hence, the mid-point of A and B is (3,3).
(ii) Mid-point of line segment.
(iii) Enjoyment and intelligence..

Let coordinates of A(5,2)=(x₁,y₁) and coordinates of B(3,1) = (x₂,y₂)
Here, m₁=2 and m₂=1

Let the coordinates of the point of trisection be C(x, y).
By section formula,
\(x=\frac { { m }_{ 1 }{ x }_{ 2 }+{ m }_{ 2 }{ x }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \) and \(y=\frac { { m }_{ 1 }{ y }_{ 2 }+{ m }_{ 2 }{ y }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \)
\(\therefore \ x=\frac { 2\times 3+1\times 5 }{ 2+1 } =\frac { 6+5 }{ 3 } =\frac { 11 }{ 3 } \quad \) and  \(y=\frac { 2\times 1+1\times 2 }{ 2+1 } =\frac { 2+2 }{ 3 } =\frac { 4 }{ 3 } \)
\(\therefore \)  Coordinates of C(x, y) =\(\left( \frac { 11 }{ 3 } ,\frac { 4 }{ 3 } \right) \)
Hence, at point \(\left( \frac { 11 }{ 3 } ,\frac { 4 }{ 3 } \right) \), the carpenter cuts the wood.
(i) Being an environment "friendly, I would concern only the usable items not go for the luxurious one.
(ii) To avoid the use of wooden furniture and awareness of forestation.

Here, from the figure,
Coordinates of A = (1, 1), coordinates of B = (6, 1) and coordinates of C = \(\left( \frac { 7 }{ 2 } ,2 \right) \)
Also, altitude of \(\Delta \)ABC = CD= 7 - 1 = 6 cm
and base of \(\Delta \)ABC, AB = 6-1 = 5 cm
\(\therefore \) Area of one banner = Area of ​​​​​​\(\Delta \)ABC
=\(\frac { 1 }{ 2 } \times \)CD\(\times \)AB
=\(\frac { 1 }{ 2 } \times \)6\(\times \)5
=3\(\times \)5 = 15 cm²
\(\therefore \) Cost of 10 banners at the rate of Rs.2 per m²
=150\(\times \)2=Rs.300
(ii) Coordinate geometry
(iii) Social awareness.

square

\(P({13\over 2},{9\over 2})\), Q(8,5), R\(({19\over 2}, {11\over2})\)

We have \(\frac{PM}{PQ} = \frac {1}{4} = \frac {PN}{PR}\)
\(\Rightarrow \frac{PQ}{PM}=4\)
\(\Rightarrow \frac { PM-MQ }{ PM } =4 \Rightarrow 1+\frac { MQ }{ PM } =4\)
\(\Rightarrow \frac { MQ }{ PM } =3 \Rightarrow \frac { PM }{ MQ } =\frac { 1 }{ 3 } \)
∴ M divides PQ in the ratio 1:3
Coordinates of M are
\(\left( \frac { 1\times 1+3\times 4 }{ 1+3 } ,\frac { 1\times 5+3\times 6 }{ 1+3 } \right) \left( \frac { 13 }{ 4 } ,\frac { 23 }{ 4 } \right) \)
Similarly, N divides PR in the ratio 1:3
∴ Coordinates of N are
\(\frac { 1\times 7+3\times 4 }{ 1+3 } ,\frac { 1\times 2+3\times 6 }{ 1+3 } =\left( \frac { 19 }{ 4 } ,5 \right) \)
Area of △PMN
=\(\frac { 1 }{ 2 } \left| 4\left( \frac { 23 }{ 4 } -5 \right) +\frac { 13 }{ 4 } (5-6)+\frac { 19 }{ 4 } \left( 6-\frac { 23 }{ 4 } \right) \right| \)
=\(\frac{15}{32}\)sq.units

Are of  \(\triangle \)PAB =10 sq units
\(\Rightarrow \) \(\frac { 1 }{ 2 } \) |x(2-8)+1(8-y)+3(y-2)| = 10
\(\Rightarrow \)  |-6x+8-y+3y-6| =20
\(\Rightarrow \) |-6x+2y+2| = 20 
\(\Rightarrow \) -6x+2y+2 = 20 or  -6x+2y+2 =- 20
\(\Rightarrow \) -6(17-3y)+2y+2 =20 or -6(17-3y)+2y+2=-20    ...(ii)
\(\Rightarrow \) -102+18y+2y+2=20  -102  -102+18y+2y+2=20  -20
 \(\Rightarrow \) 20y=120  20y=80
y=6 y=4
When y = 6, eq. (i) becomes x = 17-18 =-1 
\(\therefore \)  Point is (-1,6)
When y = 4, eq. (i) becomes x = 17-12 =5
\(\therefore \)  Point is (5,4)

find mid point O of A and C
x=(3-6)/2
= -3/2
y=(-4+2)/2 = -1
O=(-3/2,-1)
taking O as mid point find D
-3/2=(-1+x1)/2
x1=-2
-1=(-3+y1)/2
y1=1
D=-2,1