Here D is the mid-point of BC, then
\(D=\left( \frac { -2+5 }{ 2 } ,\frac { 3+6 }{ 2 } \right) =\left( \frac { 3 }{ 2 } ,\frac { 9 }{ 2 } \right) \)
Length of AD = \(\sqrt { \left( 3-\frac { 3 }{ 2 } \right) ^{ 2 }+\left( 4-\frac { 9 }{ 2 } \right) ^{ 2 } } \)
\(=\sqrt { \left( \frac { 3 }{ 2 } \right) ^{ 2 }+\left( -\frac { 1 }{ 2 } \right) ^{ 2 } } \)
\(=\sqrt { \frac { 9 }{ 4 } +\frac { 1 }{ 4 } } =\sqrt { \frac { 10 }{ 4 } } \)
\(=\frac { \sqrt { 5 } }{ 2 } \) unit
Area of \(\triangle ABD\)
\(=\frac { 1 }{ 2 } \left[ 3\left( 3-\frac { 9 }{ 2 } \right) -2\left( \frac { 9 }{ 2 } -4 \right) +\frac { 3 }{ 2 } \left( 4-3 \right) \right] \)
\(=\frac { 1 }{ 2 } \left[ -\frac { 9 }{ 2 } -1+\frac { 3 }{ 2 } \right] \)
\(=\frac { 1 }{ 2 } \left[ \frac { -9-2+3 }{ 2 } \right] =\frac { 8 }{ 4 } \)
= - 2 = 2 sq. unit.
Area of \(\triangle ACD\)
\(=\frac { 1 }{ 2 } \left[ 3\left( 6-\frac { 9 }{ 2 } \right) +5\left( \frac { 9 }{ 2 } -4 \right) +\frac { 3 }{ 2 } \left( 4-6 \right) \right] \)
\(=\frac { 1 }{ 2 } \left[ \frac { 9 }{ 2 } +\frac { 5 }{ 2 } -3 \right] =\frac { 1 }{ 2 } \left[ \frac { 9+5-6 }{ 2 } \right] \)
\(=\frac { 8 }{ 4 } =2\) sq. unit.
Hence, Area of \(\triangle ABD\) = Area of \(\triangle ACD\)

Let X-axis divides the line joining the points A(1, -5) and B(-4, 5)in the ratio k : 1 at point M.

By section formula, we get
Coordinates of M = \(\left(\frac{-4 k+1}{k+1}, \frac{5 k-5}{k+1}\right)\)   ...(i)
Since, point M lies on X-axis.
\(\therefore\) y-coordinate of M = 0
On equating y-coordinate of M to 0, we get
\(\frac{5 k-5}{k+1}=0 \Rightarrow 5 k-5=0 \Rightarrow k=1\)
On putting k = 1 in Eq. (i) we get
\(\begin{aligned} M=\left(\frac{-4 \times 1+1}{1+1}, \frac{5 \times 1-5}{1+1}\right) & =\left(\frac{-4+1}{2}, \frac{5-5}{2}\right) \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad M & =\left(\frac{-3}{2}, 0\right) \end{aligned}\)
So, the required ratio is 1 : 1 and the coordinates of point of division M is \(\left(\frac{-3}{2}, 0\right)\)

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.

Let (– 4, 6) divide AB internally in the ratio m₁ : m₂.Using the section formula, we get
\((-4,6)=\left(\frac{3 m_{1}-6 m_{2}}{m_{1}+m_{2}}, \frac{-8 m_{1}+10 m_{2}}{m_{1}+m_{2}}\right)\) 
Recall that if (x, y) = (a, b) then x = a and y = b.
So,\(-4=\frac{3 m_{1}-6 m_{2}}{m_{1}+m_{2}}\) and \(6=\frac{-8 m_{1}+10 m_{2}}{m_{1}+m_{2}}\) 
Now, \(-4=\frac{3 m_{1}-6 m_{2}}{m_{1}+m_{2}}\) gives us
– 4m₁ – 4m₂ = 3m₁ – 6m₂
i.e., 7m₁ = 2m₂
i.e., m₁ : m₂ = 2 : 7
You should verify that the ratio satisfies the y-coordinate also.
Now,\(\frac{-8 m_{1}+10 m_{2}}{m_{1}+m_{2}}=\frac{-8 \frac{m_{1}}{m_{2}}+10}{\frac{m_{1}}{m_{2}}+1}\) ,(Dividing throughout by m₂)
\(=\frac{-8 \times \frac{2}{7}+10}{\frac{2}{7}+1}=6\) 
Therefore, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in the ratio 2 : 7.
Alternatively : The ratio m₁ : m₂ can also be written as \(\frac{m_{1}}{m_{2}}: 1\) ,or k : 1. Let (– 4, 6) divide AB internally in the ratio k : 1. Using the section formula, we get
\((-4,6)=\left(\frac{3 k-6}{k+1}, \frac{-8 k+10}{k+1}\right)\) 
So, \(-4=\frac{3 k-6}{k+1}\) 
i.e., – 4k – 4 = 3k – 6
i.e., 7k = 2
i.e., k : 1 = 2 : 7
You can check for the y-coordinate also.
So, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in the ratio 2 : 7.

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

\(\because\) Points (x,y), (1,2) and (7,0) are colinear
\(\because\) Area of the triangle formed by these pointer is 0.
⇒ \(\frac{1}{2}\)[x(2-0)+1(0-y)+7(y-2)=0
⇒ [2x+1(-y)+7y-14]=0 ⇒ 2x-y+7y-14=0
⇒ 2x+6y-14=0 ⇒ x+3y-7=0

Let A(3, 0), B(4, 5), C(-1, 4) and D(-2, -1) be the vertices of the rhombus ABCD.
\(\begin{aligned} & \therefore \text { Diagonal, } A C=\sqrt{(-1-3)^2+(4-0)^2} \\ \end{aligned}\)
\(\begin{aligned} & {\left[\because \text { distance }=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\right]} \\ \end{aligned}\)
\(\begin{aligned} & =\sqrt{(-4)^2+4^2} \\ \end{aligned}\)
\(\begin{aligned} & =\sqrt{32}=4 \sqrt{2} \text { units } \end{aligned}\)
Diagonal, \(\begin{aligned} B D & =\sqrt{(-2-4)^2+(-1-5)^2} \\ \end{aligned}\)
\(\begin{aligned} & =\sqrt{(-6)^2+(-6)^2}=\sqrt{36+36} \\ \end{aligned}\)
\(\begin{aligned} & =\sqrt{72}=6 \sqrt{2} \text { units } \end{aligned}\)
\(\therefore\) Area of the rhombus ABCD
\(\begin{aligned} & =\frac{1}{2} \times \text { Product of its diagonals } \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{2} \times A C \times B D=\frac{1}{2} \times 4 \sqrt{2} \times 6 \sqrt{2} \\ \end{aligned}\)
\(\begin{aligned} & =2 \times 6 \times \sqrt{2} \times \sqrt{2} \\ \end{aligned}\)
\(\begin{aligned} & =12 \times 2=24 \text { sq units } \end{aligned}\)

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]
\(\begin{aligned} & =\sqrt{(4-2)^2+(1-3)^2} \end{aligned}\)
\(\begin{aligned} & =\sqrt{(2)^2+(-2)^2}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2} \text { units } \end{aligned}\)

Distance of the point,
\(a=\sqrt { { \left( 5-3 \right) }^{ 2 }+{ \left( 8-4 \right) }^{ 2 } } \)
\(=\sqrt { 4+16 } =\sqrt { 20 } =2\sqrt { 5 } \)
\(\because 2\sqrt { 5 } \) is less than 7
\(\therefore\)  The point lies inside the circle.

Here, given two points are A(x,y) and O(0,0)
\(\therefore \left| AO \right| =\sqrt { { (x-0) }^{ 2 }+{ (y-0) }^{ 2 } } =\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \)

2AP = AB

Let coordinates of A are (x,0) and coordinates of B are (0,y)

Using section formula, we get
\(4=\frac { 5\times 0+3\times x }{ 5+3 } \)
\(\Rightarrow \ 32=3x\ \Rightarrow \ x=\frac { 32 }{ 3 } \)
Similarly, 5=\(\frac { 5\times y+3\times 0 }{ 5+3 } \)
⇒40=5y ⇒ y=8
∴ Coordinates of A are \(\left( \frac { 32 }{ 3 } ,0 \right) \)and
Coordinates of B are (0,8).

No

Distance between P(x,y) and A(3,6) is equal to the distance between P(x,y) and B(-3,4).
⇒    PA=PB
\(\Rightarrow \sqrt { \left( x-3 \right) ^{ 2 }+\left( y-6 \right) ^{ 2 } } =\sqrt { \{ x-\left( -3 \right) ^{ 2 }+\left( y-4 \right) ^{ 2 } } \)
\(\Rightarrow \sqrt { x^{ 2 }+9-6x+y^{ 2 }+36-12y } =\sqrt { x^{ 2 }+9+6x+y^{ 2 }+16-8y } \)
Squaring both sides, we get
\(\Rightarrow \sqrt { x^{ 2 }+9-6x+y^{ 2 }+36-12y } =\sqrt { x^{ 2 }+9+6x+y^{ 2 }+16-8y } \)
\(\Rightarrow \quad 6x+6x-8y-36-12y=0\quad \Rightarrow \quad 12x+4y-20=0\quad \Rightarrow 3x+y=5\)

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

AB = \(\sqrt { { (1-5 })^{ 2 }+{ (5-6) }^{ 2 } } =\sqrt { 17 } \)
BC = \(\sqrt { { (2-1 })^{ 2 }+{ (1-5) }^{ 2 } } =\sqrt { 17 } \)
CD = \(\sqrt { { (6-2 })^{ 2 }+{ (2-1) }^{ 2 } } =\sqrt { 17 } \)
AD = \(\sqrt { { (6-5 })^{ 2 }+{ (2-6) }^{ 2 } } =\sqrt { 17 } \)
AC = \(\sqrt { { (2-5 })^{ 2 }+{ (1-6) }^{ 2 } } =\sqrt { 17 } \)
Now, AB=BC=CD=AD
and AC² = 34
AB²+BC² =17+17=34
∴ AC²=AB²+BC²
⇒ ∠ABC = 90⁰
∴ ABCD is a square.  

AB = \(\sqrt { ({ 2-7) }^{ 2 }+{ (3-5) }^{ 2 } } =\sqrt { 25+4 } =\sqrt { 29 } \)
BC = \(\sqrt { ({ 6-2) }^{ 2 }+{ (-7-3) }^{ 2 } } =\sqrt { 16+100 } =\sqrt { 116 } \)
CA = \( \sqrt { ({ 7-6) }^{ 2 }+{ (5+7) }^{ 2 } } =\sqrt { 1+144 } =\sqrt { 145 } \)
Since AB²+BC² = 29+116 = 145 =CA².
∴ △ABC is right angled at B.
Area = \(\frac{1}{2} AB \times BC = \frac {1}{2}=\sqrt{29}.\sqrt{116}=\frac {1}{2}\sqrt{29.2}.{2}\sqrt{29}=29\)