The points A(2005, 96) and B(2015, 100) are on the line AB.
Slope of AB = \(\frac { 100-96 }{ 2015-2005 } =\frac { 4 }{ 10 } =\frac { 2 }{ 5 } \)
Let the growth of population in 2030 be k crores. Assuming that the point C(2030,k) is on AB, 
we have, slope of AC = slope of AB
\(\frac { k-96 }{ 2030-2005 } =\frac { 2 }{ 5 } \) gives \(\frac { k-96 }{ 25 } =\frac { 2 }{ 5 } \)
k - 96 = 10
k = 106
Hence the estimated population in 2030 = 106 Crores.

(a) Slope of AB = \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { -2+2 }{ 6-1 } =0\)
(b) Slope of CD \(\frac { 1-1 }{ 2-5 } =\frac { 0 }{ -3 } =0\)

Area of triangle AGF
Vertices A (- 5,3), G (- 4.5,0.5) and F (- 2,3).
Area of triangle \(=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+\right. \left.x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right] \text { sq. units } \)
\(=\frac{1}{2}[-5(0.5-3)-4.5(3-3)-2(3-0.5)] \)
\(=\frac{1}{2}[12.5-5]=\frac{7.5}{2}=3.75 \text { sq. units } \)

Area of the patio = Area of the quadrilateral ABCD - Area of the swimming pool EFGH
Area of Quadrilateral ABCD
A(- 4,- 8), B (8, - 4), C (6, 10) and D ( - 10, 6)
\(\text { Area } =\frac{1}{2}\left[\left(x_{1}-x_{3}\right)\left(y_{2}-y_{4}\right)-\left(x_{2}-x_{4}\right)\left(y_{1}-y_{3}\right)\right] \)
\(= \frac{1}{2}[(-4-6)(-4-6)-(8+10)(-8-10)] \)
\(= \frac{1}{2}[100+324]=\frac{424}{2}=212 \text { sq. units } \)
Area of Quadrilateral EFGH
E (- 3, - 5), F (6,- 2),G (3, 7) and H (- 6, 4)
Area \(=\frac{1}{2}[(-3,-3)(-2-4)-(6+6)(-5-7)] \)
\(=\frac{1}{2}[36+144]=\frac{180}{2}=90 \text { sq. units } \)
Area of patio = Area of quadrilateral ABCD - Area of Quadrilateral EFGH
= 212 - 90 = 122 sq. units

Given B Q, R are the mid points of the sides of a triangle.
Let the vertices of triangle ABC be A (x₁ , y₁),
B (x₂, y₂) and C (x₃, y₃)
P is the mid point of AB
\(\Rightarrow \ \left[\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right]=(11,7)\)
Comparing the co-ordinates, we get
x₁+ x₂ = 22 and        ......(1)
y₂ + y₃ = 8             .....(2)
Q is the mid point of BC
\(\Rightarrow \ \left[\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right]=(13.5,4)\)
x₂ + x₃ = 27      ......(3)
y₂ + y₃ = 8         ......(4)
R is the mid point of AC
\(\Rightarrow \ \left[\frac{x_{1}+x_{3}}{2}, \frac{y_{1}+y_{3}}{2}\right]=(9.5,4)\)
x₁ + x₃ = 19      ......(5)
y₁ + y₃ = 8         ......(6)
Solving (1), (3) and (5), we get
x₁ = 7, x₂ = 15 and x₃ = 12.
Solving (2), (4) and (6)
we get y₁ = 7,y₂ = 7,y₃ = 1
The vertices are A (7, 7), B (15, 7) and C(12, 1)
Area of  ABC \(=\frac{1}{2}\left[\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)\right. \left.+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right] \)
\(=\frac{1}{2}[7(7-1)+15(1-7)+12(7-7)] \)
\(=\frac{1}{2}[42-90]=\frac{1}{2}(-48)=-24 \)
= 24 sq. units [ Area cannot be negative]
Area of PQR
\(=\frac{1}{2}[11(4-4)+13.5(4-7)+9.5(7-4)] \)
\(=\frac{1}{2}[0-40.5+28.5]=\frac{-12}{2}=-6 \)
Area = 6 sq. units [Area cannot be negative]
Area of ABC = 4 (Area of PQR ).

Given points A (- 3, 9), B(a, b) and C (4, - 5)
Since the points are collinear, Area of triangle ABC = 0
\(\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]=0\)
-3(b + 5) + a( -5 -9) + 4(9 - b) = 0
-3b - 15- 14a + 36 - 4b = 0
-14a - 7b + 21 = 0
2a + b = 3                           .........(1)
Given that a + b = 1            .........(2)
Solving (1) and (2)

Substituting in (2)
2 + b = 1 
b = 1 - 2 = -1
a = 2,  b = -1.

Given vertices are (- 4, - 2),(- 3, k), (3, - 2) and (2,3) and area of quadrilateral is 28 sq. units.
Area of quadrilateral \(=\frac{1}{2}\left[\left(x_{1}-x_{3}\right)\left(y_{2}-y_{4}\right)-\right. \left.\left(x_{2}-x_{4}\right)\left(y_{1}-y_{3}\right)\right] \)
\(\frac{1}{2}\) [(- 4 - 3) (k - 3) - (- 3 - 2) (- 2 + 2)] = 28
(-7) (k - 3) - (- 5) (0) = 56
-7k + 21 = 56
-7k = 56 - 21 = 35
\(k=\frac{35}{-7}=-5\)
k = -5

Given vertices are A (–9, –2), B(–8, –4), C(1, –3) and  D(2, 2)
Area of the Quadrilateral ABCD
\( =\frac{1}{2}\left[\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{4}+x_{4} y_{1}\right)-\left(x_{2} y_{1}+x_{3} y_{2}\right.\right. \left.\left.+x_{4} y_{3}+x_{1} y_{4}\right)\right] sq. units\)
= \(\frac{1}{2}\) [(- 9) (- 4) + (- 8) (-3) + (1) (2) + (2) (- 2)] - [(- 8) (- 2) + (1) (- 4) + (2) (-3) + (- 9) (2)]
= \(\frac{1}{2}\) [(36 + 24 + 2 - 4)- (16 - 4 - 6 - 18)]
= \(\frac{1}{2}\) [58 - (-12)] = \(\frac{1}{2}\) [70] = 35 sq. units
Area of quadrilateral = 35 sq. units.

The parking lot is a quadrilateral whose vertices are at A(2, 2), B(5, 5), C(4, 9) and D(1, 7).
Therefore, Area of parking lot
= \(\frac{1}{2}\) {(10 + 45 + 28 + 2) - (10 + 20 + 9 + 14)}
= \(\frac{1}{2}\) {85 - 53}

= \(\frac{1}{2}\) (32) = 16.units.
So, area of parking lot = 16 sq feet
Construction rate per square feet = Rs. 1300
Therefore, total cost for constructing the parking lot = 16 x 1300 = Rs. 20800

Before determining the area of quadrilateral, plot the vertices in a graph.
Let the vertices be A(8, 6), B(5, 11), C(-5, 12) and D(-4, 3).
Therefore, area of the quadrilateral ABCD
=\(\frac{1}{2}\) { (x₁y₂ + x₂y₃ + x₃y₁) - (x₂y₁ + x₃y₂ + x₁y₃) }
=\(\frac{1}{2}\) { (80 + 60  - 15 - 24) - (30 - 55 - 48 + 24)}
=\(\frac{1}{2}\) {109 + 49 }
=\(\frac{1}{2}\) { 158 } = 79 sq. units

Vertices of one triangular tile are at (-3, 2), (-1, -1) and (1, 2)

(-3, 2), (-1, -1) and (1, 2)
Area of this tile = \(\frac12\) {(3 - 2 + 2) - (- 2 - 1 - 6)} sq. units
= \(\frac12\) (12) = 6 sq. units
Since the floor is covered by 110 triangle shaped identical tiles,
Area of floor = 110 x 6 = 660 sq. units

Since the three points P(-1, -4), Q(b, c) and R(5, -1) are collinear,
Area of triangle PQR = 0
\(\frac{1}{2}\) { (x₁y₂ + x₂y₃ + x₃y₁) - (x₂y₁ + x₃y₂ + x₁y₃) } = 0
\(\frac{1}{2}\) { (- c - b - 20) - (- 4b + 5c + 1) } = 0
- c - b - 20 + 4b - 5c - 1 = 0
b - 2c = 7 ...(1)
Also, 2b + c = 4 .....(2) (from given information)
Solving (1) and (2) we get b = 3, c = -2

The vertices are A(-1, 2), B(k, -2) and C(7, 4)
Area of triangle ABC is 22 sq. units
\(\frac{1}{2}\) { (x₁y₂ + x₂y₃ + x₃y₁) - (x₂y₁ + x₃y₂ + x₁y₃) } = 22
\(\frac{1}{2}\) { (2 + 4k + 14) - (2k - 14 - 4) } = 22
2k + 34 = 44 gives 2k = 10 so k = 5.

The points are P(-1, 5, 3), Q(6, -2) , R(-3, 4)
Area of Δ PQR = \(\frac{1}{2}\) { (x₁y₂ + x₂y₃ + x₃y₁) - (x₂y₁ + x₃y₂ + x₁y₃) }
= \(\frac{1}{2}\) { (3 + 24 - 9) - (18 + 6 - 6) }
= \(\frac{1}{2}\) { 18 - 18 } = 0
Therefore, the given points are collinear.

Plot the points in a rough diagram and take them in counter-clockwise order.
Let the vertices be


The area of Δ ABC is 
= \(\frac{1}{2}\) { (x₁y₂ + x₂y₃ + x₃y₁) - (x₂y₁ + x₃y₂ + x₁y₃) }
= \(\frac{1}{2}\) { (6 + 30 + 25) - (25 - 10 18) }
= \(\frac{1}{2}\) { 61 + 3 }
= \(\frac{1}{2}\) (64) = 32 sq. units