Let A(−2, −1) and B(4, 8) are the given points.

Let P(a,b) and Q(c,d) be the points of trisection of AB, so that AP = PQ = QB .
By the formula proved above,
P is the point.
\(\left( \frac { { x }_{ 2 }+{ 2x }_{ 1 } }{ 3 } ,\frac { { y }_{ 2 }+{ 2y }_{ 1 } }{ 3 } \right) =\left( \frac { 4+2(-2) }{ 3 } ,\frac { 8+2(-1) }{ 3 } \right) \) = (0, 2)
Q is the point
\(\left( \frac { { 2x }_{ 2 }+{ x }_{ 1 } }{ 3 } ,\frac { {2 y }_{ 2 }+{ y }_{ 1 } }{ 3 } \right) =\left( \frac { 2(4)-2 }{ 3 } ,\frac { 2(8)-1 }{ 3 } \right) \) = (2, 5)

Let us form table of values for each line and then fix the ordered pairs to be plotted.
Graph of y = 2x + 1
Graph of -4x + 2y = 2
2y = 4x + 2
y = 2x + 1

Points to be plotted : (−2, −3), (−1, −1), (0, 1), (1, 3), (2, 5)

Points to be plotted : (−2, −3), (−1, −1), (0, 1), (1, 3), (2, 5)
Here both the equations are identical; they were only represented in different forms. Since they are identical, their solutions are same. All the points on one line are also on the other!
This means we have an infinite number of solutions which are the ordered pairs of all the points on the line.

\(\frac { x }{ 3 } +\frac { x }{ 5 } \) = 1
\(\frac { 5x+3x }{ 15 } \) = 1
\(\frac { 8x }{ 15 } \) = 1
8x = 15
x = \(\frac { 15 }{ 8 } \)

Faces of a dice (S) = {1, 2, 3, 4,5, 6}
n(S) = 6
Event of throwing an even number
A = {2, 4, 6}, n(A) = 3
∴ Probability of throwing an even number
P(A) = \(\frac { n(A) }{ n(S) } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \).

Area of the quadrilateral ABCD
= Area of the Δ ABD + Area of the Δ BCD
Sides of the triangle ABD are 13 cm, 14 cm, 15 cm.
s = \(\frac { 13+14+15 }{ 2 } \)cm
= \(\frac { 42 }{ 2 } \) = 21 cm
Area =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt { 21(21-13)(21-14)(21-15) } \)
=\(\sqrt { 21\times 8\times 7\times 6 } =\sqrt { 7056 } \)=84 cm²
sides of the triangle
BCD are 12 cm, 9 cm, 15 cm
∴ s = \(\frac { 12+9+15 }{ 2 } =\frac { 36 }{ 2 } \) = 18 cm
Area =\(\sqrt { 18(18-2)(18-9)(18-15) } \)
=\(\sqrt { 18\times 6\times 9\times 3 } =\sqrt { 2916 } \) = 54 cm²
∴ Area of the quadrilateral
= 84 cm² + 54 cm² =138 cm²

side: 22 m, 120 m, 122 m
Using Heron's formula
s = \(\frac { 22+120+122 }{ 2 } =\frac { 264 }{ 2 } \) = 12 m
Area = \(\sqrt { s(s-a)(s-b)(s-c) } \)
= \(\sqrt { 132(132-22)(132-120)(132-122) } \)
= \(\sqrt { 132\times 110\times 12\times 10 } \)
= \(\sqrt { 1742400 } =\sqrt { 11\times 11\times 3\times 3\times 2\times 2\times 2\times 2\times 10\times 10 } \)
= 11 \(\times\) 3 \(\times\) 2 \(\times\) 10 = 1320 m²
cost of levelling 1 m² = Rs. 20
∴ cost oflevelling 1320 m²= 1320 \(\times\) 20 = Rs. 26400

8 sin 2(15°). cos 4 (15°). sin 6(15°)
= 8 sin 30° cos 60° sin 90°
= \(8\times \cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } \times 1=2\)

Let the man's present age = x
Five years ago his age is = x - 5
Let his son's age be = y
5 years ago his son's age = y - 5
\(\therefore\) x - 5 = 7(y- 5)
x - 5 = 7y-35
x - 7y = -35 + 5
x - 7y = -30    ....(1)
After 5 years, man's age will be = x + 5
His son's age will be = y + 5
\(\therefore\) x + 5 = 4(y + 5)
x+ 5 = 4y+ 20
x-4y = 20-5
\(\Rightarrow\) x-4y = 15
(1) \(\Rightarrow\) x-7y = -30
(2) \(\Rightarrow\) \(\cfrac { x-4y=15 }{ 3y\quad =\quad 45 } \)
y= 15
Substitute y = 15 in (1)
x -7 (15) = -30
x-105 =- 30
x = -30 + 105
x = 75
\(\therefore\) Man's Age = 75, His son's Age = 15

Let the monthly income of A and B be 3x and 4x respectively.
Let the monthly expenditure of A and B be 5y and 7y respectively.
\(\therefore\) 3x - 5y = 5000 --------- (1)
4x - 7y = 5000 --------- (2)
(1) \(\times\)4 \(\Rightarrow\) 12x - 20y = 20000
(2) \(\times\) 37 \(\Rightarrow\) \(\cfrac { 12-21y=15000 }{ 1y\quad =\quad 5000 } \)
Substitute y = 5000 in (1)
3x - 5 (5000) = 5000
3x - 25000 = 5000
3x = 5000 + 25000
3x = 30000
x = 10000
\(\therefore\) Monthly income of A is 3x = 3\(\times\)1 0000 = Rs. 30000
Monthly income of B is 4x = 4 \(\times\) 10000 = Rs. 40000

(i)sin 30⁰ + cos 30⁰ = \(\frac { 1 }{ 2 } +\frac { \sqrt { 3 } }{ 2 } =\frac { 1+\sqrt { 3 } }{ 2 } \)
(ii) tan 60⁰ cot 60⁰ = \(\sqrt { 3 } \times \frac { 1 }{ \sqrt { 3 } } =1\)
(iii) \(\frac { tan45° }{ tan30°+tan60° } \) = \(\frac { 1 }{ \frac { 1 }{ \sqrt { 3 } } +\frac { \sqrt { 3 } }{ 1 } } =\frac { 1 }{ \frac { 1+\left( \sqrt { 3 } \right) ^{ 2 } }{ \sqrt { 3 } } } =\frac { 1 }{ \frac { 1+3 }{ \sqrt { 3 } } } = \frac { \sqrt { 3 } }{ 4 } \)
(iv) sin²45⁰ + cos²45⁰ = \(\left( \frac { 1 }{ \sqrt { 2 } } \right) ^{ 2 }+\left( \frac { 1 }{ \sqrt { 2 } } \right) ^{ 2 }=\frac { 1^{ 2 } }{ \left( \sqrt { 2 } \right) ^{ 2 } } +\frac { 1^{ 2 } }{ \left( \sqrt { 2 } \right) ^{ 2 } } =\frac { 1 }{ 2 } +\frac { 1 }{ 2 } =1\)

 x₁ y₁ x₂ y₂  x₃y₃
(2, -4) (-3, -7) (7, 2)
(i) Centroid G (x, y) \(=\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) \)
\(=\left( \frac { (2)+(-3)+7 }{ 3 } ,\frac { \left( -4 \right) +\left( -7 \right) +\left( 2 \right) }{ 3 } \right) \)
\(=\left( \frac { 6 }{ 3 } ,\frac { -9 }{ 3 } \right) =(2,-3)\)
(ii)  x₁ y₁  x₂ y₂   x₃y₃
(2, -4) (1, -4) (-4, -2)
Centroid G (x, y) \(=\left( \frac { (-5)+1+(-4) }{ 3 } ,\frac { \left( -5 \right) +\left( -4 \right) +\left( -2 \right) }{ 3 } \right) \)
\(=\left( \frac { -8 }{ 3 } ,\frac { -11 }{ 3 } \right) \)

x₁ y₁ x₂ y₂
A(7, -5) B(9, -3)
\(\bar { AB } =\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } \)
\(=\sqrt { { (9-7) }^{ 2 }+{ (-3-(-5)) }^{ 2 } } =\sqrt { { 2 }^{ 2 }+{ (-3+5) }^{ 2 } } =\sqrt { 4+4 } \)
\(=\sqrt { 8 } =\sqrt { 4\times 2 } =2\sqrt { 2 } \)
\(\bar { BC } =\sqrt { { (13-7 })^{ 2 }+{ (1-(-5)) }^{ 2 } } =\sqrt { { 4 }^{ 2 }+{ 4 }^{ 2 } } =\sqrt { 16+16 } \)
\(=\sqrt { 32 } =\sqrt { 16\times 2 } =4\sqrt { 2 } \)
\(\bar { AC } =\sqrt { { (13-7) }^{ 2 }+{ (1-(-5)) }^{ 2 } } \)
\(=\sqrt { { 6 }^{ 2 }+{ 6 }^{ 2 } } =\sqrt { 36+36 } \)
\(=\sqrt { { 6 }^{ 2 }+{ 6 }^{ 2 } } =\sqrt { 36+36 } \)
\(2\sqrt { 2 } +4\sqrt { 2 } =6\sqrt { 2 } \)
∴ AB + BC = AC, Here B is the common Point
∴ A, B, C are collinear

Let ‘a’ be the side of the cube.
Given that, total surface area = 864 cm²
6a²= 864
a² = \(\frac{864}{6}\)
a² = 144
Therefore, side (a) = 12 cm
Now, volume of the cube = a³
= 123 = 12 ×12 ×12 = 1728 cm³

Let E be the event that it will rain tomorrow. Then E′ is the event that it will not rain tomorrow.
Since P(E) = 0.91, we have P(E′) = 1−0.91 (how?)
= 0.09
Therefore, the probability that it will not rain tomorrow
= 0.09

Total number of staff members = 42
The relative frequencies:
Car users \(=\frac { 7 }{ 42 } =\frac { 1 }{ 6 } \)
Two-wheeler users \(=\frac { 20 }{ 42 } =\frac { 10 }{ 21 } \)
Cycle users \(=\frac { 15 }{ 42 } =\frac { 5 }{ 14 } \)

Here, Total Surface Area of the cube = 486 cm²
6a² = 486 \(\Rightarrow\) a² = \(\frac{486}{6}\) and so, a² = 81 . This gives a = 9.
The side of the cube = 9 cm
Lateral Surface Area = 4a² = 4 × 9² = 4 × 81 = 324 cm²

(i) sin 38⁰36' + tan 12⁰12'
sin 38⁰36' = 0.6239
tan12⁰12' = 0.2162
sin 38⁰36' + tan 12⁰12' = 0.8401
(ii) tan 60⁰25^' - cos 49⁰20^'
tan 60⁰25' = 1.7603 + 0.0012 = 1.7615
cos 49⁰20^' = 0.6521 - 0.0004 = 0.6517
tan 60⁰25' - cos 49⁰20' = 1.1098

write 64⁰34' = 64⁰30' + 4'
From the table we have, sin64⁰30' = 0.9026 

Given two linear equations are
2x - 3y = 7;
(k + 2)x - (2k + 1)y = 3(2k - 1)
\(\left[ \begin{matrix} { a }_{ 1 }x+{ b }_{ 1 }y+{ c }_{ 1 }=0 \\ { a }_{ 2 }x+{ b }_{ 2 }y+{ c }_{ 2 }=0 \end{matrix} \right] \)
Here a₁ = 2, b₁ = −3, a₂ = (k + 2), b₂ = −(2k +1), c₁ = 7, c₂ = 3(2k −1)
For infinite number of solution we consider \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
\(\frac { 2 }{ k+2 } =\frac { -3 }{ -(2k+1) } =\frac { 7 }{ 3(2k-1) } \)
\(\frac { 2 }{ k+2 } =\frac { -3 }{ -(2k+1) } \)
2(2k +1) = 3(k + 2)
4k + 2 = 3k + 6
k = 4
\(\frac { -3 }{ -(2k+1) } =\frac { 7 }{ 392k-1) } \)
9(2k −1) = 7(2k +1)
18k − 9 = 14k + 7
4k = 16
k = 4

(i) sin74° = sin(90⁰ -16⁰)    (since, 90⁰ -16⁰ = 74⁰ )
RHS is of the form sin(90⁰ - \(\theta\)) = cos\(\theta\)
Therefore sin74° =  cos16⁰
(ii)  tan12° =  tan(90⁰ - 78⁰) (since, 12⁰= 90⁰ = 78⁰ )
RHS is of the form tan(90⁰-  \(\theta\)) = cot \(\theta\)
Therefore tan12° = cot 78⁰
(iii) cosec 39° = cosec(90⁰ -  51⁰) (since, 39⁰ = 90⁰ - 51⁰ )
RHS is of the form cosec(90⁰ - \(\theta\)) = sec\(\theta\)
Therefore cosec39° = sec51⁰

Let a = 28 m, b = 15 m and c = 41 m
Then, s = \(\frac{a+b+c}{2}=\frac{28+15+41}{2}=\frac{84}{2}\) = 42m
Area of triangular field =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt { 42(42-28)(42-15)(42-41) } \)
=\(\sqrt { 42\times 14​​\times 17\times 1 } \)
=\(\sqrt { 2\times 3\times 7\times 7\times 2\times 3\times 3\times 3\times 1 } \)
\(=2 \times 3 \times 7 \times 3\)
= 126 m²
Given the cost of levelling is Rs. 20 per m².
The total cost of levelling the field = 20 \(\times\)126 = Rs. 2520.

(i) 8x- 3y = 12      ...(1)
5x-2y = 7      ..(2)
8x- 3y-12 = 0
5x-2y- 7 = 0
For cross multiplication method, we write the co-efficients as

\(\cfrac { x }{ (-3)(-7)(-2)(-12) } =\cfrac { y }{ (-12)(5)-(-7)(8) } =\cfrac { 1 }{ (8)(-2)-(5)(-3) } \)
\(\cfrac { x }{ 21-24 } =\cfrac { y }{ -60+56 } =\cfrac { 1 }{ -16+15 } \)
\(\cfrac { x }{ -3 } =\cfrac { y }{ -60+56 } =\cfrac { 1 }{ -16+15 } \)
\(\therefore \ \cfrac { x }{ 3 } =\cfrac { 1 }{ -1 } \ \cfrac { y }{ -4 } =\cfrac { 1 }{ -1 } \)
x = 3 , y = 4
\(\therefore\)  Solutions: x = 3; y = 4
(ii) 6x+ 7y-11 = 0
5x+ 2y-13 = 0
For cross multiplication method, we write the co-efficients as
\(\cfrac { x }{ -91-(-22) } =\cfrac { y }{ -55-(-68) } =\cfrac { 1 }{ 12-35 } \)
\(\cfrac { x }{ -91+22 } =\cfrac { y }{ -55+78 } =\cfrac { 1 }{ -23 } \)
\(\cfrac { x }{ -69 } =\cfrac { y }{ 23 } =\cfrac { 1 }{ -23 } \)
\(\cfrac { x }{ -69 } =\cfrac { 1 }{ -23 } \quad \cfrac { y }{ 23 } =\cfrac { 1 }{ -23 } \)

x = 3, y= -1
\(\therefore\)  x = 3; y= -1
(iii) 

\(\cfrac { 2 }{ x } +\cfrac { 3 }{ y } -5=0\)
\(\cfrac { 3 }{ x } -\cfrac { 1 }{ y } +9=0\)
In (1), (2) Put \(\cfrac { 1 }{ x } =a,\cfrac { 1 }{ y } =b\)
(1) \(\Rightarrow\) 2a + 3b - 5 = 0
(2) \(\Rightarrow\) 3a - b + 9 = 0
For cross multiplication method, we write the co-efficients as
\(\cfrac { a }{ (3)(9)-(-1)(-5) } =\cfrac { b }{ (-5)(3)-(9)(2) } =\cfrac { 1 }{ (2)(-1)-(3)(3) } \)
\(\cfrac { a }{ 27-5 } =\cfrac { b }{ -15-18 } =\cfrac { 1 }{ -2-9 } \)
\(\cfrac { a }{ 22 } =\cfrac { b }{ -33 } =\cfrac { 1 }{ -11 } \)
\(\therefore \ \cfrac { a }{ 22 } =\cfrac { 1 }{ -11 } \cfrac { b }{ -33 } =\cfrac { 1 }{ -11 } \)

a = -2    b = 3
\(a=\cfrac { 1 }{ x } =-2\quad b=\cfrac { 1 }{ y } =3\)
\(\therefore \ x=\cfrac { -1 }{ 2 } \ \ y=\cfrac { 1 }{ 3 } \)
solution \(\ x=\cfrac { -1 }{ 2 } \ \ y=\cfrac { 1 }{ 3 } \)

The centroid G(x, y) of a triangle whose vertices are (x₁, y₁), (x₂ , y₂ ) and (x₃ , y₃) is given by

G(x,y)=G\(\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 }3 }{ } \right) \)
We have (x₁, y₁) = (6, −1); (x₂ , y₂ ) = (8, 3); (x₃ , y₃) = (10, −5)
The centroid of the triangle
G(x, y) =G\(\left( \frac { 6+8+10 }{ 3 } ,\frac { -1+3-5 }{ 3 } \right) \)
= G \((\frac{24}{3},\frac{-3}{3})\)= G(8, -1)