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Ex 12.1 Class 11 Maths Question 1.
A point is on the x-axis. What are its y-coordinate and z-coordinate?
Solution:
The coordinates of any point on the x-axis will be (x, 0, 0). Thus y-coordinate and z-coordinate of the point are zero.

Ex 12.1 Class 11 Maths Question 2.
A point is in the XZ-plane. What can you say about its y-coordinate?
Solution:
The coordinates of any point in XZ-plane will be (x, 0, z). Thus y-coordinate of the point is zero.

Ex 12.1 Class 11 Maths Question 3.
Name the octants in which the following points lie:
(1, 2, 3), (4, -2, 3), (4, -2, -5), (4, 2, -5), (-4, 2, -5), (-4, 2, 5), (-3, -1, 6), (2, -4, -7)
Solution:
Point (1, 2, 3) lies in Octant I.
Point (4, -2, 3) lies in Octant IV.
Point (4, -2, -5) lies in Octant VIII.
Point (4, 2, -5) lies in Octant V.
Point (- 4, 2, -5) lies in Octant VI.
Point (- 4, 2, 5) lies in Octant II.
Point (- 3, -1, 6) lies in Octant III.
Point (2, – 4, -7) lies in Octant VIII.

Ex 12.1 Class 11 Maths Question 4.
Fill in the blanks:
(i) The x-axis and y-axis taken together determine a plane known as ______
(ii) The coordinates of points in the XY-plane are of the form _______
(iii) Coordinate planes divide the space into ______ octants.
Solution:
(i) XY-plane
(ii) (x, y, 0)
(iii) Eight

We hope the NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.1 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.1, drop a comment below and we will get back to you at the earliest.

Ex 12.2 Class 11 Maths Question 1.
Find the distance between the following pairs of points:
(i) (2, 3, 5) and (4, 3, 1)
(ii) (-3, 7, 2) and (2, 4, -1)
(iii) (-1, 3, -4) and (1, -3, 4)
(iv) (2, -1, 3) and (-2, 1, 3)
Solution:
(i) The distance PQ between the points P(2, 3, 5) and Q(4, 3, 1) is

= 

(ii) The distance PQ between the points P(-3, 7, 2) and Q(2, 4, -1) is

(iii) The distance PQ between the points P(-1, 3, -4) and Q(1, -3, 4) is

(iv) The distance PQ between the points P(2, -1, 3) and Q(-2, 1, 3) is

Ex 12.2 Class 11 Maths Question 2.
Show that the points (-2, 3, 5), (1, 2, 3) and (7, 0, -1) are collinear.
Solution:
Let A(-2, 3, 5), B(1, 2, 3) and C(7, 0, -1) be three given points.

Now AC = AB + BC
Thus, points A, B and C are collinear.

Ex 12.2 Class 11 Maths Question 3.
Verify the following:
(i) (0, 7, -10), (1, 6, -6) and (4, 9, -6) are the vertices of an isosceles triangle.
(ii) (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of a right angled triangle.
(iii) (-1, 2, 1), (1, -2, 5), (4, -7,8) and (2, -3,4) are the vertices of a parallelogram.
Solution:
(i) Let A(0, 7, -10), B(l, 6, -6) and C(4, 9, -6) be three vertices of triangle ABC. Then

Now, AB = BC
Thus, ABC is an isosceles triangle.

(ii) Let A(0, 7,10), B(-l, 6, 6) and C(-A, 9, 6) be three vertices of triangle ABC. Then

Now, AC² = AB² + BC²
Thus, ABC is a right angled triangle.

(iii) Let A(-1, 2, 1), B(1, -2, 5) and C(4, -7, 8) and D(2, -3,4) be four vertices of quadrilateral ABCD. Then

Now AB = CD, BC = AD and AC ≠ BD
Thus A, B, C and D are vertices of a parallelogram ABCD.

Ex 12.2 Class 11 Maths Question 4.
Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, -1).
Solution:
Let A(x, y, z) be any point which is equidistant from points B(1, 2, 3) and C(3, 2, -1).

Ex 12.2 Class 11 Maths Question 5.
Find the equation of the set of points P, the sum of whose distances from A(4, 0, 0) and B(-4,0,0) is equal to 10.
Solution:
Let P(x, y, z) be any point.

We hope the NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2, drop a comment below and we will get back to you at the earliest.

Ex 12.3 Class 11 Maths Question 1.
Find the coordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) in the ratio
(i) 2 : 3 internally,
(ii) 2 : 3 enternally
Solution:
(i) Let P(x, y, z) be any point which divides the line segment joining the points A(-2, 3, 5) and B(1, -4, 6) in the ratio 2 : 3 internally.

(ii) Let P(x, y, z) be any point which divides the line segment joining the points 71 (-2, 3, 5) and B(1, -4, 6) in the ratio 2 : 3 externally. Then

Ex 12.3 Class 11 Maths Question 2.
Given that P(3, 2, -4), Q(5, 4, -6) and R(9, 8, -10) are collinear. Find the ratio in which Q divides PR.
Solution:
Let Q(5, 4, -6) divides the line segment joining the points P(3, 2, -4) and R(9, 8, -10) in the ratio k : 1 internally.

Ex 12.3 Class 11 Maths Question 3.
Find the ratio in which the YZ-plane divides the line segment formed by joining the points (-2, 4, 7) and (3, -5, 8).
Solution:
Let the line segment joining the points A(-2, 4, 7) and B(3, -5, 8) be divided by the YZ -plane at a point C in the ratio k : 1.

Ex 12.3 Class 11 Maths Question 4.
Using section formula, show that the points A(2, -3, 4), B(-1, 2, 1) and C are collinear.
Solution:
Let the points A(2, -3, 4), B(-l, 2,1) and C be the given points. Let the point P divides AB in the ratio k : 1. Then coordinates of P are 
Let us examine whether for some value of k, the point P coincides with point C.

AB internally in the ratio 2:1. Hence A, B, C are collinear.

Ex 12.3 Class 11 Maths Question 5.
Find the coordinates of the points which trisect the line segment joining the points P(4, 2, -6) and Q(10, -16, 6).
Solution:
Let R and S be two points which trisect the line segment PQ.

We hope the NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.3 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.3, drop a comment below and we will get back to you at the earliest.

In each of the following Exercises 1 to 5, find the equation of the circle with

Ex 11.1 Class 11 Maths Question 1.
centre (0, 2) and radius 2
Solution:
Here h = 0,k = 2 and r = 2
The equation of circle is,
(x-h)² + (y- k)² = r²
∴ (x – 0)² + (y – 2)² = (2)²
⇒ x² + y² + 4 – 4y = 4
⇒ x² + y² – 4y = 0

Ex 11.1 Class 11 Maths Question 2.
centre (-2,3) and radius 4
Solution:
Here h=-2,k = 3 and r = 4
The equation of circle is,
(x – h)² + (y – k)² = r²
∴(x + 2)² + (y – 3)² = (4)²
⇒ x² + 4 + 4x + y² + 9 – 6y = 16
⇒ x² + y² + 4x – 6y – 3 = 0

Ex 11.1 Class 11 Maths Question 3.
centre  and radius 
Solution:
here h = , k =  and r = 
The equation of circle is,

Ex 11.1 Class 11 Maths Question 4.
centre (1, 1) and radius 
Solution:
Here h = l, k=l and r = 
The equation of circle is,
(x – h)² + (y – k)² = r²
∴ (x – 1)² + (y – 1)² = ²
⇒ x² + 1 – 2x + y² +1 – 2y = 2
⇒ x² + y² – 2x – 2y = 0

Ex 11.1 Class 11 Maths Question 5.
centre (-a, -b) and radius .
Solution:
Here h=-a, k = -b and r = 
The equation of circle is, (x – h)² + (y – k)² = r²
∴ (x + a)² + (y + b)² = 
⇒ x² + a² + 2ax + y² + b² + 2by = a² -b²
⇒ x² + y² + 2ax + 2 by + 2b² = 0

In each of the following exercises 6 to 9, find the centre and radius of the circles.

Ex 11.1 Class 11 Maths Question 6.
(x + 5)² + (y – 3)² = 36
Solution:
The given equation of circle is,
(x + 5)² + (y – 3)² = 36
⇒ (x + 5)² + (y – 3)² = (6)²
Comparing it with (x – h)2² + (y – k)² = r², we get
h = -5, k = 3 and r = 6.
Thus the co-ordinates of the centre are (-5, 3) and radius is 6.

Ex 11.1 Class 11 Maths Question 7.
x² + y² – 4x – 8y – 45 = 0
Solution:
The given equation of circle is
x² + y² – 4x – 8y – 45 = 0
∴ (x² – 4x) + (y² – 8y) = 45
⇒ [x² – 4x + (2)²] + [y² – 8y + (4)²] = 45 + (2)² + (4)²
⇒ (x – 2)² + (y – 4)² = 45 + 4 + 16
⇒ (x – 2)² + (y – 4)² = 65
⇒ (x – 2)² + (y – 4)²= 
Comparing it with (x – h)² + (y – k)² = r², we
have h = 2,k = 4 and r = .
Thus co-ordinates of the centre are (2, 4) and radius is .

Ex 11.1 Class 11 Maths Question 8.
x² + y² – 8x + 10y – 12 = 0
Solution:
The given equation of circle is,
x² + y² – 8x + 10y -12 = 0
∴ (x² – 8x) + (y² + 10y) = 12
⇒ [x² – 8x + (4)²] + [y² + 10y + (5)²] = 12 + (4)² + (5)²
⇒ (x – 4)² + (y + 5)² = 12 + 16 + 25
⇒ (x – 4)² + (y + 5)² = 53
⇒ (x – 4)² + (y + 5)² = 
Comparing it with (x – h)² + (y – k)² = r², we have h = 4, k = -5 and r = 
Thus co-ordinates of the centre are (4, -5) and radius is .

Ex 11.1 Class 11 Maths Question 9.
2x² + 2y² – x = 0
Solution:
The given equation of circle is,
2x² + 2y² – x = 0

Ex 11.1 Class 11 Maths Question 10.
Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.
Solution:
The equation of the circle is,
(x – h)² + (y – k)² = r² ….(i)
Since the circle passes through point (4, 1)
∴ (4 – h)² + (1 – k)² = r²
⇒ 16 + h² – 8h + 1 + k² – 2k = r²
⇒ h²+ k² – 8h – 2k + 17 = r² …. (ii)
Also, the circle passes through point (6, 5)
∴ (6 – h² + (5 – k)² = r²
⇒ 36 + h² -12h + 25 + k² – 10k = r²
⇒ h² + k² – 12h – 10kk + 61 = r² …. (iii)
From (ii) and (iii), we have h² + k² – 8h – 2k +17
= h² + k²– 12h – 10k + 61
⇒ 4h + 8k = 44 => h + 2k = ll ….(iv)
Since the centre (h, k) of the circle lies on the line 4x + y = 16
∴ 4h + k = 16 …(v)
Solving (iv) and (v), we get h = 3 and k = 4.
Putting value of h and k in (ii), we get
(3)² + (4)² – 8 x 3 – 2 x 4 + 17 = r²
∴ r² = 10
Thus required equation of circle is
(x – 3)² + (y – 4)² = 10
⇒ x² + 9 – 6x + y² +16 – 8y = 10
⇒ x² + y² – 6x – 8y +15 = 0.

Ex 11.1 Class 11 Maths Question 11.
Find the equation of the circle passing through the points (2, 3) and (-1, 1) and whose centre is on the line x – 3y – 11 = 0.
Solution:
The equation of the circle is,
(x – h)² + (y – k)² = r² ….(i)
Since the circle passes through point (2, 3)
∴ (2 – h)² + (3 – k)² = r²
⇒ 4 + h² – 4h + 9 + k² – 6k = r²
⇒ h²+ k² – 4h – 6k + 13 = r² ….(ii)
Also, the circle passes through point (-1, 1)
∴ (-1 – h)² + (1 – k)² = r²
⇒ 1 + h² + 2h + 1 + k² – 2k = r²
⇒ h² + k² + 2h – 2k + 2 = r² ….(iii)
From (ii) and (iii), we have
h² + k² – 4h – 6k + 13 = h² + k² + 2h – 2k + 2
⇒ -6h – 4k = -11 ⇒ 6h + 4k = 11 …(iv)
Since the centre (h, k) of the circle lies on the line x – 3y-11 = 0.
∴ h – 3k – 11 = 0 ⇒ h -3k = 11 …(v)
Solving (iv) and (v), we get
h =  and k = 
Putting these values of h and k in (ii), we get

⇒  ⇒ 
Thus required equation of circle is
⇒ 
⇒ 
⇒ 4x² + 49 – 28x + 4y² + 25 + 20y = 130
⇒ 4x² + 4y² – 28x + 20y – 56 = 0
⇒ 4(x² + y² – 7x + 5y -14) = 0
⇒ x² + y² – 7x + 5y -14 = 0.

Ex 11.1 Class 11 Maths Question 12.
Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).
Solution:
Since the centre of the circle lies on x-axis, the co-ordinates of centre are (h, 0). Now the circle passes through the point (2, 3).
∴ Radius of circle

Ex 11.1 Class 11 Maths Question 13.
Find the equation of the circle passing through (0, 0) and making intercepts a and b on the co-ordinate axes.
Solution:
Let the circle makes intercepts a with x-axis and b with y-axis.
∴ OA = a and OB = b
So the co-ordinates of A are (a, 0) and B are (0,b)
Now, the circle passes through three points 0(0, 0), A(a, 0) and B(0, b).

Ex 11.1 Class 11 Maths Question 14.
Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).
Solution:
The equation of circle is
(x – h)² + (y – k)² = r² ….(i)
Since the circle passes through point (4, 5) and co-ordinates of centre are (2, 2)
∴ radius of circle

Ex 11.1 Class 11 Maths Question 15.
Does the point (-2.5, 3.5) lie inside, outside or on the circle x² + y² = 25?
Solution:
The equation of given circle is x² + y² = 25
⇒ (x – 0)² + (y – 0)² = (5)²
Comparing it with (x – h)² + (y – k)² = r², we
get
h = 0,k = 0, and r = 5
Now, distance of the point (-2.5, 3.5) from the centre (0, 0)

 = 4.3 < 5.
Thus the point (-2.5, 3.5) lies inside the circle.

We hope the NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections EX 11.1, drop a comment below and we will get back to you at the earliest.

In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

Ex 11.2 Class 11 Maths Question 1.
y²= 12x
Solution:
The given equation of parabola is y² = 12x which is of the form y² = 4ax.
∴ 4a = 12 ⇒ a = 3
∴ Coordinates of focus are (3, 0)
Axis of parabola is y = 0
Equation of the directrix is x = -3 ⇒ x + 3 = 0
Length of latus rectum = 4 x 3 = 12.

Ex 11.2 Class 11 Maths Question 2.
x² = 6y
Solution:
The given equation of parabola is x² = 6y which is of the form x² = 4ay.

Ex 11.2 Class 11 Maths Question 3.
y² = – 8x
Solution:
The given equation of parabola is
y² = -8x, which is of the form y² = – 4ax.
∴ 4a = 8 ⇒ a = 2
∴ Coordinates of focus are (-2, 0)
Axis of parabola is y = 0
Equation of the directrix is x = 2 ⇒ x – 2 = 0
Length of latus rectum = 4 x 2 = 8.

Ex 11.2 Class 11 Maths Question 4.
x² = -16y
Solution:
The given equation of parabola is
x² = -16y, which is of the form x² = -4ay.
∴ 4a = 16 ⇒ a = 4
∴ Coordinates of focus are (0, -4)
Axis of parabola is x = 0
Equation of the directrix is y = 4 ⇒ y – 4 = 0
Length of latus rectum = 4 x 4 = 16.

Check the Parabola Calculator to solve Parabola Equation.

Ex 11.2 Class 11 Maths Question 5.
y²= 10x
Solution:
The given equation of parabola is y² = 10x, which is of the form y² = 4ax.

Ex 11.2 Class 11 Maths Question 6.
x² = -9y
Solution:
The given equation of parabola is
x² = -9y, which is of the form x² = -4ay.

In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:

Ex 11.2 Class 11 Maths Question 7.
Focus (6, 0); directrix x = -6
Solution:
We are given that the focus (6, 0) lies on the x-axis, therefore x-axis is the axis of parabola. Also, the directrix is x = -6 i.e. x = -a and focus (6, 0) i.e. (a, 0). The equation of parabola is of the form y² = 4ax.
The required equation of parabola is
y² = 4 x 6x ⇒ y² = 24x.

Ex 11.2 Class 11 Maths Question 8.
Focus (0, -3); directri xy=3
Solution:
We are given that the focus (0, -3) lies on the y-axis, therefore y-axis is the axis of parabola. Also the directrix is y = 3 i.e. y = a and focus (0, -3) i.e. (0, -a). The equation of parabola is of the form x² = -4ay.
The required equation of parabola is
x² = – 4 x 3y ⇒ x² = -12y.

Ex 11.2 Class 11 Maths Question 9.
Vertex (0, 0); focus (3, 0)
Solution:
Since the vertex of the parabola is at (0, 0) and focus is at (3, 0)
∴ y = 0 ⇒ The axis of parabola is along x-axis
∴ The equation of the parabola is of the form y² = 4ax
The required equation of the parabola is
y² = 4 x 3x ⇒ y² = 12x.

Ex 11.2 Class 11 Maths Question 10.
Vertex (0, 0); focus (-2, 0)
Solution:
Since the vertex of the parabola is at (0, 0) and focus is at (-2, 0).
∴ y = 0 ⇒ The axis of parabola is along x-axis
∴ The equation of the parabola is of the form y² = – 4ax
The required equation of the parabola is
y² = – 4 x 2x ⇒ y² = -8x.

Ex 11.2 Class 11 Maths Question 11.
Vertex (0, 0), passing through (2, 3) and axis is along x-axis.
Solution:
Since the vertex of the parabola is at (0, 0) and the axis is along x-axis.
∴ The equation of the parabola is of the form y² = 4ax
Since the parabola passes through point (2, 3)

Ex 11.2 Class 11 Maths Question 12.
Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis.
Solution:
Since the vertex of the parabola is at (0, 0) and it is symmetrical about the y-axis.
∴ The equation of the parabola is of the form x² = 4ay
Since the parabola passes through point (5, 2)

We hope the NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections EX 11.2, drop a comment below and we will get back to you at the earliest.

In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

Ex 11.3 Class 11 Maths Question 1.

Solution:
Given equation of ellipse of 
Clearly, 36 > 16
The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 2.

Solution:
Given equation of ellipse is 
Clearly, 25 > 4
The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 3.

Solution:
Given equation of ellipse is 
Clearly, 16 > 9
The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 4.

Solution:
Given equation of ellipse is 
Clearly, 100 > 25
The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 5.

Solution:
Given equation of ellipse is 
Clearly, 49 > 36
The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 6.

Solution:
Given equation of ellipse is 
Clearly, 400 > 100
The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 7.
36x² + 4y² = 144
Solution:
Given equation of ellipse is 36x² + 4y² = 144

Ex 11.3 Class 11 Maths Question 8.
16x² + y² = 16
Solution:
Given equation of ellipse is16x² + y² = 16

Ex 11.3 Class 11 Maths Question 9.
4x² + 9y² = 36
Solution:
Given equation of ellipse is4x² + 9y² = 36

In each 0f the following Exercises 10 to 20, find the equation for the ellipse that satisfies the given conditions:

Ex 11.3 Class 11 Maths Question 10.
Vertices (±5, 0), foci (±4,0)
Solution:
Clearly, The foci (±4, o) lie on x-axis.
∴ The equation of ellipse is standard form is

Ex 11.3 Class 11 Maths Question 11.
Vertices (0, ±13), foci (0, ±5)
Solution:
Clearly, The foci (0, ±5) lie on y-axis.
∴ The equation of ellipse is standard form is

Ex 11.3 Class 11 Maths Question 12.
Vertices (±6, 0), foci (±4,0)
Solution:
Clearly, The foci (±4, 0) lie on x-axis.
∴ The equation of ellipse is standard form is

Ex 11.3 Class 11 Maths Question 13.
Ends of major axis (±3, 0), ends of minor axis (0, ±2)
Solution:
Since, ends of major axis (±3, 0) lie on x-axis.
∴ The equation of ellipse in standard form

Ex 11.3 Class 11 Maths Question 14.
Ends of major axis (0, ), ends of minor axis (±1, 0)
Solution:
Since, ends of major axis (0, ) lie on i-axis.
∴ The equation of ellipse in standard form

Ex 11.3 Class 11 Maths Question 15.
Length of major axis 26, foci (±5, 0)
Solution:
Since the foci (±5, 0) lie on x-axis.
∴ The equation of ellipse in standard form

Ex 11.3 Class 11 Maths Question 16.
Length of major axis 16, foci (0, ±6)
Solution:
Since the foci (0, ±6) lie on y-axis.
∴ The equation of ellipse in standard form

Ex 11.3 Class 11 Maths Question 17.
Foci (±3, 0) a = 4
Solution:
since the foci (±3, 0) on x-axis.
∴ The equation of ellipse in standard form

Ex 11.3 Class 11 Maths Question 18.
b = 3, c = 4, centre at the origin; foci on the x axis.
Solution:
Since the foci lie on x-axis.
∴ The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 19.
Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6)
Solution:
Since the major axis is along y-axis.
∴ The equation of ellipse in standard form

Ex 11.3 Class 11 Maths Question 20.
Major axis on the x-axis and passes through the points (4, 3) and (6, 2).
Solution:
Since the major axis is along the x-axis.
∴ The equation of ellipse in standard form is

We hope the NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections EX 11.3, drop a comment below and we will get back to you at the earliest.

In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, eccentricity and the length of the latus rectum of the hyperbolas.

Ex 11.4 Class 11 Maths Question 1.

Solution:
Given equation of hyperbola is

Ex 11.4 Class 11 Maths Question 2.

Solution:
Given equation of hyperbola is

Ex 11.4 Class 11 Maths Question 3.
9y² – 4x² = 36
Solution:
Given equation of hyperbola is9y² – 4x² = 36

Ex 11.4 Class 11 Maths Question 4.
16x² – 9y² = 576
Solution:
Given equation of hyperbola is16x² – 9y² = 576

Ex 11.4 Class 11 Maths Question 5.
5y² – 9x² = 36
Solution:
Given equation of hyperbola is
5y² – 9x² = 36

Ex 11.4 Class 11 Maths Question 6.
49y² – 16x² = 784
Solution:
Given equation of hyperbola is49y² – 16x² = 784

In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions.

Ex 11.4 Class 11 Maths Question 7.
Vertices (±2,0), foci (±3,0)
Solution:
Vertices are (±2, 0) which lie on x-axis. So the equation of hyperbola in standard form

Ex 11.4 Class 11 Maths Question 8.
Vertices (0, ±5), foci (0, ±8)
Solution:
Vertices are (0, ±5) which lie on x-axis. So the equation of hyperbola in standard form

Ex 11.4 Class 11 Maths Question 9.
Vertices (0, ±3), foci (0, ±5)
Solution:
Vertices are (0, ±3) which lie on x-axis. So the equation of hyperbola in standard form

Ex 11.4 Class 11 Maths Question 10.
Foci (±5, 0), the transverse axis is of length 8.
Solution:
Here foci are (±5, 0) which lie on x-axis. So the equation of the hyperbola in standard

Ex 11.4 Class 11 Maths Question 11.
Foci (0, ±13), the conjugate axis is of length 24.
Solution:
Here foci are (0, ±13) which lie on y-axis.
So the equation of hyperbola in standard

Ex 11.4 Class 11 Maths Question 12.
Foci (,0) , the latus rectum is of length 8.
Solution:
Here foci are (, 0) which lie on x-axis.
So the equation of the hyperbola in standard

Ex 11.4 Class 11 Maths Question 13.
Foci (±4, 0), the latus rectum is of length 12.
Solution:
Here foci are (±4, 0) which lie on x-axis.
So the equation of the hyperbola in standard

Ex 11.4 Class 11 Maths Question 14.
Vertices (+7, 0), e = 
Solution:
Here vertices are (±7, 0) which lie on x-axis.
So, the equation of hyperbola in standard

Ex 11.4 Class 11 Maths Question 15.
Foci (0, ), passing through (2, 3).
Solution:
Here foci are (0, ) which lie on y-axis.
So the equation of hyperbola in standard form

We hope the NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections EX 11.4, drop a comment below and we will get back to you at the earliest.

Ex 6.1 Class 11 Maths Question 1.
Solve 24x < 100, when
(i) x is a natural number
(ii) x is an integer.
Solution.
Given inequality is 24x < 100
Dividing both sides by 24, we get

This inequality is true when
(i) x is a natural number, {1, 2, 3, 4} satisfies this inequality.
(ii) x is an integer, {…., -4, -3,-2, -1, 0, 1, 2, 3, 4} satisfies this inequality.

Ex 6.1 Class 11 Maths Question 2.
Solve – 12x > 30, when
(i) x is a natural number
(ii) x is an integer
Solution.
Given inequality is -12x > 30 Dividing both sides by -12, we get

(i) This inequality is not true for any natural number.
(ii) Integers that satisfy this inequality are (…, -5, -4, -3}.

Ex 6.1 Class 11 Maths Question 3.
Solve 5x – 3 < 7, when
(i) x is an integer
(ii) x is a real number
Solution.
Given inequality is 5x – 3 < 7
Transposing 3 to R.H.S., we get
5x < 7 + 3 or 5x < 10
Dividing both sides by 5, we get
x < 2
(i) When x is an integer, {…. -2, -1, 0, 1} satisfies this inequality.
(ii) When x is real number, the solution is (-∞, 2).

Ex 6.1 Class 11 Maths Question 4.
Solve 3x + 8 > 2, when
(i) x is an integer
(ii) x is a real number.
Solution.
Given inequality is 3x + 8 > 2
Transposing 8 to R.H.S., we get
3x > 2 – 8 = -6
Dividing both sides by 3, we get
x > -2
(i) When x is an integer, the solution is (-1, 0, 1, 2, 3,…}
(ii) When x is real, the solution is (-2, ∞).

Solve the inequalities in Exercises 5 to 16 for real x.
Ex 6.1 Class 11 Maths Question 5.
4x + 3 < 5x + 7
Solution.
The inequality is 4x + 3 < 5x + 7
Transposing 5x to L.H.S. and 3 to R.H.S., we get
4x – 5x < 7 – 3 or -x < 4 Dividing both sides by -1, we get x > -4
∴ The solution is (- 4, ∞).

Ex 6.1 Class 11 Maths Question 6.
3x – 7 > 5x – 1
Solution.
The inequality is 3x – 7 > 5x -1
Transposing 5x to L.H.S. and -7 to R.H.S., we get
3x – 5x > -1 + 7 or -2x > 6
Dividing both sides by -2, we get
x < -3
∴ The solution is (-∞, -3).

Ex 6.1 Class 11 Maths Question 7.
3(x – 1) < 2(x – 3)
Solution.
The inequality is 3(x – 1) < 2(x – 3) or 3x – 3 < 2x – 6
Transposing 2x to L.H.S. and -3 to R.H.S., we get
3x – 2x < – 6 + 3
⇒ x<-3
∴ The solution is (- ∞, -3],

Ex 6.1 Class 11 Maths Question 8.
3(2 -x) > 2(1 -x)
Solution.
The inequality is 3(2 – x) > 2(1 – x) or 6 – 3x > 2 – Zx
Transposing -2x to L.H.S. and 6 to R.H.S., we get
-3x + 2x > 2 – 6 or -x > -4
Multiplying both sides by -1, we get
x ≤ 4
∴ The solution is (- ∞, 4],

Ex 6.1 Class 11 Maths Question 9.

Solution.
The inequality is 
Simplifying, 
Multiplying both sides by , we get
x < 6
∴ The solution is (- ∞, 6),

Ex 6.1 Class 11 Maths Question 10.

Solution.
The inequality is 
Transposing  to L.H.S., we get

Simplifying, 
Multiplying both sides by -6, we get
x < -6
∴ The solution is (- ∞, – 6).

Ex 6.1 Class 11 Maths Question 11.

Solution.
The inequality is 
Multiply both sides by the L.C.M. of 5, 3 i.e., by 15.
3 x 3(x – 2) ≤ 5 x 5(2 – x)
or, 9(x – 2) ≤ 25(2 – x)
Simplifying, 9x – 18 ≤ 50 – 25x
Transposing -25x to L.H.S. and -18 to R.H.S.
∴ 9x + 25x ≤ 50 + 18 or 34x ≤ 68
Dividing both sides by 34, we get
x < 2
∴ Solution is (- ∞, 2].

Ex 6.1 Class 11 Maths Question 12.

Solution.
The inequality is 
or, 
Multiplying both sides by 30,
3(3x + 20) ≥ 10(x – 6) or, 9x + 60 ≥ 10x-60
Transposing 10 x to L.H.S. and 60 to R.H.S., we get
∴ 9x-10x ≥ -60 – 60 or -x ≥-120
Multiplying both sides by -1, we get
x < 120
∴ The solution is (- ∞, 120].

Ex 6.1 Class 11 Maths Question 13.
2(2x + 3) – 10 < 6(x – 2)
Solution.
The inequality is 2(2x + 3) – 10 < 6(x – 2)
Simplifying, 4x + 6 -10 < 6x -12
or, 4x – 4 < 6x – 12
Transposing 6x to L.H.S. and – 4 to R.H.S., we get
∴ 4x – 6x < -12 + 4 or -2x < – 8 Dividing both sides by -2, we get x>4
∴ The solution is (4, ∞).

Ex 6.1 Class 11 Maths Question 14.
37 – (3x + 5) ≥ 9x – 8(x – 3)
Solution.
The inequality is 37 – (3x + 5) ≥ 9x – 8(x – 3)
Simplifying, 37 – 3x – 5 ≥ 9x – 8x + 24
or 32 – 3x ≥ x + 24
Transposing x to L.H.S. and 32 to R.H.S., We get
-3x – x ≥ 24 – 32 or -4x ≥ -8
Dividing both sides by – 4, we get
x < 2
∴ The solution is (- ∞, 2].

Ex 6.1 Class 11 Maths Question 15.

Solution.
The inequality is 
Multiplying each term by the L.C.M. of 4, 3, 5, i.e., by 60, we get
15x < 100x – 40 – 84x + 36
or, 15x < 100x – 84x -40 + 36
or, 15x < 16x – 4
Tranposing 16x to L.H.S., we get
15x – 16x < -4 or -x < -4 Multiplying both sides by -1, we get x >4
∴ The solution is (4, ∞).

Ex 6.1 Class 11 Maths Question 16.

Solution.
The inequality is 
Multiplying each term by L.C.M. of 3,4, 5, i.e., by 60

or, 20(2x – 1) ≥ (3x – 2) x 15 – (2 – x) x 12
or, 40x – 20 ≥ 45x – 30 – 24 + 12x
or, 40x – 20 ≥ 57x – 54
Transposing 57x to L.H.S. and -20 to R.H.S., we get
40x – 57x ≥ -54 + 20 or -17x ≥ -34
Dividing both sides by -17, we get
x <2
∴ The solution is (- ∞, 2].

Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line.
Ex 6.1 Class 11 Maths Question 17.
3x – 2 < 2x + 1
Solution.
The inequality is 3x – 2 < 2x + 1
Transposing 2x to L.H.S. and -2 to R.H.S, we get
3x- 2x < 1 + 2 or, x <3

Ex 6.1 Class 11 Maths Question 18.
5x – 3 ≥ 3x – 5
Solution.
The inequality is 5x – 3 ≥ 3x – 5
Transposing 3x to L.H.S. and -3 to R.H.S., we get
∴ 5x – 3x ≥ -5 + 3 or, 2x ≥ -2
Dividing both sides by 2, we get

Ex 6.1 Class 11 Maths Question 19.
3(1 – x) < 2(x + 4)
Solution.
3(1-x) < 2(x + 4)
Simplifying 3 – 3x < 2x + 8
Transposing 2x to L.H.S. and 3 to R.H.S., we get
-3x – 2x < 8 – 3 or -5x < 5
Dividing both sides by -5, we get

Ex 6.1 Class 11 Maths Question 20.

Solution.
The inequality is 

Ex 6.1 Class 11 Maths Question 21.
Ravi obtained 70 and 75 marks in first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
Solution.
Let Ravi gets x marks in third unit test.
∴ Average marks obtained by Ravi

He has to obtain atleast 60 marks,
∴ 
Multiplying both sides by 3,
145 + x ≥ 60 x 3 = 180
Transposing 145 to R.H.S., we get
x ≥ 180 – 145 = 35
∴ Ravi should get atleast 35 marks in the third unit test.

Ex 6.1 Class 11 Maths Question 22.
To receive Grade ‘A’ in the course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade ‘A’ in the course.
Solution.
Let Sunita obtained x marks in the fifth examination.
∴ Average marks of 5 examinations

This average must be atleast 90
∴ 
Multiplying both sides by 5
368 + x ≥ 5 x 90 = 450
Transposing 368 to R.H.S., we get
x ≥ 450 – 368 = 82
∴ Sunita should obtain atleast 82 marks in the fifth examination.

Ex 6.1 Class 11 Maths Question 23.
Find all pairs of consecutive odd positive integers both of which are smaller than 10, such that their sum is more than 11.
Solution.
Let x be the smaller of the two odd positive integers. Then the other integer is x + 2. We should havex + 2< 10 and x + (x + 2) > 11 or, 2x + 2 > 11
or, 2x > 11 – 2 or, 2x > 9 or, 
Hence, if one number is 5 (odd number), then the other is 7. If the smaller number is 7, then the other is 9. Hence, possible pairs are (5, 7) and (7, 9).

Ex 6.1 Class 11 Maths Question 24.
Find all pairs of consecutive even positive integers, both of which are larger than 5, such that their sum is less than 23.
Solution.
Let x be the smaller of the two positive even integers then the other one is x + 2, then we should have x > 5
and x + x + 2 < 23 or, 2x + 2 < 23
or, 2x < 21 or, 
Thus, the value of x may be 6,8,10 (even integers) Hence, the pairs may be (6, 8), (8,10), (10,12).

Ex 6.1 Class 11 Maths Question 25.
The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.
Solution.
Let the shortest side measures x cm
The longest side will be 3x cm.
Third side will be (3x – 2) cm.
According to the problem, x + 3x + 3x – 2 ≥ 61
or, 7x – 2 ≥ 61 or, 7x ≥ 63 or, x ≥ 9
Hence, the minimum length of the shortest side is 9 cm.

Ex 6.1 Class 11 Maths Question 26.
A man wants to cut three lengths from a single piece of board of length 91 cm. The second length is to be 3 cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths for the shortest board if the third piece is to be atleast 5 cm longer than the second?
[Hint: If x is the length of the shortest board, then x, (x + 3) and 2x are the lengths of the second and third piece, respectively. Thus, x + (x + 3) + 2x ≤ 91 and 2x ≥ (x + 3) + 5].
Solution.
Let x be the length of the shortest board, then x + 3 is the second length and 2x is the third length. Thus, x + (x + 3) + 2x ≤ 91
or 4x + 3 ≤ 91 or 4x ≤ 88 or x ≤ 22
According to the problem, 2x ≥ (x + 3) + 5 or x ≥ 8
∴ Atleast 8 cm but not more than 22 cm are the possible lengths for the shortest board.

We hope the NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.1 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.1, drop a comment below and we will get back to you at the earliest.

Solve the following inequalities graphically in two-dimensional plane
Ex 6.2 Class 11 Maths Question 1.
x + y < 5
Solution.
Consider the equation x + y = 5. It passes through the points (0, 5) and (5, 0). The line x + y = 5 is represented by AB. Consider the inequality x + y < 5

Put x = 0, y = 0
0 + 0 = 0 < 5, which is true. So, the origin O lies in the plane x + y < 5
∴ Shaded region represents the inequality x + y < 5

Ex 6.2 Class 11 Maths Question 2.
2x + y ≥ 6
Solution.
Consider the equation 2x + y = 6
The line passes through (0, 6), (3, 0).
The line 2x + y = 6 is represented by AB.

Now, consider 2x + y ≥ 6
Put x = 0, y = 0
0 + 0 ≥ 6, which does not satisfy this inequality.
∴ Origin does not lie in the region of 2x + y ≥ 6.
The shaded region represents the inequality 2x + y ≥ 6

Ex 6.2 Class 11 Maths Question 3.
3x + 4y ≤ 12
Solution.
We draw the graph of the equation 3x + 4y = 12. The line passes through the points (4, 0), (0, 3). This line is represented by AB. Now consider the inequality 3x + 4y ≤ 12
Putting x = 0, y = 0 0 + 0 = 0 ≤ 12, which is true

∴ Origin lies in the region of 3x + 4y ≤ 12 The shaded region represents the inequality 3x + 4y ≤ 12

Ex 6.2 Class 11 Maths Question 4.
y + 8 ≥ 2x
Solution.
Given inequality is y + 8 ≥ 2x
Let us draw the graph of the line, y+ 8 = 2x
The line passes through the points (4, 0), (0, -8).
This line is represented by AB.

Now, consider the inequality y + 8 ≥ 2x.
Putting x = 0, y = 0
0 + 8 ≥ 0, which is true
∴ Origin lies in the region of y + 8 ≥ 2x
The shaded region represents the inequality y + 8 ≥ 2x.

Ex 6.2 Class 11 Maths Question 5.
x – y ≤ 2
Solution.
Given inequality is x – y ≤ 2
Let us draw the graph of the line x – y = 2
The line passes through the points (2, 0), (0, -2)
This line is represented by AB.

∴ Origin lies in the region of x – y ≤ 2
The shaded region represents the inequality x – y ≤ 2.

Ex 6.2 Class 11 Maths Question 6.
2x – 3y > 6
Solution.
We draw the graph of line 2x – 3y = 6.
The line passes through (3, 0), (0, -2)
AB represents the equation 2x – 3y = 6
Now consider the inequality 2x – 3y > 6
Putting x = 0, y = 0
0 – 0 > 6, which is not true
∴ Origin does not lie in the region of 2x – 3y > 6.

The shaded region represents the inequality 2x – 3y > 6

Ex 6.2 Class 11 Maths Question 7.
-3x + 2y ≥ -6.
Solution.
Let us draw the line -3x + 2y = -6
The line passes through (2, 0), (0, -3)
The line AB represents the equation -3x + 2y = -6
Now consider the inequality -3x+ 2y ≥ -6
Putting x = 0, y = 0
0 + 0 ≥ -6, which is true.
∴ Origin lies in the region of -3x + 2y ≥ -6

The shaded region represents the inequality -3x + 2y ≥ – 6

Ex 6.2 Class 11 Maths Question 8.
3y- 5x < 30
Solution.
Given inequality is 3y – 5x < 30
Let us draw the graph of the line 3y – 5x = 30
The line passes through (-6, 0), (0, 10)
The line AB represents the equation 3y – 5x = 30
Now, consider the inequality 3y – 5x < 30
Putting x = 0, y = 0
0 – 0 < 30, which is true.
∴ Origin lies in the region of 3y – 5x < 30

The shaded region represents the inequality 3y – 5x < 30

Ex 6.2 Class 11 Maths Question 9.
y<- 2
Solution.
Given inequality is y < -2 ………(1)
Let us draw the graph of the line y = -2
AB is the required line.
Putting y = 0 in (1), we have
0 < -2, which is not true.

The solution region is the shaded region below the line.
Hence, every point below the line (excluding the line) is the solution area.

Ex 6.2 Class 11 Maths Question 10.
x > -3
Solution.
Let us draw the graph of x = -3
∴ AB represents the line x = -3
By putting x = 0 in the inequality x > -3
We get, 0 > -3, which is true.
∴ Origin lies in the region of x > -3.

Graph of the inequality x > -3 is shown in the figure by the shaded area

We hope the NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.2 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.2, drop a comment below and we will get back to you at the earliest.

Solve the following system of inequalities graphically:
Ex 6.3 Class 11 Maths Question 1.
x ≥ 3, y ≥ 2
Solution.
x ≥ 3, y ≥ 2
(i) AB represents the line x = 3
Putting x = 0 in x ≥ 3
0 ≥ 3, which is not true.
Therefore, origin does not lie in the region of x ≥ 3
Its graph is shaded on the right side of AB.

(ii) CD represents the line y = 2
Putting y = 0 in y ≥ 2 0 ≥ 2, which is not true.
Therefore, origin does not lie in the region of y ≥ 2.
Its graph is shaded above the CD.
Solution of system x ≥ 3 and y ≥ 2 is shown as the shaded region.

Ex 6.3 Class 11 Maths Question 2.
3x + 2y ≤ 12, x ≥ 1, y ≥ 2
Solution.
Inequalities are 3x + 2y ≤ 12, x ≥ 1, y ≥ 2
(i) The line l₁ : 3x + 2y = 12 passes through (4, 0), (0, 6)
AB represents the line, 3x + 2y = 12.
Consider the inequality 3x + 2y ≤ 12
Putting x = 0, y = 0 in 3x + 2y ≤ 12
0 + 0 ≤ 12, which is true.
Therefore, origin lies in the region 3x + 2y ≤ 12
∴ The region lying below the line AB including all the points lying on it.

(ii) The line l₂ : x = 1 passes through (1, 0). This line is represented by EF. Consider the inequality x ≥ 1
Putting x = 0, 0 ≥ 1, which is not true.
Therefore, origin does not lie in the region of x ≥ 1
The region lies on the right of EF and the points on EF from the inequality x ≥ 1.

(iii) The line l₃ : y = 2 passes through (0, 2). This line is represented by CD.
Consider the inequality y ≥ 2
Putting y = 0 in y ≥ 2, we get 0 ≥ 2 which is false.
∴ Origin does not lie in the region of y ≥ 2. y ≥ 2 is represented by the region above CD and all the points on this line. Hence, the region satisfying the inequalities.
3x + 2y ≥ 12, x ≥ 1, y ≥ 2 is the APQR.

Ex 6.3 Class 11 Maths Question 3.
2x + y ≥ 6, 3x + 4y ≤ 12
Solution.
The inequalities are 2x + y ≥ 6, 3x + 4y ≤ 12
(i) The line l₁ : 2x + y = 6 passes through (3, 0), (0, 6)
AB represents the line 2x + y = 6
Putting x = 0, y = 0 in 2x + y ≥ 6 0 ≥ 6, which is false.
∴ Origin does not lie in the region of 2x + y ≥ 6 Therefore, the region lying above the line AB and all the points on AB represents the inequality 2x + y ≥ 6

(ii) The line l₂ : 3x + 4y = 12 passes through (4, 0) and (0, 3).
This line is represented by CD.

Consider the inequality 3x + 4y ≤ 12
Putting x = 0, y = 0 in 3x + 4y ≤ 12, we get 0 ≤ 12, which is true.
∴ 3x + 4y ≤ 12 represents the region below the line CD (towards origin) and all the points lying on it.
The common region is the solution of 2x + 3y ≥ 6 are 3x + 4y ≤ 12 represented by the
shaded region in the graph.

Ex 6.3 Class 11 Maths Question 4.
x + y ≥ 4, 2x – y > 0
Solution.
The inequalities are , x + y ≥ 4, 2x – y > 0
(i) The line l₁: x + y = 4 passes through (4, 0) and (0, 4). This line is represented by AB. Consider the inequality x + y ≥ 4
Putting x = 0, y = 0 in x + y ≥ 4, we get 0 ≥ 4, which is false.
Origin does not lie in this region.

Therefore, ,r + y > 4 is represented by the region above the line x + y = 4 and all points lying on it.

(ii) The line l₂ : 2x – y = 0 passes through (0, 0) and (1, 2).
This line is represented by CD.
Consider the inequality 2x – y > 0
Putting x = 1, y = 0, we get 2 > 0, which is true
This shows (1, 0) lies in the region.
i.e. region lying below the line 2x – y = 0
represents 2x — y > 0
∴ The common region to both inequalities is shaded region as shown in the figure.

Ex 6.3 Class 11 Maths Question 5.
2x – y> 1, x – 2y < -1
Solution.
The inequalities are 2x – y > 1 and x – 2y < -1

(i) Let us draw the graph of line
l₁ : 2x – y = 1, passes through  and
(0, -1) which is represented by AB. Consider the inequality 2x – y > 1.
Putting x = y = 0, we get 0 > 1, which is false.
Therefore, origin does not lie in region of 2x – y > 1 i.e., 2x – y > 1 represents the area below the line AB excluding all the points lying on 2x – y = 1.

(ii) Let us draw the graph of the line
l₂ : x – 2y = -1, passes through (-1, 0) and (0,1/2) which is represented by CD.
Consider the inequality x – 2y < -1
Putting x = y = 0, we have 0 < -1, which is false.
Therefore, origin does not lie in region of x – 2y < -1 i.e., x – 2y < -1 represents the area above the line CD excluding all the points lying on x – 2y = -1
⇒ The common region of both the inequality is the shaded region as shown in figure.

Ex 6.3 Class 11 Maths Question 6.
x + y ≤ 6, x + y ≥ 4
Solution.
The inequalities are
x + y ≤ 6 and x + y ≥ 4
(i) The line l₁: x + y = 6 passes through (6, 0) and (0, 6). It is represented by AB.
Consider the inequality x + y ≤ 6
Putting x = 0, y = 0 in x + y ≤ 6 0 ≤ 6, which is true.

∴ Origin lies in the region of x + y ≤ 6
∴ x + y ≤ 6 is represented by the region below the line x + y = 6 and all the points lying on it.

(ii) The line l₂ : x + y = 4 passes through (4, 0) and (0, 4). It is represented by CD.
Consider the inequality x + y ≥ 4
Putting x = 0, y = 0 in x + y ≥ 4 or, 0 + 0 ≥ 4, which is false.
∴ Origin does not lie in the region of x + y ≥ 4
∴ x + y ≥ 4 is represented by the region above the line x + y = 4 and all the points lying on it.
∴ The solution region is the shaded region between AB and CD as shown in the figure.

Ex 6.3 Class 11 Maths Question 7.
2x + y ≥ 8, x + 2y ≥ 10
Solution.
The inequalities are 2x + y ≥ 8 and x + 2y ≥ 10
(i) Let us draw the graph of the line
l₁ : 2x + y = 8 passes through (4, 0) and (0, 8) which is represented by AB.
Consider the inequality 2x + y ≥ 8

Putting x = y = 0, we get 0 ≥ 8, which is false.
∴ Origin does not lie in the region of 2x + y ≥ 8.
i.e., 2x + y ≥ 8 represents the area above the line AB and all the points lying on 2x + y = 8.

(ii) Let us draw the graph of line
l₂ : x + 2y = 10, passes through (10, 0) and (0, 5) which is represented by CD. Consider the inequality x + 2y ≥ 10
Putting x = 0, y = 0, we have 0 ≥ 10, which is false.
∴ The origin does not lie in region of x + 2y ≥ 10
i.e. x + 2y ≥ 10 represents the area above the line CD and all the points lying on x + 2y = 10.
⇒ The common region of both the inequality is the shaded region as shown in the figure.

Ex 6.3 Class 11 Maths Question 8.
x + y ≤ 9, y > x, x ≥ 0
Solution.
The inequalities are x + y ≤ 9, y > x and x ≥ 0
(i) Consider the inequality x + y ≤ 9
The line l₁ : x + y = 9 passes through (9, 0) and (0, 9). AB represents this line.
Putting x = 0, y = 0 in x + y ≤ 9
0 + 0 = 0 ≤ 9, which is true.

Origin lies in this region. i.e., x + y ≤ 9 represents the area below the line AB and all the points lying on x + y = 9.

(ii) The line l₂ : y = x, passes through the origin and (2, 2).
∴ CD represents the line y = x
Consider the inequality y – x > 0
Putting x = 0,y = 1 in y – x > 0
1 – 0 > 0, which is true.
∴ (0, 1) lies in this region.
The inequality y > x is represented by the region above the line CD, excluding all the points lying on y – x = 0.

(iii) The region x ≥ 0 lies on the right of y-axis.
∴ The common region of the inequalities is the region bounded by ΔPQO is the solution of x + y ≤ 9, y > x, x ≥ 0.

Ex 6.3 Class 11 Maths Question 9.
5x + 4y ≤ 20, x ≥ 1, y ≥ 2
Solution.
The inequalities are 5x + 4y ≤ 20, x ≥ 1, y ≥ 2
(i) The line l₁ : 5x + 4y = 20 passes through (4, 0) and (0, 5). This line is represented by AB.
Consider the inequality 5x + 4y ≤ 20
Putting x = 0, y = 0
0 + 0 = 0 ≤ 20, which is true.
The origin lies in this region, i.e., region below the line 5x + 4y = 20 and all the points lying on it belong to 5x + 4y ≤ 20.

(ii) The line l₂ : y = 2, line is parallel to x-axis at a distance 2 from the origin. It is represented by EF. Putting y = 0, 0 ≥ 2 is not true.
Origin does not lie in this region.
Region above y = 2 represents the inequality y ≥ 2 including the points lying on it.

(iii) The line l₃ : x = 1, line parallel to y-axis at a distance 1 from the origin. It is represented by CD. Putting x = 0in x – 1 ≥ 0
-1 ≥ 0, which is not true.
Origin does not lie in this region.
∴ The region on the right of x = 1 and all the points lying on it belong to x ≥ 1.

∴ Shaded area bounded by ΔPQR is the solution of given inequalities.

Ex 6.3 Class 11 Maths Question 10.
3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0
Solution.
The inequalities are

We first draw the graphs of lines
l₁ : 3x + 4y = 60, l₂ : x + 3y – 30, x = 0 and y = 0.
(i) The line 3x + 4y = 60 passes through (20, 0) and (0,15) which is represented by AB. Consider the inequality 3x + 4y ≤ 60, putting x = 0, y = 0 in 3x + 4y ≤ 60, we get 0 + 0 ≤ 60, which is true.
∴ 3x + 4y ≤ 60 represents the region below AB and all the points on AB.

(ii) Further, x + 3y = 30 passes through (0,10) and (30, 0), CD represents this line.
Consider the inequality x + 3y ≤ 30
Putting x = 0, y = 0 in x + 3y < 30, w’e get 0 < 30 is true.
∴Origin lies in the region x + 3y ≤ 30. This inequality represents the region below it and the line itself.
Thus, we note that inequalities (1) and (2) represent the two regions below the respective lines (including the lines).
Inequality (3) represents the region on the right of y-axis and the i/-axis itself.
Inequality (4) represents the region above x-axis and the x-axis itself.
∴ Shaded area in the figure is the solution area.

Ex 6.3 Class 11 Maths Question 11.
2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6
Solution.
We have the inequalities :
We first draw the graphs of lines
l₁ : 2x + y = 4, l₂ : x + y = 3 and l₃ : 2x – 3y = 6
(i) 2x + y = 4, passes through (2, 0) and (0, 4) which represents by AB
Consider the inequality 2x + y ≥ 4
Putting x = 0, y = 0 in 2x + y ≥ 4, we get 0 ≥ 4 is false.
∴ Origin does not lie in the region of 2x + y ≥ 4 This inequality represents the region above the line AB and all the points on the line AB.

(ii) Again, x + y = 3 is represented by the line CD, passes through (3, 0) and (0, 3). Consider the inequality x + y ≤ 3, putting x = 0,y = 0 in x + y ≤ 3, we get 0 ≤ 3 is true.
∴ Origin lies in the region of x + y ≤ 3
∴ x + y ≤ 3 represents the region below the line CD and all the points on the line CD.

(iii) Further, 2x – 3y = 6 is represented by EF passes through (0, -2) and (3, 0).
Consider the inequality 2x – 3y ≤ 6, putting x = 0, y = 0 in 2x – 3y ≤ 6, we get 0 ≤ 6, which is true.
∴ Origin lies in it.
∴ 2x – 3y ≤ 6 represents the region above the line EF and all the points on the line EF.
∴ Shaded triangular area in the figure is the solution of given inequalities.

Ex 6.3 Class 11 Maths Question 12.
x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
Solution.
The inequalities are
x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
(i) The line l₁: x – 2y = 3 passes through (3, 0) and 
This is represented by AB. Consider the inequality x – 2y ≤ 3, putting x = 0, y = 0 we get 0 ≤ 3, which is true.
⇒ Origin lies in the region of x – 2y ≤ 3.
Region on the above of this line and including its points represents x – 2y ≤ 3

(ii) The line l₂ : 3x + 4y = 12 passes through (4, 0) and (0, 3). CD represents this line. Consider the inequality 3x + 4y ≥12 putting x = 0, y = 0, we get 0 ≥ 12 which is false.
∴ Origin does not lie in the region of 3x + 4y ≥ 12.
The region above the line CD and including points of the line CD represents 3x + 4y ≥ 12.

(iii) x ≥ 0 is the region on the right of Y-axis and all the points lying on it.

(iv) The line l₃ : y = 1 is the line parallel to X-axis at a distance 1 from it. Consider y ≥ 1 or y – 1 ≥ 0, putting y = 0 in y -1 ≥ 0
We get -1 ≱ 0, origin does not lie in the region.
y ≥ 1 is the region above y = 1 and the points lying on it.

∴ The shaded region shown in figure represents the solution of the given inequalities.

Ex 6.3 Class 11 Maths Question 13.
4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0
Solution.
The inequalities are 4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0
(i) The line l₁ : 4x + 3y = 60 passes through (15, 0), (0, 20) and it is represented by AB. Consider the inequality 4x + 3y ≤ 60
Putting x = 0, y = 0.
0 + 0 = 0 ≤ 60 which is true,
therefore, origin lies in this region.
Thus, region is below the line AB and the points lying on the line AB represents the inequality 4x + 3y ≤ 60.

(ii) The line l₂ : y = 2x passes through (0, 0). It is represented by CD.
Consider the inequality y ≥ 2x. Putting x = 0, y = 5 in y – 2x ≥ 0 5 ≥ 0 is true.
∴ (0, 5) lies in this region.

Region lying above the line CD and including the points on the line CD represents y ≥ 2x

(iii) x ≥ 3 is the region lying on the right of line l₃ : x = 3 and points lying on x = 3 represents the inequality x ≥ 3.
∴ The shaded area APQR in which x ≥ 0 and y ≥ 0 is true for each point, is the solution of given inequalities.

Ex 6.3 Class 11 Maths Question 14.
3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0
Solution.
The inequalities are 3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0
(i) The line l₁ : 3x + 2y = 150 passes through the points (50,0) and (0, 75). AB represents the line. Consider the inequality 3x + 2y ≤ 150.
Putting x = 0, y = 0 in 3x + 2y ≤ 150
⇒ 0 ≤ 150 which is true, shows that origin lies in this region.

The region lying below the line AB and the points lying on AB represents the inequality 3x + 2y < 150.

(ii) The line l₂ : x + 4y = 80 passes through the points (80, 0), (0, 20). This is represented by CD.
Consider the inequality x + 4y ≤ 80 putting x = 0, y = 0, we get 0 ≤ 80, which is true.
⇒ Region lying below the line CD and the points on the line CD represents the inequality x + 4y ≤ 80
(iii) x ≤ 15 is the region lying on the left to
l₃ : x = 15 represented by EF and the points lying on EF.

(iv) x ≥ 0 is the region lying on the right side of Y-axis and all the points on Y-axis.

(v) y ≥ 0 is the region lying above the X-axis and all the points on X-axis.
Thus, the shaded region in the figure is the solution of the given inequalities.

Ex 6.3 Class 11 Maths Question 15.
x+2y ≤ 10 , x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0
Solution.
The inequalities are x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0
(i) l₁ : x + 2y = 10 passes through (10, 0) and (0, 5). The line AB represents this equation. Consider the inequality x + 2y ≤ 10 putting x = 0, y = 0, we get 0 ≤ 10 which is true.
∴ Origin lies in the region of x + 2y ≤ 10.
∴ Region lying below the line AB and the points lying on it represents x + 2y ≤ 10

(ii) l₂ : x + y = 1 passes through (1, 0) and (0, 1). Thus line CD represents this equation. Consider the inequality x + y ≥ 1 putting x = 0, y = 0, we get 0 ≥ 1, which is not true. Origin does not lie in the region of x + y ≥ 1.
∴ The region lying above the line CD and the points lying on it represents the inequality x + y ≥1

(iii) l₃ : x – y = 0, passes through (0, 0). This is being represented by EF.
Consider the inequality x – y ≤0, putting x = 0, y = 1, We get 0 – 1 ≤ 0 which is true
⇒ (0,1) lies on x – y ≤ 0
The region lying above the line EF and the points lying on it represents the inequality x – y ≤ 0.

(iv) x ≥ 0 is the region lying on the right of Y-axis and the points lying on x = 0.

(v) y ≥ 0 is the region above X-axis, and the points lying on y = 0.
∴ The shaded area in the figure represents the given inequalities.

We hope the NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.3 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.3, drop a comment below and we will get back to you at the earliest.

Ex 10.1 Class 11 Maths Question 1.
Draw a quadrilateral in the Cartesian plane, whose vertices are (- 4, 5), (0, 7), (5, -5) and (-4, -2). Also, find its area.
Solution:
The figure of quadrilateral whose vertices are A(- 4, 5), B(0, 7), C(5, -5) and D(-4, -2) is shown in the below figure.

Ex 10.1 Class 11 Maths Question 2.
The base of an equilateral triangle with side 2a lies along they-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.
Solution:
Since base of an equilateral triangle lies along y-axis.

Ex 10.1 Class 11 Maths Question 3.
Find the distance between P(x₁ y₁) and Q(x₂, y₂) when :
(i) PQ is parallel to the y-axis,
(ii) PQ is parallel to the x-axis.
Solution:
We are given that co-ordinates of P is (x₁, y₁) and Q is (x₂, y₁).
Distance between the points P(x₁, y₁) and Q(x₂, y₁) is

Ex 10.1 Class 11 Maths Question 4.
Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
Solution:
Let the point be P(x, y). Since it lies on the x-axis ∴ y = 0 i.e., required point be (x, 0).
Since the required point is equidistant from points A(7, 6) and B(3, 4) ⇒ PA = PB

Ex 10.1 Class 11 Maths Question 5.
Find the slope of a line, which passes through the origin and the mid-point of the line segment joining the points P(0, -4) and B(8,0).
Solution:
We are given that P(0, -4) and B(8, 0).
Let A be the midpoint of PB, then

Ex 10.1 Class 11 Maths Question 6.
Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (-1, -1) are the vertices of a right angled triangle.
Solution:
Let A(4, 4), B(3, 5) and C(-1, -1) be the vertices of ∆ABC.
Let m₁ and m₂ be the slopes of AB and AC respectively.

Ex 10.1 Class 11 Maths Question 7.
Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
Solution:
The given line makes an angle of 90° + 30° = 120° with the positive direction of x-axis.
Hence, m = tan 120° = – .

Ex 10.1 Class 11 Maths Question 8.
Find the value of x for which the points (x, -1), (2, 1) and (4,5) are collinear.
Solution:
Let A(x, -1), B(2, 1) and C(4, 5) be the given collinear points. Then by collinearity of A, B, C, we have slope of AB = slope of BC

Ex 10.1 Class 11 Maths Question 9.
Without using distance formula, show that points (-2, -1), (4, 0), (3, 3) and (-3, 2) are the vertices of a parallelogram.
Solution:
Let A(-2, -1), B(4, 0), C(3, 3) and D(-3, 2) be the vertices of the given quadrilateral ABCD. Then,

Ex 10.1 Class 11 Maths Question 10.
Find the angle between the x-axis and the line joining the points (3, -1) and (4, -2).
Solution:
We are given that the points are A(3, -1) and B(4, -2)

Ex 10.1 Class 11 Maths Question 11.
The slope of a line is double of the slope of another line. If tangent of the angle between them is , find the slopes of the lines.
Solution:
Let m₁ and m₂ be the slopes of two lines.

Ex 10.1 Class 11 Maths Question 12.
A line passes through (x₁, y₂) and (h, k). If slope of the line is m, show that k – y₁ = m (h – x₁).
Solution:
A line passes through (x₁, y₁) and (h, k). Also, the slope of the line is m.

Ex 10.1 Class 11 Maths Question 13.
If three points (h, 0), (a, b) and (0, k) lie on a line, show that 
Solution:
Let A(h, 0), B(o, b) and C(0, k) be the given collinear points.
∴ Slope of AB = Slope of BC

Ex 10.1 Class 11 Maths Question 14.
Consider the following population and year graph, find the slope of the line AB and using it, find what will be the population in the year 2010?
Solution:
Slope of AB + 
Let the population in year 2010 is y, and co-ordinate of C is (2010, y) then, slope of AB = slope of BC

We hope the NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.1, help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.1, drop a comment below and we will get back to you at the earliest.

Ex 10.2 Class 11 Maths Question 1.
Write the equations for the x-and y-axes.
Solution:
We know that the ordinate of each point on the x-axis is 0.
If P(x, y) is any point on the x-axis, then y = 0.
∴ Equation of x-axis is y = 0.
Also, we know that the abscissa of each point on the y-axis is 0. If P(x, y) is any point on the y-axis, then x = 0.
∴ Equation of y-axis is x = 0.

Ex 10.2 Class 11 Maths Question 2.
Passing through the point (-4,3) with slope .
Solution:
We know that the equation of a line with slope m and passing through the point (x₀, y₀) is given by (y – y₀) = m (x – x₀).

Ex 10.2 Class 11 Maths Question 3.
Passing through (0, 0) with slope m.
Solution:
We know that the equation of a line with slope m and passing through the point (x₀, y₀) is given by (y – y₀) = m(x – x₀)
Here, slope = m, x₀ = 0, y₀ = 0 Required equation is (y – 0) = m(x – 0)
⇒ y = mx.

Ex 10.2 Class 11 Maths Question 4.
Passing through (2,2^3) and inclined with the x-axis at an angle of 75°.
Solution:
We know that the equation of a line with slope m and passing through the point (X₀, y₀) is given by (y – y₀) = m(x – x₀)

Ex 10.2 Class 11 Maths Question 5.
Intersecting the x-axis at a distance of 3 units to the left of origin with slope -2.
Solution:
We know that the equation of a line with slope m and passing through the point
(x₀, y₀) is given by (y – y₀) = m(x – x₀).
Here, m = – 2, x₀ = – 3, y₀ = 0
y-0 = -2(x + 3) ⇒ 2x + y + 6 = 0

Ex 10.2 Class 11 Maths Question 6.
Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with positive direction of the x-axis.
Solution:
We know that the equation of line with slope m and passing through the point (x₀, y₀) is given by (y – y₀) = m(x – x₀)

Ex 10.2 Class 11 Maths Question 7.
Passing through the points (-1,1) and (2, -4).
Solution:
Let the given points be A(-1, 1) and B(2, -4).
We know that the equation of a line passing through the given points (xx, y,) and (x2, y2) is given by

Ex 10.2 Class 11 Maths Question 8.
Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 30°.
Solution:
Here, we are given that p = 5 and ⍵ = 30°.

Ex 10.2 Class 11 Maths Question 9.
The vertices of ∆PQR are P(2, 1), Q(-2, 3) and ff(4, 5). Find equation of the median through the vertex R.
Solution:
The vertices of ∆PQR are P( 2, 1), Q(-2, 3) and R(4, 5).
Let S be the midpoint of PQ.

Ex 10.2 Class 11 Maths Question 10.
Find the equation of the line passing through (-3, 5) and perpendicular to the line through the points (2,5) and (-3,6).
Solution:
Let M(2, 5) and N(-3, 6) be the end points of the given line segment.

Ex 10.2 Class 11 Maths Question 11.
A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line.
Solution:
Let A(1, 0) and B( 2, 3) be the given points and D divides the line segment in the ratio 1 : n.

Ex 10.2 Class 11 Maths Question 12.
Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2,3).
Solution:
Let the required line make intercepts a on the x-axis and y-axis.
Then its equation is 
⇒ x + y = a … (i)
Since (i) passes through the point (2, 3), we have
2 + 3 = a ⇒ a = 5
So, required equation of the line is
 ⇒ x + y = 5.

Ex 10.2 Class 11 Maths Question 13.
Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.
Solution:
Let the intercepts made by the line on the x-axis and y-axis be o and 9 – a respectively.
Then its equation is

Since it passes through point (2, 2), we have 
⇒ 2(9 – a) + 2a = a(9 – a)
⇒ 18 – 2a + 2a = 9a – 9a²
⇒ 18 = 9a – a² v a² – 9a + 18 = 0
⇒ a² – 6a – 3a + 18 = 0
⇒ a(a – 6) – 3 (a – 6) = 0 ⇒ a = 3, 6
Now, if a = 3 ⇒ b = 9 – 3 = 6 and if a = 6 ⇒ b = 9 – 6 = 3
So, required equation is

i.e., 2x + y – 6 = 0 or x + 2y – 6 = 0.

Ex 10.2 Class 11 Maths Question 14.
Find equation of the line through the point (0, 2) making an angle  with the positive x-axis. Also, find the equation of the line parallel to it and crossing the y-axis at a distance of 2 units below the origin.
Solution:
Here, m = tan  = 
The equation of the line passing through point (0, 2) is y -2 = (x – 0)
⇒  + y – 2 = 0
The slope of line parallel to
 + y – 2 = 0 is .
Since, it passes through (0, -2).
So, the equation of line is
y + 2= (x – 0)
⇒  + y + 2 = 0.

Ex 10.2 Class 11 Maths Question 15.
The perpendicular from the origin to a line meets it at the point (-2, 9), find the equation of the line.
Solution:

Ex 10.2 Class 11 Maths Question 16.
The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms C.
Solution:
Assuming L along x-axis and C along y-axis, we have two points (124.942, 20) and (125.134, 110). By two point form, the point (L, C) satisfies the equation

Ex 10.2 Class 11 Maths Question 17.
The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs. 14/litre and 1220 litres of milk each week at Rs. 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs. 17/litre?
Solution:
Assuming L (litres) along x-axis and R(rupees) along y-axis, we have two points (980,14) and (1220,16).
By two point form, the point (L, R) satisfies the equation.

Ex 10.2 Class 11 Maths Question 18.
P(a, b) is the mid-point of a lone segment between axes. Show that equation of the line is .
Solution:
Let the line AB makes intercepts c and d on the x-axis and y-axis respectively.

Ex 10.2 Class 11 Maths Question 19.
Point R(h, k) divides a line segment between the axes in the ratio 1:2. Find equation of the line.
Solution:
Let AB be the given line segment making intercepts a and b on the x-axis & y-axis respectively.
Then, the equation of line AB is 

So, these points are A(a, 0) and B(0, b).
Now, R(h,k) divides the line segment Ab in the ratio 1 : 2.

Ex 10.2 Class 11 Maths Question 20.
By using the concept of equation of a line, prove that the three points (3, 0), (- 2, – 2) and (8, 2) are collinear.
Solution:
Let the given points be A(3, 0), B(-2, -2) and C(8, 2). Then the equation of the line passing through A and B is

Clearly the point C(8, 2) satisfy the equation 2x – 5y – 6 = 0.
(∵ 2(8) – 5(2) – 6 = 16 – 10 – 6 = 0)
Hence, the given points lie on the same straight line whose equation is 2x – 5y – 6 = 0.

We hope the NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2, help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2, drop a comment below and we will get back to you at the earliest.

Ex 10.3 Class 11 Maths Question 1.
Reduce the following equations into slope- intercept form and find their slopes and the y-intercepts.
(i) x + 7y = 0,
(ii) 6x + 3y-5 = 0,
(iii) y=0
Solution:
(i) We have given an equation x + 7y = 0, which can be written in the form
⇒ 7y = – x ⇒ y = x + 0 … (1)
Also, the slope intercept form is y-mx + c …(2)
On comparing (1) and (2), we get
m = , c = 0
Hence the slope is  and the y-intercept = 0.

(ii) We have given an equation 6x + 3y – 5 = 0, which can be written in the form 3y = – 6x + 5
⇒ y = – 2x +  …(1)
Also, the slope intercept form is y = mx + c … (2)
On comparing (1) and (2), we get
m = – 2 and c = 
i.e. slope = – 2 and the y-intercept = 

(iii) We have given an equation y = 0
y = 0·x + 0 … (1)
Also, the slope intercept form is y = mx + c … (2) On comparing (1) and (2), we get
m = 0, c = 0.
Hence, slope is 0 and the y-intercept is 0.

Ex 10.3 Class 11 Maths Question 2.
Reduce the following equations into intercept form and find their intercepts on the axes.
(i) 3x + 2y – 12 = 0,
(ii) 4x – 3y = 6,
(iii) 3y + 2 = 0
Solution:
(i) Given equation is 3x + 2y – 12 = 0 We have to reduce the given equation into intercept form, i.e.,  …(1)
Now given, 3x + 2y = 12
⇒  ⇒  …(2)
On comparing (1) and (2), we get a = 4, b = 6 Hence, the intercepts of the line are 4 and 6.

(ii) Given equation is 4x – 3y = 6
We have to reduce the given equation into intercept form, i.e.,  …(1)
 …(2)
On comparing (1) and (2), we get
a =  and b = – 2
Hence, the intercepts of the line are  and -2.

(iii) Given equation is 3y + 2 = 0
We have to reduce the given equation into intercept form, i.e., 
3y = -2
⇒ y = 
The above equation shows that, it is not the required equation of the intercept form as it is parallel to x-axis.
We observe that y-intercept of the line is , but there is no intercept on x-axis.

Ex 10.3 Class 11 Maths Question 3.
Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.
(i) x –  + 8 = 0,
(ii) y-2 = 0,
(iii) x-y = 4.
Solution:
(i) Given equation is x –  + 8 = 0
 x –  = -8
 -x +  = 8 … (i)
Also, 

Now dividing both the sides of (1) by 2, we get

⇒ – cos 60°x + sin 60° y = 4.
⇒ {cos (180° – 60°)) x + {sin (180° – 60°)|y = 4
⇒ cos 120° x + sin 120° y = 4
∴ x cos 120° + y sin 120° = 4 is the required equation in normal form
∵ The normal form is x coso) + y sin⍵ = p
So, ⍵ = 120° and p = 4
⍵ Distance of the line from origin is 4 and the angle between perpendicular and positive x-axis is 120°.

(ii) Given equation is y – 2 = 0
⇒ y = 2
⇒ 0 · x + l · y = 2
⇒ x cos 90° + y sin 90° = 2 is the required equation in normal form
∵ The normal form is x cos⍵ + y sin⍵ = p
So, ⍵ = 90° and p = 2
⍵ Distance of the line from origin is 2 and the angle between perpendicular and positive x-axis is 90°.

(iii) Given equation is x – y = 4 … (1)
Also 

Now dividing both the sides of (1) bt , we get

is the required equation in normal form.
∵ The normal form is x cos⍵ + y sin⍵ = p
So, P =  and ⍵ = 315°
∴ Distance of the line from the origin is  and the angle between perpendicular and the positive x-axis is 315°.

Ex 10.3 Class 11 Maths Question 4.
Find the distance of the point (-1, 1) from the line 12 (x+ 6) = 5(y — 2).
Solution:
The equation of line is 12(x + 6) = 5(y – 2) …(i)
⇒ 12x + 72 = 5y-10
⇒ 12x – 5y + 82 = 0
∴ Distance of the point (-1, 1) from the line (i)

Ex 10.3 Class 11 Maths Question 5.
Find the points on the x-axis, whose distances from the line  are 4 units.
Solution:
We have a equation of line , which can be written as
4x + 3y – 12 = 0 … (i)
Let (a, 0) be the point on x-axis whose distance from line (i) is 4 units.

Ex 10.3 Class 11 Maths Question 6.
Find the distance between parallel lines
(i) 15x+8y-34 = 0and 15x + 8y+31 =0
(ii) |(x + y) + p = 0 and |(x + y) – r = 0.
Solution:
If lines are Ax + By + Q = 0
and Ax + By + C₂ = 0, then distance between

Ex 10.3 Class 11 Maths Question 7.
Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (-2, 3).
Solution:
We have given an equation of line 3x – 4y + 2 = 0
Slop of the line(i) = 
Thus, slope of any line parallel to the given line (i) is  and passes through (-2, 3), then its equation is

Ex 10.3 Class 11 Maths Question 8.
Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.
Solution:
Given equation is x – 7y + 5 = 0 … (i)
Slope of this line = 
∴ Slope of any line perpendicular to the line (i) is -7 and passes through (3, 0) then
y – 0 = -7(x – 3)
[∵ Product of slope of perpendicular lines is -1]
⇒ y = -7x + 21
⇒ 7x + y – 21 = 0, is the required equation of line.

Ex 10.3 Class 11 Maths Question 9.
Find angles between the lines  + y = 1 and x +  = 1.
Solution:
The given equations are
 + y = 1 … (i)
x +  = 1 … (ii)
Since we have to find an angle between the two lines i.e., firstly we have to find the slopes of (i) and (ii).

Ex 10.3 Class 11 Maths Question 10.
The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0 at right angle. Find the value of h.
Solution:
Given points are (h, 3) and (4,1).
∴ Slope of the line joining (h, 3) & (4,1)

Ex 10.3 Class 11 Maths Question 11.
Prove that the line through the point (x₁ y₁) and parallel to the line Ax + By + C = 0 is A(x-x₁) + B(y-y₁) = 0.
Solution:
Given equation of a line is Ax + By + C = 0
∴ Slope of the above line = 
i.e. slope of any line parallel to given line and passing through (x₁, y₁) is 
Then equation is (y – y₂) =  (x – x₁)
=> B(y – y₁) = -A(x – x₁)
=> A(x – x₁) + B(y – y₁) = 0.
Hence proved.

Ex 10.3 Class 11 Maths Question 12.
Two lines passing through the point (2, 3) intersects each other at an angle of 60°. If slope of one line is 2, find equation of the other line.
Solution:
We have given a point (2, 3), through which two lines are passing and intersects at an angle of 60°.
Let m be the slope of the other line

Ex 10.3 Class 11 Maths Question 13.
Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2).
Solution:
suppose the given points are A and B.
Let M be the mid point of AB.

Ex 10.3 Class 11 Maths Question 14.
Find the coordinates of the foot of perpendicular from the point (-1, 3) to the line 3x – 4y – 16 = 0.
Solution:
We have, 3x – 4y – 16 = 0
Slope of the kine(i) = 
Then equation of any line ⊥ from (-1, 3) to the given line(i) is

Ex 10.3 Class 11 Maths Question 15.
The perpendicular from the origin to the line y = mx + c meets it at the point (-1,2). Find the values of m and c.
Solution:
Given, the perpendicular from the origin to the line y = mx + c meets it at the point (-1, 2)
∴ 2 = m (-1) + c … (i)
⇒ c – m = 2

Ex 10.3 Class 11 Maths Question 16.
If p and q are the lengths of perpendiculars from the origin to the lines x cosθ – y sinθ = k cos 2θ and x secθ + y cosecθ = k, respectively, prove that p² + 4q² = k².
Solution:
Given p and q are the lengths of perpendiculars from the origin to the lines x cos θ – ysinθ=k cos 2θ and xsecθ+y cosec θ = k.

Ex 10.3 Class 11 Maths Question 17.
In the triangle ABC with vertices A(2, 3), 8(4, -1) and C( 1, 2), find the equation and length of altitude from the vertex A.
Solution:
We have given a AABC with the vertices, A (2, 3), B (4, -1) and C (1, 2)

Ex 10.3 Class 11 Maths Question 18.
If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that .
Solution:
Given, p be the length of perpendicular from the origin to the line whose intercepts

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Ex 9.1 Class 11 Maths Question 1.
a_n = n(n + 2)
Solution:
We haven a_n = n(n + 2)
subtituting n = 1, 2, 3, 4, 5, we get
a₁ = 9(1 + 2) = 1 x 3 = 3
a₂ = 2(2 + ) = 2 x 4 = 8
a₃ = 3(3 + 2) = 3 x 5 = 15
a₄ = 4(4 + 2) = 4 x 6 = 24
a₅ = 5(5 + 2) = 5 x 7 = 35
∴ The first five terms are 3, 8, 15, 24, 35.

Ex 9.1 Class 11 Maths Question 2.
a_n = 
Solution:

Ex 9.1 Class 11 Maths Question 3.
a_n = 2ⁿ
Solution:

Ex 9.1 Class 11 Maths Question 4.
a_n = 
Solution:

Ex 9.1 Class 11 Maths Question 5.
a_n = (- 1)^n-1 5ⁿ⁺¹
Solution:
We have, a_n = (- 1)^n-1 5ⁿ⁺¹
Substituting n = 1, 2, 3, 4, 5, we get
a₁ =(-1)^1-1 5¹⁺¹ = (-1)^° 5² = 25
a₂ =(-1)^2-1 5²⁺¹ = (-1)¹ 5³ = 125
a₃ =(-1)^3-1 5³⁺¹ = (-1)² 5⁴ = 625
a₄ =(-1)^4-1 5⁴⁺¹ = (-1)³ 5⁵ = -3125
a₅ =(-1)^5-1 5⁵⁺¹ = (-1)⁴ 5⁶ = 15625
∴ The first five terms are 25, – 125, 625, -3125, 15625.

Ex 9.1 Class 11 Maths Question 6.
a_n = n
Solution:
Substituting n = 1, 2, 3, 4, 5, we get


Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

Ex 9.1 Class 11 Maths Question 7.
a_n = 4n – 3; a₁₇, a₂₄
Solution:
We have a_n = 4n – 3

Ex 9.1 Class 11 Maths Question 8.
a_n = ; a₇
Solution:
We have, a_n = ; a₇

Ex 9.1 Class 11 Maths Question 9.
a_n = (-1)^{n – 1} n³; a₉
Solution:
We have, a_n = (-1)^{n – 1} n³

Ex 9.1 Class 11 Maths Question 10.
a_n = ; a ₂₀
Solution:
We have, a_n = 

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:

Ex 9.1 Class 11 Maths Question 11.
a₁ = 3, a_n = 3a_n-1+2 for all n>1
Solution:
We have given a₁ = 3, a_n = 3a_n-1+2
⇒ a₁ = 3, a₂ = 3a₁ + 2 = 3.3 + 2 = 9 + 2 = 11,
a₃ = 3a₂ + 2 = 3.11 + 2 = 33 + 2 = 35,
a₄ = 3a₃ + 2 = 3.35 + 2 = 105 + 2 = 107,
a₅ = 3a₄ + 2 = 3.107 + 2 = 321 + 2 = 323,
Hence, the first five terms of the sequence are 3, 11, 35, 107, 323.
The corresponding series is 3 + 11 + 35 + 107 + 323 + ………..

Ex 9.1 Class 11 Maths Question 12.
a₁ = -1, a_n = , n ≥ 2
Solution:
We have given

Hence the first five terms of the given sequence are – 1, -1/2, -1/6, -1/24, -1/120.
The corresponding series is

Ex 9.1 Class 11 Maths Question 13.
a₁ = a₂ = 2, a_n = a_n-1 – 1, n >2
Solution:
We have given a₁ = a₂ = 2, a_n = a_n-1 – 1, n >2
a₁ = 2, a₂ = 2, a₃= a₂ – 1 = 2 – 1 = 1,
a₄ = a₃ – 1 = 1 – 1 = 0 and a₅ = a₄ – 1 = 0 – 1 = -1
Hence the first five terms of the sequence are 2, 2, 1, 0, -1
The corresponding series is
2 + 2 + 1 + 0 + (-1) + ……

Ex 9.1 Class 11 Maths Question 14.
Find Fibonacci sequence is defined by 1 = a₁ = a₂ and a_n = a_n-1 + a_n-2, n > 2
Find , for n = 1, 2, 3, 4, 5
Solution:
We have,

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Ex 9.2 Class 11 Maths Question 1.
Find the sum of odd integers from 1 to 2001.
Solution:
We have to find 1 + 3 + 5 + ……….. + 2001
This is an A.P. with first term a = 1, common difference d = 3-1 = 2 and last term l = 2001
∴ l = a + (n-1 )d ⇒ 2001 = 1 + (n -1)2 ⇒ 2001 = 1 + 2n – 2 ⇒ 2n = 2001 + 1

Ex 9.2 Class 11 Maths Question 2.
Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
Solution:
We have to find 105 +110 +115 + ……..+ 995
This is an A.P. with first term a = 105, common difference d = 110 -105 = 5 and last term 1 = 995

Ex 9.2 Class 11 Maths Question 3.
In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is -112.
Solution:
Let a = 2 be the first term and d be the common difference.

Ex 9.2 Class 11 Maths Question 4.
How many terms of the A.P. -6, , -5, ……. are needed to give the sum – 25 ?
Solution:
Let a be the first term and d be the common difference of the given A.P., we have

Ex 9.2 Class 11 Maths Question 5.
In an A.P., if pth term is  and qth term is  prove that the sum of first pq terms is , where p±q.
Solution:
Let a be the first term & d be the common difference of the A.P., then

Ex 9.2 Class 11 Maths Question 6.
If the sum of a certain number of terms of the A.P. 25, 22, 19, …. is 116. Find the last term.
Solution:
Let a be the first term and d be the common difference.
We have a = 25, d = 22 – 25 = -3, S_n = 116

Ex 9.2 Class 11 Maths Question 7.
Find the sum to n terms of the A.P., whose k^th term is 5k + 1.
Solution:
We have a_k = 5k +1
By substituting the value of k = 1, 2, 3 and 4,
we get

Ex 9.2 Class 11 Maths Question 8.
If the sum of n terms of an A.P. is (pn + qn²), where p and q are constants, find the common difference.
Solution:
We have S_n = pn + qn², where S„ be the sum of n terms.

Ex 9.2 Class 11 Maths Question 9.
The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18^th terms.
Solution:
Let a₁, a₂ & d₁ d₂ be the first terms & common differences of the two arithmetic progressions respectively. According to the given condition, we have

Ex 9.2 Class 11 Maths Question 10.
If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.
Solution:
Let the first term be a and common difference be d.
According to question

Ex 9.2 Class 11 Maths Question 11.
Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.
Prove that
Solution:
Let the first term be A & common difference be D. We have

Ex 9.2 Class 11 Maths Question 12.
The ratio of the sums of m and n terms of an A.P. is m²: n². Show that the ratio of m^th and n^th term is (2m -1): (2n -1).
Solution:
Let the first term be a & common difference be d. Then

Ex 9.2 Class 11 Maths Question 13.
If the sum of n terms of an A.P. is 3n² + 5n and its mth term is 164, find the value of m.
Solution:
We have S_n = 3n² + 5n, where Sn be the sum of n terms.

Ex 9.2 Class 11 Maths Question 14.
Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Solution:
Let A₁, A₂, A₃, A₄, A₅ be numbers between 8 and 26 such that 8, A₁, A₂, A₃, A₄, A₅, 26 are in A.P., Here a = 8, l = 26, n = 7

Ex 9.2 Class 11 Maths Question 15.
If  is the A.M. between a and b, then find the value of n.
Solution:
We have 

Ex 9.2 Class 11 Maths Question 16.
Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7^thand (m – 1 )^th numbers is 5:9. Find the value of m.
Solution:
Let the sequence be 1, A₁, A₂, ……… A_m, 31 Then 31 is (m + 2)^th term, a = 1, let d be the common difference

Ex 9.2 Class 11 Maths Question 17.
A man starts repaying a loan as first instalment of Rs. 100. If he increases the instalment by Rs. 5 every month, what amount he will pay in the 30th instalment?
Solution:
Here, we have an A.P. with a = 100 and d = 5
∴ a_n= a + 29d = 100 + 29(5) = 100 + 145 = 245
Hence he will pay Rs. 245 in 30^th instalment.

Ex 9.2 Class 11 Maths Question 18.
The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.
Solution:
Let there are n sides of a polygon.

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Ex 9.3 Class 11 Maths Question 1.
Find the 20^th and n^th terms of the G.P. , ….
Solution:

Ex 9.3 Class 11 Maths Question 2.
Find the 12^th term of a G.P. whose 8^th term is 192 and the common ratio is 2.
Solution:
We have, a_s = 192, r = 2

Ex 9.3 Class 11 Maths Question 3.
The 5^th, 8^th and 11th terms of a G.P. are p, q and s, respectively. Show that q² = ps.
Solution:
We are given

Ex 9.3 Class 11 Maths Question 4.
The 4th term of a G.P. is square of its second term, and the first term is – 3. Determine its 7^th term.
Solution:
We have a= -3, a₄ = (a₂)²
Ex 9.3 Class 11 Maths Question 5.
Which term of the following sequences:


Solution:

Ex 9.3 Class 11 Maths Question 6.
For what values of x, the numbers , x,  are in G.P.?
Solution:
Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10:

Ex 9.3 Class 11 Maths Question 7.
0.14, 0.015, 0.0015, …. 20 items.
Solution:
In the given G.P.

Ex 9.3 Class 11 Maths Question 8.
, …. n terms
Solution:
In the given G.P.

Ex 9.3 Class 11 Maths Question 9.
1, -a, a²,- a³ … n terms (if a ≠ -1)
Solution:
In the given G.P.. a = 1, r = -a

Ex 9.3 Class 11 Maths Question 10.
x³, x⁵, ⁷, ….. n terms (if ≠±1).
Solution:
In the given G.P., a = x³, r = x²

Ex 9.3 Class 11 Maths Question 11.
Evaluate
Solution:

Ex 9.3 Class 11 Maths Question 12.
The sum of first three terms of a G.P. is  and 10 their product is 1. Find the common ratio and the terms.
Solution:
Let the first three terms of G.P. be ,a,ar, where a is the first term and r is the common ratio.

Ex 9.3 Class 11 Maths Question 13.
How many terms of G.P. 3, 3², 3³, … are needed to give the sum 120?
Solution:
Let n be the number of terms we needed. Here a = 3, r = 3, S_n = 120

Ex 9.3 Class 11 Maths Question 14.
The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.
Solution:
Let a₁, a₂, a₃, a₄, a₅, a₆ be the first six terms of the G.P.

Ex 9.3 Class 11 Maths Question 15.
Given a G.P. with a = 729 and 7^th term 64, determine S₇.
Solution:
Let a be the first term and the common ratio be r.

Ex 9.3 Class 11 Maths Question 16.
Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.
Solution:
Let a₁ a₂ be first two terms and a₃ a₅ be third and fifth terms respectively.
According to question

Ex 9.3 Class 11 Maths Question 17.
If the 4^th, 10^th and 16^th terms of a G.P. are x, y and z respectively. Prove that x, y, z are in G.P.
Solution:
Let a be the first term and r be the common ratio, then according to question

Ex 9.3 Class 11 Maths Question 18.
Find the sum to n terms of the sequence, 8, 88, 888, 8888 ………
Solution:
This is not a G.P., however we can relate it to a G.P. by writing the terms as S_n= 8 +88 + 888 + 8888 + to n terms

Ex 9.3 Class 11 Maths Question 19.
Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 
Solution:
On multiplying the corresponding terms of sequences, we get 256, 128, 64, 32 and 16, which forms a G.P. of 5 terms

Ex 9.3 Class 11 Maths Question 20.
Show that the products of the corresponding terms of the sequences a, ar, ar², ………… ar^n-1 and A, AR, AR², …….. , AR^n-1 form a G.P., and find the common ratio.
Solution:
On multiplying the corresponding terms, we get aA, aArR, aAr²R²,…… aAr^n-1R^n-1. We can see that this new sequence is G.P. with first term aA & the common ratio rR.

Ex 9.3 Class 11 Maths Question 21.
Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4^th by 18.
Solution:
Let the four numbers forming a G.P. be a, ar, ar², ar³
According to question,

Ex 9.3 Class 11 Maths Question 22.
If the p^th, q^th and r^th terms of a G.P. are a, b and c, respectively. Prove that a^q-r b^r-p c^p-q = 1.
Solution:
Let A be the first term and R be the common ratio, then according to question

Ex 9.3 Class 11 Maths Question 23.
If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P² – (ab)ⁿ.
Solution:
Let r be the common ratio of the given G.P., then b = n^th term = ar^n-1

Ex 9.3 Class 11 Maths Question 24.
Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)^th to (2n)^th term is 
Solution:
Let the G.P. be a, ar, ar², ……
Sum of first n terms = a + ar + ……. + ar^n-1

Ex 9.3 Class 11 Maths Question 25.
If a, b,c and d are in G.P., show that
(a² + b² + c²) (b² + c² + d²) = (ab + bc + cd)²
Solution:
We have a, b, c, d are in G.P.
Let r be a common ratio, then

Ex 9.3 Class 11 Maths Question 26.
Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Solution:
Let G₁, G₂ be two numbers between 3 and 81 such that 3, G₁ G₂,81 is a G.P.

Ex 9.3 Class 11 Maths Question 27.
Find the value of n so that  may be the geometric mean between a and b.
Solution:

Ex 9.3 Class 11 Maths Question 28.
The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio .
Solution:
Let a and b be the two numbers such that a + b = 6 

Ex 9.3 Class 11 Maths Question 29.
If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are .
Solution:
Let a and b be the numbers such that A, G are A.M. and G.M. respectively between them.

Ex 9.3 Class 11 Maths Question 30.
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour?
Solution:
There were 30 bacteria present in the culture originally and it doubles every hour. So, the number of bacteria at the end of successive hours form the G.P. i.e., 30, 60, 120, 240, …….

Ex 9.3 Class 11 Maths Question 31.
What will Rs. 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
Solution:
We have, Principal value = Rs. 500 Interest rate = 10% annually

Ex 9.3 Class 11 Maths Question 32.
If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
Solution:
Let α & β be the roots of a quadratic equation such that A.M. & G.M. of α, β are 8 and 5 respectively.

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Find the sunt to n terms of each of the series in Exercises 1 to 7.

Ex 9.4 Class 11 Maths Question 1.
1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ………
Solution:
In the given series, there is a sum of multiple of corresponding terms of two A.P’s. The two A.P’s are

Ex 9.4 Class 11 Maths Question 2.
1 x 2 x 3 + 2 x 3 x 4 + 3 x 4 x 5 + ……
Solution:
In the given series, there is a sum of multiple of corresponding terms of two A.P’s. The three A.P’s are

Ex 9.4 Class 11 Maths Question 3.
3 x 1² + 5 x 2² + 7 x 3² + …..
Solution:
In the given series there is sum of multiple of corresponding terms of two A.P’s. The two A.P’s are
(i) 3, 5, 7, …………… and
(ii) 1², 2², 3², ………………….
Now the n^th term of sum is an = (nth term of the sequence formed by first A.P.) x (n^th term of the sequence formed by second A.P.) = (2 n + 1) x n² = 2n³ + n² Hence, the sum to n terms is,

Ex 9.4 Class 11 Maths Question 4.
 …….
Solution:
In the given series there is sum of multiple of corresponding terms of two A.P’s. The two A.P’s are

Ex 9.4 Class 11 Maths Question 5.
5² + 6² + 7² + ………….. + 20²
Solution:
The given series can be written in the following way

Ex 9.4 Class 11 Maths Question 6.
3 x 8 + 6 x 11 + 9 x 25 + ………….
Solution:
In the given series, there is sum of multiple of corresponding terms of two A.P/s. The two A.P/s are
(i) 3, 6, 9, ………….. and
(ii) 8, 11, 14, ……………….
Now the nth term of sum is an = (n^th term of the sequence formed by first A.P.) x (n^th term of the sequence formed by second A.P.)

Ex 9.4 Class 11 Maths Question 7.
1² + (1² + 2²) + (1² + 2² + 3²) + ………….
Solution:
In the given series
a_n = 1² + 2² + …………….. + n²

Find the sum to n terms of the series in Exercises 8 to 10 whose n^th terms is given by

Ex 9.4 Class 11 Maths Question 8.
n(n + 1)(n + 4)
Solution:
We have

Ex 9.4 Class 11 Maths Question 9.
n² + 2ⁿ
Solution:
We have a_n = n² + 2ⁿ
Hence, the sum to n terms is,

Ex 9.4 Class 11 Maths Question 10.
(2n – 1)²
Solution:
We have

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