Bhagya

Students can download Maths Chapter 3 Algebra Ex 3.8 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.8

Question 1.
Find the square root of the following polynomials by division method
(i) x⁴ – 12x³ + 42x² – 36x + 9
Answer:

(ii) 31 x² – 28x³ + 4x⁴ + 42x + 9
Answer:
Rearrange the order we get
4x⁴ – 28x³ + 37x² + 42x + 9

(iii) 16x⁴ + 8x² + 1
Answer:

(iv) 121 x⁴ – 198x³ – 183x² + 216x + 144
Answer:

Question 2.
Find the square root of the expression
\(\frac{x^{2}}{y^{2}}\) – \(\frac { 10x }{ y } \) + 27 – \(\frac { 10y }{ x } \) + \(\frac{y^{2}}{x^{2}}\)
Answer:

Question 3.
Find the values of a and b if the following polynomials are perfect squares.
(i) 4x⁴ – 12x³ + 37x² + bx + a
Answer:

Since it is a perfect square
b + 42 = 0
b = – 42
a – 49 = 0
a = 49
∴ The value of a = 49 and b = – 42

(ii) ax⁴ + bx³ + 361x² + 220x + 100
Answer:
Re-arrange the order we get
100 + 220x + 361x² + bx³ + ax⁴

Since it is a perfect square
b – 264 = 0
b = 264
a – 144 = 0
a = 144
∴ The value of a = 144 and b = 264

Question 4.
Find the values of m and n if the following expressions are perfect sqaures.
(i) \(\frac{1}{x^{4}}\) – \(\frac{6}{x^{3}}\) + \(\frac{13}{x^{2}}\) + \(\frac { m }{ x } \) + n
Answer:

Since it is a perfect square
\(\frac { 1 }{ x } \) (m + 12) = 0
m + 12 = 0
m = -12
n – 4 = 0
n = 4
∴ The value of m = -12 and n = 4

(ii) x⁴ – 8x³ + mx² + nx + 16
Answer:

Since it is a perfect square
m – 16 – 8 = 0
m – 24 = 0
m = 24
n + 32 = 0
n = -32
∴ The value of m = 24 and n = -32

Students can download Maths Chapter 3 Algebra Ex 3.16 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.16

1. In the

write (i) The number of elements
(ii) The order of the matrix
(iii) Write the elements a₂₂, a₂₃, a₂₄, a₃₄, a₄₃, a₄₄.
Answer:
(i) The number of elements is 16
(ii) The order of the matrix is 4 × 4
(iii) Elements corresponds to

Question 2.
If a matrix has 18 elements, what are the possible orders it can have? What if it has 6 elements?
Answer:
The possible orders of the matrix having 18 elements are

The possible orders of the matrix having 6 elements are

Question 3.
Construct a 3 × 3 matrix whose elements are given by
(i) a_ij = |i – 2j|
Answer:
a_ij = |i – 2j|
The general 3 × 3 matrices is

a₁₁ = |1 – 2(1)| = |1 – 2| = | – 1| = 1
a₁₂ = |1 – 2(2)| = |1 – 4| = | – 3| = 3
a₁₃ = |1 – 2(3)| = |1 – 6| = | – 5| = 5
a₂₁ = |2 – 2(1)| = |2 – 2| = 0 = 0
a₂₂ = |2 – 2(2)| = |2 – 4| = | – 2| = 2
a₂₃ = |2 – 2(3)| = |2 – 6| = | – 4| = 4
a₃₁ = |3 – 2(1)| = |3 – 2| = | 1 | = 1
a₃₂ = |3 – 2(2)| = |3 – 4| = | – 1 | = 1
a₃₃ = |3 – 2(3)| = |3 – 6| = | – 3 | = 3
The required matrix

(ii) a_ij = \(\frac{(i+j)^{3}}{3}\)
Answer:
a₁₁ = \(\frac{(1+1)^{3}}{3}\) = \(\frac{2^{3}}{3}\) = \(\frac { 8 }{ 3 } \)
a₁₂ = \(\frac{(1+2)^{3}}{3}\) = \(\frac { 27 }{ 3 } \) = 9
a₁₃ = \(\frac{(1+3)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₂₁ = \(\frac{(2+1)^{3}}{3}\) = \(\frac { 27 }{ 3 } \) = 9
a₂₂ = \(\frac{(2+2)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₂₃ = \(\frac{(2+3)^{3}}{3}\) = \(\frac { 125 }{ 3 } \) = \(\frac { 125 }{ 3 } \)
a₃₁ = \(\frac{(3+1)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₃₂ = \(\frac{(3+2)^{3}}{3}\) = \(\frac { 125 }{ 3 } \) = \(\frac { 125 }{ 3 } \)
a₃₃ = \(\frac{(3+3)^{3}}{3}\) = \(\frac { 216 }{ 3 } \) = 72
The required matrix

Question 4.
If  then find the tranpose of A.
Answer:

transpose of A = (A^T)

Question 5.
If then find the tranpose of – A
Answer:

Transpose of – A = (-A^T) =

Question 6.
If A = then verify (AT)T = A
Answer:

Hence it is verified

Question 7.
Find the values of x, y and z from the following equations
(i)

Answer:
Since the given matrices are equal then all the corresponding elements are equal.
y = 12, z = 3, x = 3
The value of x = 3, y = 12 and z = 3

(ii)

Answer:
x + y = 6 ……(1)
5 + z = 5
z = 5 – 5 = 0
xy = 8
y = \(\frac { 8 }{ x } \)
Substitute the value of y = \(\frac { 8 }{ x } \) in (1)
x + \(\frac { 8 }{ x } \) = 6
x² + 8 = 6x
x² – 6x + 8 = 0
(x – 4) (x – 2) = 0
∴ x – 4 = 0 or x – 2 = 0
x = 4 or x = 2
y = \(\frac { 8 }{ 4 } \) = 2 or y = \(\frac { 8 }{ 2 } \) = 4

∴ The value of x, y and z are 4, 2, 0 (or) 2, 4, 0

(iii)

Solution:
x + y + z = 9 ……….(1)
x + z = 5 ……….(2)
y + z = 7 ……….(3)

Substitute the value of y = 4 in (3)
y + z = 7
4 + z = 7
z = 7 – 4
= 3
Substitute the value of z = 3 in (2)
x + 3 = 5
x = 5 – 3
= 2
∴ The value of x = 2 , y = 4 and z = 3

Students can download Maths Chapter 4 Geometry Ex 4.1 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Ex 4.1

Question 1.
In the figure, AB is parallel to CD, find x.

Solution:
(i) Through T draw TE || AB.

∴ ∠BAT + ∠ATE = 180° (AB || TE)
140° + ∠ATE = 180°
∠ATE = 180°- 140° = 40°
Similarly ∠ETC + ∠TCD = 180° (TE || CD)
∠ETC+150° = 180°
∠ETC = 180°- 150° = 30°
x = ∠ATE + ∠ETC
= 40°+ 30° = 70°
x = 70°

(ii) Draw TE || AB.

∠ABT + ∠ETB = 180° (AB || TE)
48° + ∠ETB = 180°
∠ETB = 180° – 48° = 132°
Similarly ∠CDT + ∠DTE = 180°
24° + ∠DTE = 180°
∴ ∠DTE = 180° – 24°
= 156°
∴ ∠BTE + ∠ETD = 132° + 156°
= 288°
x = 288°

(iii) In the given figure AB || CD, AD is the transversal.
∠CDA = ∠BAD
= 53° (alternate angles are equal)
In ΔECD, ∠D = ∠A = 53° (Alternate angles are equal)
∠E + ∠C + ∠D = 180° (sum of the angles of a triangle)
x° + 38° + 53° = 180°
x° = 180°- 91°
= 89°
x = 89°

Question 2.
The angles of a triangle are in the ratio 1 : 2 : 3, find the measure of each angle of the triangle.
Solution:
The ratio of the angles of a triangle = 1 : 2 : 3.
Let the angles of a triangle be x, 2x and 3x.
x + 2x + 3x = 180° (Total angle of a triangle is 180°)
6x = 180°
x = \(\frac{180°}{6}\)
= 30°
x = 30°; 2x = 2 × 30° = 60°; 3x = 3 × 30° = 90°
Measures of the angles of a triangle = 30°, 60° and 90°.

Question 3.
Consider the given pairs of triangles and say whether each pair is that of congruent triangles. If the triangles are congruent, say ‘how’; if they are not congruent say ‘why’ and also say if a small modification would make them congruent:

(i) In ΔPQR and ΔABC
PQ = AB (Given)
RQ = BC (Given)
ΔABC is not congruent to ΔPQR.
If PR = AC then ΔABC ≅ ΔPQR

(ii) In ΔABD and ΔCDB
AB = CD (Given)
AD = BC (Given)
BD is common
By SSS congruency
ΔABD ≅ ΔCDB

(iii) In ΔPXY and ΔPXZ
PX is common.
XY = XZ (Given)
PY = PZ (Given)
By SSS congruency
ΔPXY ≅ ΔPXZ

(iv) In the given figure BD bisect AC
In ΔAOB and ΔOCD
OA = OC (Given)
∠AOB = ∠DOC (vertically opposite angles)
∠B = ∠D (Given)
By ASA congruency ΔAOB ≅ ΔOCD

(v) In the given figure AC and BD bisect each other at O.
∴ OA = OC (Given); OB = OD (Given)
∠AOB = ∠COD (vertically opposite angles)
By SAS congruency
ΔAOB ≅ ΔOCD

(vi) In the given figure
AB = AC (Given)
BM = MC (AM is the median of the ΔABC)
AM is common (By SSS congruency)
∴ ΔABM ≅ ΔACM

Question 4.
ΔABC and ΔDEF are two triangles in which AB = DF, ∠ACB = 70°, ∠ABC = 60°; ∠DEF = 70° and ∠EDF = 60°. Prove that the triangles are congruent.
Solution:

In ΔABC ∠B = 60° and ∠C = 70°
∴ ∠A = 180° – (60° + 70°)
= 180° – 130°
= 50°
In ΔDEF ∠E = 70° and ∠D = 60°
∠F = 180° – (70° + 60°)
= 180° – 130°
= 50°
∠A = ∠F = 50°
∠B = ∠D = 60°
∠C = ∠E = 70°
By AAA congruency
ΔABC ≅ ΔFDE
(or)
∠B = ∠D = 60°
∠C = ∠E = 70°
AB = FE
By ASA congruency
ΔABC ≅ ΔFDE

Question 5.
Find all the three angles of the ΔABC.

Solution:
∠A + ∠B = ∠ACD (An exterior angle of a triangle is sum of its interior opposite angles)
x + 35 + 2x – 5 = 4x – 15
3x + 30 = 4x – 15
30 + 15 = 4x – 3x
45° = x
∠A = x + 35°
= 45° + 35°
= 80°
∠B = 2x – 5
= 2(45°) – 5°
= 90° – 5°
= 85°
∠ACD = 4x – 15
= 4 (45°) – 15°
= 180° – 15°
= 165°
∠ACB = 180° – ∠ACD
= 180° – 165°
= 15°
∠A = 80°, ∠B = 85° and ∠C = 15°.

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.3 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.3

Question 1.
Find the mid-points of the line segment joining the points.
(i) (-2, 3) and (-6, -5)
(ii) (8, -2) and (-8, 0)
(iii) (a, b) and (a + 2b, 2a – b)
(iv) (\(\frac{1}{2},\frac{-3}{7}\)) and (\(\frac{3}{2},\frac{-11}{7}\))
Solution:

Question 2.
The centre of a circle is (-4, 2). If one end of the diameter of the circle is (-3, 7) then find the other end.
Solution:
Let the other end of the diameter B be (a, b)

∴ \(\frac{-3+a}{2}\) = -4
-3 + a = -8
a = -8 + 3
a = -5
\(\frac{7+b}{2}\)
7 + b = 4
b = 4 – 7 ⇒ b = -3
The other end of the diameter is (-5, -3).

Question 3.
If the mid-point (x, y) of the line joining (3, 4) and (p, 7) lies on 2x + 2y + 1 = 0, then what will be the value of p?
Solution:

3 + p + 11 + 1 = 0 ⇒ p + 15 = 0 ⇒ p = -15
The value of p is -15.

Question 4.
The mid-point of the sides of a triangle are (2, 4), (-2, 3) and (5, 2). Find the coordinates of the vertices of the triangle.
Solution:
Let the vertices of the ΔABC be A (x₁ y₁), B (x₂, y₂) and C (x₃, y₃)

\(\frac{x_{2}+x_{3}}{2}\) = -2 ⇒ x₂ + x₃ = -4 → (3)
\(\frac{y_{2}+y_{3}}{2}\) = 3 ⇒ y₂ + y₃ = 6 → (4)

\(\frac{x_{1}+x_{3}}{2}\) = 5 ⇒ x₁ + x₃ = 10 → (5)
\(\frac{y_{1}+y_{3}}{2}\) = 2 ⇒ y₁ + y₃ = 4 → (6)
By adding (1) + (3) + (5) we get
2x₁ + 2x₂ + 2x₃ = 4 – 4 + 10
2(x₁ + x₂ + x₃) = 10 ⇒ x₁ + x₂ + x₃ = 5
From (1) x₁ + x₂ = 4 ⇒ 4 + x₃ = 5
x₃ = 5 – 4 = 1
∴ The vertices of the ΔABC are
A (9, 3)
B (-5, 5), C (1, 1)
From (3) x₂ + x₃ = -4 ⇒ x₁ + (-4) = 5
x₁ = 5 + 4 = 9
From (5) ⇒ x₁ + x₃ = 8
x₂ + 10 = 5
x₂ = 5 – 10 = -5
∴ x₁ = 9, x₂ = -5, x₃ = 1
By adding (2) + (4) + (6) we get
2y₃ + 2y₂ + 2y₃ = 8 + 6 + 4
2(y₁ +y₂ + y₃) = 18 ⇒ y₁ + y₂ + y₃ = 9
From (2) ⇒ y₁ + y₂ = 8
8 + y₃ = 9 ⇒ y₃ = 9 – 8 = 1
From (4)
y₂ + y₃ = 6 ⇒ y₁ + 6 = 9
y₁ = 9 – 6 = 3
From (6)
y₁ + y₃ = 4 ⇒ y₂ + 4 = 9
y₂ = 9 – 4 = 5
∴ y₁ = 3, y₂ = 5, y₃ = 1

Question 5.
O (0, 0) is the centre of a circle whose one chord is AB, where the points A and B (8, 6) and (10, 0) respectively. OD is the perpendicular from the centre to the chord AB. Find the coordinates of the mid-point of OD.
Solution:
Note: Since OD is perpendicular to AB, OD bisect the chord
D is the mid-point of AB

Question 6.
The points A(-5, 4) , B(-1, -2) and C(5, 2) are the vertices of an isosceles right-angled triangle where the right angle is at B. Find the coordinates of D so that ABCD is a square.
Solution:
Since ABCD is a square
Mid-point of AC = mid-point of BD
Let the point D be (a, b)

But mid-point of BD = Mid-point of AC
(\(\frac{-1+a}{2}, \frac{-2+b}{2}\)) = (0, 3)
\(\frac{-1+a}{2}\) = 0
-1 + a = 0
a = 1
(\(\frac{-2+b}{2}\))
-2 + b = 6
b = 6 + 2 = 8
∴ The vertices D is (1, 8).

Question 7.
The points A (-3, 6), B (0, 7) and C (1, 9) are the mid points of the sides DE, EF and FD of a triangle DEF. Show that the quadrilateral ABCD is a parallellogram.
Solution:
Let D be (x₁ y₁), E (x₂, y₂) and F (x₃, y₃)


x₁ + x₂ = -6 → (1)
y₁ + y₂ = 12 → (2)

x₁ + x₃ = 2 → (5)
y₁ + y₃ = 18 → (6)
Add (1) + (3) + (5)
2x₁ + 2x₂ + 2x₃ = -6 + 0 + 2
2 (x₁ + x₂ + x₃) = -4
x₁ + x₂ + x₃ = -2
From (1) x₁ + x₂ = -6
x3 = -2 + 6 = 4
From (3) x₂ + x₃ = 0
x₁ = -2 + 0
x₁ = -2
From (5) x₁ + x₃ = 2
x₂ + 2 = -2
x₂ = -2 -2 = -4
∴ x₁ = -2, x₂ = -4, x₃= 4
Add (2) + (4) + 6
2y₁ + 2y₂ + 2y₃ = 12 + 14 + 18
2(y₁ + y₂ + y₃) = 44
y₁ + y₂+ y₃ = 44/2 = 22
From (2) y₁ + y₂ = 12
12 + y₃ = 22 ⇒ y₃ = 22 – 12 = 11
From (4) y₂ + y₃ = 14
y₁ + 14 = 22 ⇒ y₁ = 22 – 14
y₁ = 8
From (6)
y₁ +y₃ = 18
y2 + 18 = 22 ⇒ = 22 – 18
= 4
∴ y₁ = 8, y₂ = 4, y₃ = 11
The vertices D is (-2, 8)

Mid-point of diagonal AC = Mid-point of diagonal BD
The quadrilateral ABCD is a parallelogram

Question 8.
A(-3, 2) , B(3, 2) and C(-3, -2) are the vertices of the right triangle, right angled at A. Show that the mid point of the hypotenuse is equidistant from the vertices.
Solution:

AD = CD = BD = \(\sqrt{13}\)
Mid-point of BC is equidistance from the centre.

Students can download Maths Chapter 3 Algebra Ex 3.7 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.7

Question 1.
Find the square root of the following.
(i) \(\frac{400 x^{4} y^{12} z^{16}}{100 x^{8} y^{4} z^{4}}\)
Answer:

(ii) \(\frac{7 x^{2}+2 \sqrt{14} x+2}{x^{2}-\frac{1}{2} x+\frac{1}{16}}\)
Answer:

(iii) \(\frac{121(a+b)^{8}(x+y)^{8}(b-c)^{8}}{81(b-c)^{4}(a-b)^{12}(b-c)^{4}}\)
Answer:

Question 2.
Find the square root of the following
(i) 4x² + 20x + 25
Answer:

(ii) 9x² – 24xy + 30xz – 40yz + 25z² + 16y²
Answer:

(iii) \(1+\frac{1}{x^{6}}+\frac{2}{x^{3}}\)
Answer:

(iv) (4x² – 9x + 2)(7x² – 13x – 2)(28x² – 3x – 1)
Answer:
4x² – 9x +2 = 4x² – 8x – x + 2
= 4x(x – 2)-1 (x – 2)
= (x – 2)(4x – 1)

7x² – 13x – 2 = 7x² – 14x + x – 2
= 7x (x – 2) + 1 (x – 2)
= (x – 2) (7x + 1)

28x² – 3x – 1 = 28x² – 7x + 4x – 1
= 7x (4x – 1) + 1 (4x – 1)
= (4x – 1) (7x + 1)

(v) (2x² + \(\frac { 17 }{ 6 } \)x + 1) (\(\frac { 3 }{ 2 } \) x² + 4x + 2) (\(\frac { 4 }{ 3 } \) x² + \(\frac { 11 }{ 3 } \) x + 2)
Answer:

Students can download Maths Chapter 3 Algebra Ex 3.18 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.18

Question 1.
Find the order of the product matrix AB if

Answer:
Given A = [a_ij]_p×q and B = [a_ij]_q×r
Order of product of AB = p × r
Order of product of BA is not defined. Number columns in r is not equal to the number of rows in P.
∴ Product BA is not defined.

Question 2.
A has ‘a’ rows and ‘a + 3 ’ columns. B has ‘6’ rows and ‘17 – b’ columns, and if both products AB and BA exist, find a, b?
Solution:
A has a rows, a + 3 columns.
B has b rows, 17 – b columns
If AB exists a × a + 3
b × 17 – b
a + 3 = 6 ⇒ a – 6 = -3 ………… (1)
If BA exists 6 × 17-6
a × a + 3
17 – 6 = a ⇒ a + 6 = 17 …………. (2)
(1) + (2) ⇒ 2a = 14 ⇒ a = 7
Substitute a = 7 in (1) ⇒ 7 – b = -3 ⇒ b = 10
a = 7, b = 10

Question 3.
A has ‘a’ rows and ‘a + 3 ’ columns. B has rows and ‘b’ columns, and if both products AB and BA exist, find a,b?
Answer:

1. Order of matrix AB = 3 × 3
2. Order of matrix AB = 4 × 2
3. Order of matrix AB = 4 × 2
4. Order of matrix AB = 4 × 1
5. Order of matrix AB = 1 × 3

Question 4.

find AB, BA and check if AB = BA?
Answer:

Question 5.
Given that


verify that A(B + C) = AB + AC
Answer:

From (1) and (2) we get
A (B + C) = AB + AC

Question 6.
Show that the matrices

satisfy commutative property AB = BA
Answer:

From (1) and (2) we get
AB = BA. It satisfy the commutative property.

Question 7.

Show that (i) A(BC) = (AB)C
(ii) (A-B)C = AC – BC
(iii) (A-B)^T = A^T – B^T
Answer:



From (1) and (2) we get
A(BC) = (AB)C

From (1) and (2) we get
(A – B) C = AC – BC


From (1) and (2) we get
(A-B)^T = A^T – B^T

Question 8.

then snow that A² + B² = I.
Answer:

Question 9.

prove that AA^T = I.
Answer:

AA^T = I
∴ L.H.S. = R.H.S.

Question 10.
Verify that A² = I when

Answer:

∴ L.H.S. = R.H.S.

Question 11.

show that A² – (a + d)A = (bc – ad)I₂.
Answer:

L.H.S. = R.H.S.
A² – (a + d) A = (bc – ad)I₂

Question 12.

verify that (AB)^T = B^T A^T
Answer:


From (1) and (2) we get, (AB)^T = B^T A^T

Question 13.

show that A² – 5A + 7I₂ = 0
Answer:


L.H.S. = R.H.S.
∴ A² – 5A + 7I₂ = 0

Students can download Maths Chapter 3 Algebra Additional Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Additional Questions

I. Multiple choice questions

Question 1.
Which of the following is a monomial?
(a) 4x²
(b) a + b
(c) a + b + c
(d) a + b + c + d
Solution:
(a) 4x²

Question 2.
Which of the following is trinomial?
(a) -7z
(b) z² – 4y²
(c) x²y – xy² + y
(d) 12a – 9ab + 5b – 3
Solution:
(c) x²y – xy² + y

Question 3.
The sum of 5x²; -7x²; 8x²; 11x² and -9x² is ………
(a) 2x²
(b) 4x²
(c) 6x²
(d) 8x²
Solution:
(d) 8x²

Question 4.
The area of a rectangle with length 2l²m and breadth 3lm² is ………
(a) 6l³m³
(b) l³m³
(c) 2l³m³
(d) 4l³m³
Solution:
(a) 6l³m³

Question 5.
The coefficient of x² and x in 2x³ – 5x² + 6x – 3 are respectively ………
(a) 2, -5
(b) 2, 6
(c) – 5, 6
(d) -5, -3
Solution:
(c) – 5, 6

Question 6.
In the system 6x -2y = 3; kx – y = 2 has a unique solution then ………
(a) k = 3
(b) k ≠ 3
(c) k = 4
(d) k ≠ 4
Solution:
(b) k ≠ 3

Question 7.
A system of two linear equation in two variables is inconsistent. If their graphs ………
(a) coincide
(b) intersect only at a point
(c) do not intersect at any point
(d) cut the x-axis
Solution:
(c) do not intersect at any point

Question 8.
The system of equation x – 4y = 8; 3x – 12y = 24 ……….
(a) has infinitely many solution
(b) has no solution
(c) has a unique solution
(d) may or may not have a solution
Solution:
(a) has infinitely many solution

Question 9.
The solution set of x – ay = 4 and x + y = 0 is (1, -1) the value of a is ………
(a) -1
(b) 1
(c) -3
(d) 3
Solution:
(d) 3

Question 10.
The solution set of x + y = 7; x – y = 3 is ………
(a) (-5, -2)
(b) (-5, 2)
(c) (5, 2)
(d) (2, 5)
Solution:
(c) (5, 2)

II. Answer following Questions

Question 1.
What must be added to x⁴ – 3x² + 2x + 6 to get x⁴ – 2x³ – x + 8?
Solution:
Let A be the required number to be added.
(x⁴ – 3x² + 2x + 6) + A = x⁴ – 2x³ – x + 8
A = x⁴ – 2x³ – x + 8 – (x⁴ – 3x² + 2x + 6)
= x⁴ – 2x³ – x + 8 – x⁴ + 3x² – 2x – 6
= -2x³ + 3x² – 3x + 2
Hence -2x³ + 3x² – 3x + 2 must be added.

Question 2.
What must be subtracted to y⁴ + 2y³ – 3y + 8 to get y⁴ – 2y³ + 6?
Solution:
Let A be the required number to be subtracted.
(y⁴ + 2y³ – 3y² + 8) – A = y⁴ – 2y³ + 6
y⁴ + 2y³ – 3y² + 8 – (y⁴ – 2y³ + 6) = A
y⁴ + 2y³ – 3y² + 8 – y⁴ + 2y³ – 6 = A
4y³ – 3y² + 2 = A
Hence 4y³ – 3y² + 2 must be subtracted.

Question 3.
The area of a rectangle is x⁴ + 9x² + 20 sq.units and its length is x² + 4 units. Find its breadth in term of x.
Solution:
Let the breadth of a rectangle be “b”
Length of the rectangle = x² + 4
Area of the rectangle = x⁴ + 9x² + 20
Length × Breadth = x⁴ + 9x² + 20
(x² + 4) × b = x⁴ + 9x² + 20
b = \(\frac{x^{4}+9x^{2}+20}{x^{2}+4}\)
= x² + 5
breadth of a rectangle = x² + 5

Question 4.
Solve 3x + 4y = 24; 20x – 11y = 47 using cross multiplication method.
Solution:
3x + 4y – 24 = 0 → (1)
20x – 11y – 47 = 0 → (2)

\(\frac{x}{-452}\) = \(\frac{1}{-113}\)
-113 = -452
x = \(\frac{452}{113}\)
= 4
But \(\frac{y}{-339}\) = \(\frac{1}{-113}\)
-113y = -339
y = \(\frac{339}{113}\)
= 3
∴ The solution set is (4, 3)

Question 5.
A fraction such that if the numerator is multiplied by 3 and the denominator is reduced by 3, we get \(\frac{18}{11}\), but if the numerator is increased by 8 and the denominator is doubled, we get \(\frac{2}{5}\). Find the fraction.
Solution:
Let the numerator be x and the denominator be y
∴ The fraction is \(\frac{x}{y}\)
According to the given condition
\(\frac{3x}{y-3}\) = \(\frac{18}{11}\)
33x = 18(y – 3)
33x = 18y – 54
33x – 18y – 54 = 0
11x – 6y – 18 = 0 ……. (1)
According to the second condition
\(\frac{x+8}{2y}\) = \(\frac{2}{5}\)
5x + 40 = 4y
5x – 4y + 40 = 0 ……..(2)

∴ The fraction is = \(\frac{12}{25}\)

Question 6.
One number is greater than the thrice the other number by 2. If 4 times the smaller number exceeds the greater by 5, find the number.
Solution:
Let the greater number be x and the smaller number be “y” By the given first condition
x = 3y + 2
x – 3y = 2 ……(1)
Again by the given second condition
4y = x + 5
-x + 4y = 5 …….(2)
Add (1), (2) ⇒ y = 7
Substitute the value of y = 7 in (1)
x – 3(7) = 2
x = 2 + 21
= 23
The greater number is 23 and the smaller number is 7.

Question 7.
The cost of 11 pencils and 3 erasers is Rs 50 and the cost of 8 pencils and 3 erasers is Rs 38. Find the cost of 5 pencils and 5 erasers.
Solution:
Let the cost of a pencil be Rs x and the cost of an eraser be Rs y. According to the first condition.
11x + 3y = 50 …….(1)
According to the second condition
8x + 3y = 38 ……..(2)
(1) – (2) ⇒ 3x = 12
x = \(\frac{12}{3}\)
= 4
Substitute the value of x = 4 in (1)
11 (4) + 3y = 50
3y = 50 – 44
3y = 6
y = \(\frac{6}{3}\)
= 2
Cost of 5 pencils + 5 erasers = 5(4) + 5(2)
= 20 + 10
= 30
The required cost is Rs 30

Students can download Maths Chapter 3 Algebra Ex 3.15 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.15

Question 1.
Graph the following quadratic equations and state their nature of solutions.
(i) x² – 9x + 20 = 0
(ii) x² – 4x + 4 = 0
(iii) x² + x + 7 = 0
(iv) x² – 9 = 0
(v) x² – 6x + 9 = 0
(vi) (2x – 3) (x + 2) = 0

(i) x² – 9x + 20 = 0
Answer:
Let y = x² – 9x + 20
(i) Prepare the table of values for y = x² – 9x + 20

(ii) Plot the points (-1, 30) (0,20) (1, 12) (2, 6) (3,2), (4, 0), (5, 0), (6,2) (omit the high value)
(iii) Join the points by a free hand smooth curve.
(iv) The roots of the equation are the X-coordinates of the intersecting points of the curve with X-axis (4, 0) and (5, 0)
There are two points of intersection with the X-axis at 4 and 5. The solution set is 4 and 5. The quadratic equation has real and unequal roots.
(v) Since there is two point of intersection with X-axis (different solution)
∴ The equation x² – 9x + 20 = 0 has real and unequal roots.

(ii) x² – 4x + 4 = 0
Answer:
Let y = x² – 4x + 4
(i) Prepare the table of values for y = x² – 4x + 4

(ii) Plot the points (-3,25) (-2,16) (-1, 9) (0,4) (1,-1) (2, 0), (3,1) and (4, 4)
(iii) Join the points by a free hand smooth curve.
(iv) The roots of the equation are the X-coordinates of the intersecting points of the curve with X-axis (2, 0) which is 2.
(v) Since there is only one point of intersection with X-axis (2, 0).
∴ The solution set is 2.
The Quadratic equation x² – Ax + 4 = 0 has real and equal roots.

(iii) x² + x + 7 = 0
Answer:
Let y = x² + x + 7
(i) Prepare the table of values for y = x² + x + 7

(ii) Plot the points (-4,19) (-3,13) (-2, 9) (-1, 7) (0, 7) (1, 9), (2,13) (3,19) and (4,27)
(iii) Join the points by a free hand smooth curve.
(iv) The solution of the given quadratic equation are the X-coordinates of the intersecting points of the parabola with the X-axis.
(v) The curve does not intersecting the X-axis. There is no solution set.
The equation x² + x + 7 = 0 has no real roots.

(iv) x² – 9 = 0
Answer:
Let y = x² – 9
(i) Prepare the table of values for y = x² – 9

(ii) Plot the points (-4, 7) (-3, 0) (-2, -5) (-1, -8) (0, -9) (1, -8), (2, -5) (3, 0) (4, 7)
(iii) Join the points by a free hand smooth curve.
(iv) The curve intersect the X-axis at -3 and 3.
The solution is (-3, 3).
(v) Since there are two points of intersection -3 and 3 with the X-axis the quadratic equation has real and unequal roots.

(v) x² – 6x + 9 = 0
Answer:
Let y = x² – 6x + 9
(i) Prepare a table of values for y = x² – 6x + 9

(ii) Plot the points (-2,25) (-1,16) (0,9) (1,4) (2,1) (3,0), (4,1) and (5,4) on the graph using suitable scale (omit the points (-4, 49) and (-3, 36)
(iii) Join the points by a free hand smooth curve.
(iv) The X – coordinates of the point of intersection of the curve with X-axis are the roots of the , given equation, provided they intersect.
The solution is 3.
(v) Since there is only one point of intersection with X-axis the quadratic equation x² – 6x + 9 = 0 has real and equal roots.

(vi) (2x – 3) (x + 2) = 0
Answer:
y = (2x – 3) (x + 2)
= 2x² + 4x – 3x – 6
= 2x² + x – 6

(i) Prepare a table of values for y from x – 4 to 4

(ii) Plot the points (-4, 22) (-3, 9) (-2, 0) (-1, -5) (0, -6) (1, -3), (2, 4), (3, 15) and (4, 30).
(iii) Join the points by a free hand smooth curve.
(iv) The curve intersect the X – axis at (-2, 0) and (1\(\frac { 1 }{ 2 } \), 0)
∴ The solution set is (-2,1\(\frac { 1 }{ 2 } \))
(v) Since there are two points of intersection with X – axis, the quadratic equation has real and un – equal roots.

Question 2.
Draw the graph of y = x² – 4 and hence solve x² – x – 12 = 0
Answer:
(i) Draw the graph of y = x² – 4 by preparing the table of values as below.

(ii) Plot the points for the ordered pairs (-4, 12) (-3, 5) (-2, 0) (-1, -3) (0, -4) (1, -3), (2, 0), (3, 5) and (4, 12). Draw the curve with the suitable scale.
(iii) To solve x² – x – 12 = 0 subtract x² – x – 12 from y = x² – 4

The equation y = x + 8 represents a straight line. Prepare a table for y = x + 8

(iv) Mark the point of intersection of the curve and the straight line is (-3, 5) and 4, 12)
∴ The solution set is (-3, 4) for x² – x – 12 = 0.

Question 3.
Draw the graph of y = x² + x and hence solve x² + 1 = 0
Answer:
Let y = x² + x
(i) Draw the graph of y = x² + x by preparing the table.

(ii) Plot the points (-4, 12), (-3, 6), (-2, 2), (-1, 0), (0, 0), (1, 2), (2, 6), (3, 12) and (4, 20).
(iii) Join the points by a free hand to get smooth curve.
(iv) To solve x² + 1 = 0, subtract x² + 1 = 0 from x² + x we get.

The equation represent a straight line. Draw a line y = x – 1

Observe the graph of y = x² + 1 does not interset the parabola y = x² + x.
This x² + 1 has no real roots.

Question 4.
Draw the graph of y = x² + 3x + 2 and use it to solve x² + 2x + 1 = 0.
(i) Draw the graph of y = x² + 3x + 2 preparing the table of values as below.

(ii) Plot the points (-4, 6), (-3, 2), (-2, 0), (-1, 0), (0, 2), (1, 6), (2, 12), (3, 20) (4, 30).
(iii) To solve x² + 2x + 1 = 0 subtract x² + 2x + 1 = 0 from y = x² + 3x + 2

(iv) Draw the graph of y = x + 1 from the table

The equation y = x + 1 represent a straight line.
This line intersect the curve at only one point (-1, 0). The solution set is (-1).

Question 5.
Draw the graph of y = x² + 3x – 4 and hence use it to solve x² + 3x – 4 = 0
Answer:
Let y = x² + 3x – 4
(i) Draw the graph of y = x² + 3x – 4

(ii) Plot the points (-5, 6), (-4, 0), (-3, -4), (-2, -6), (-1, -6), (0, -4), (1, 0), (2, 6), (3, 14) on the graph using suitable scale.
(iii) Join the points by a free hand smooth curve.
The smooth curve is the graph of y = x² – 4x + 4
(iv) To solve x + 3x – 4 = 0, subtract x² + 3x – 4 = 0 from y = x² + 3x – 4.
y = 0
∴ The point of intersection with the x – axis is the solution set.
The solution set is -4 and 1.

Question 6.
Draw the graph of y = x² – 5x – 6 and hence solve x², – 5x – 14 = 0
Answer:
Let y = x² – 5x – 6
(i) Draw the graph of y = x² – 5x – 6 by preparing the table of values as below.

(ii) Plot the points (-3, 18), (-2, 8), (-1, 0), (0, -6), (1, -10), (2, -12), (3, -12), (4, -10), (5,-6), (6, 0) and (7, 8).
(iii) Join the points by a free hand to get smooth curve.
(iv) To solve x² – 5x – 14 = 0, subtract x² – 5x – 14 = 0 from y = x² – 5x – 6.

The equation y = 8 represent a straight line draw a straight line through y = 8 intersect the curve at two places. From the two points draw perpendicular line to the X – axis it will intersect at -2 and 7.
The solution is -2 and 7

Question 7.
Draw the graph of y = 2x² – 3x – 5 and hence solve 2x² – 4x – 6 = 0
Answer:
(i) Draw the graph of y = 2x² – 3x – 5 by preparing the table of values given below.

(ii) Plot the points (-3, 22), (-2, 9), (-1, 0), (0, -5), (1,-6), (2, -3), (3, 4), (4, 15) on the graph sheet using suitable scale.
(iii) To solve 2x² – 4x – 6 = 0 subtract 2x² – 4x – 6 = 0 from y = 2x² – 3x – 5

(iv) y = x + 1 represent a straight line.

The straight line intersect the curve at (-1, 0) and (3, 4). From the two point draw perpendicular lines to the X – axis it will intersect at -1 and 3.
The solution set is (-1, 3)

Question 8.
Draw the graph of y = (x – 1) (x + 3) and hence solve x² – x – 6 = 0
Answer:
y = (x – 1) (x + 3)
y = x² + 2x – 3
(i) Draw the graph of y = x² + 2x – 3 by preparing the table of values given below

(ii) Plot the points (-4, 5), (-3, 0), (-2, -3), (-1, -4), (0, -3), (1, 0), (2, 5), (3, 12) and (4, 21) on the graph sheet using suitable scale.
(iii) To solve x² – x – 6 = 0 subtract x² – x – 6 = 0 from y = x2 + 2x – 3
(iv) Draw the graph of y = 3x + 3 by preparing the table.

(v) The straight line cuts the curve at (-2, -3) and (3, 12). Draw perpendicular lines from the point to X – axis.
The line cut the X – axis at -2 and 3.
The solution set is (-2, 3)

Students can download Maths Chapter 3 Algebra Ex 3.15 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.15

Multiple Choice Questions.

Question 1.
If x³ + 6x² + kx + 6 is exactly divisible by x + 2, then k = 2
(a) 6
(b) -7
(c) -8
(d) 11
Solution:
(d) 11
Hint:
p(x) = x³ + 6x² + kx + 6
Given p(-2) = 0
(-2)³ + 6(-2)² + k(-2) + 6 = 0
-8 + 24 – 2k + 6 = 0
22 – 2k = 0
k = \(\frac{22}{2}\)
= 11

Question 2.
The root of the polynomial equation 2x + 3 = 0 is…….
(a) \(\frac{1}{3}\)
(b) –\(\frac{1}{3}\)
(c) –\(\frac{3}{2}\)
(d) –\(\frac{2}{3}\)
Solution:
(c) –\(\frac{3}{2}\)
Hint:
2x + 3 = 0
2x = – 3 ⇒ x = –\(\frac{3}{2}\)

Question 3.
The type of the polynomial 4 – 3x³ is ……..
(a) constant polynomial
(b) linear polynomial
(c) quadratic polynomial
(d) cubic polynomial
Solution:
(d) cubic polynomial

Question 4.
If x⁵¹ + 51 is divided by x + 1, then the remainder is …….
(a) 0
(b) 1
(c) 49
(d) 50
Solution:
(d) 50
Hint:
p(x) = x⁵¹ + 51
p(-1)= (-1)⁵¹ + 51
= -1 + 51
= 50

Question 5.
The zero of the polynomial 2x + 5 is ……..
(a) \(\frac{5}{2}\)
(b) –\(\frac{5}{2}\)
(c) \(\frac{2}{5}\)
(d) –\(\frac{2}{5}\)
Solution:
(b) –\(\frac{5}{2}\)
Hint:
2x + 5 = 0 ⇒ 2x = -5 ⇒ x = –\(\frac{5}{2}\)

Question 6.
The sum of the polynomials p(x) = x³ – x² – 2, q(x) = x² – 3x + 1
(a) x³ – 3x – 1
(b) x³ + 2x² – 1
(c) x³ – 2x² – 3x
(d) x³ – 2x² + 3x – 1
Solution:
(a) x³ – 3x – 1
Hint:
p(x) + q(x) = (x³ – x² – 2) + (x² – 3x + 1) = x³ – x² – 2 + x² – 3x + 1
= x³ – 3x – 1

Question 7.
Degree of the polynomial (y³ – 2) (y³ + 1) is
(a) 9
(b) 2
(c) 3
(d) 6
Solution:
(d) 6
(y³ – 2) (y³ + 1) = y⁶ + y³ – 2y³ – 2
= y⁶ – y³ – 2

Question 8.
Let the polynomials be
(A) -13q⁵ + 4q² + 12q
(B) (x² + 4)(x² + 9)
(C) 4q⁸ – q⁶ + q²
(D) –\(\frac{5}{7}\) y¹² + y³ + y⁵.
Then ascending order of their degree is
(a) A, B, D, C
(b) A, B, C, D
(c) B, C, D, A
(d) B, A, C, D
Solution:
(d) B, A, C, D

Question 9.
If p(a) = 0 then (x – a) is a …….. of p(x)
(a) divisor
(b) quotient
(c) remainder
(d) factor
Solution:
(d) factor

Question 10.
Zeros of (2 – 3x) is ……..
(a) 3
(b) 2
(c) \(\frac{2}{3}\)
(d) \(\frac{3}{2}\)
Solution:
(c) \(\frac{2}{3}\)

Question 11.
Which of the following has x -1 as a factor?
(a) 2x – 1
(b) 3x – 3
(c) 4x – 3
(d) 3x – 4
Solution:
(b) 3x – 3

Question 12.
If x – 3 is a factor of p(x), then the remainder is ……..
(a) 3
(b) -3
(c) p(3)
(d) p(-3)
Solution:
(c) p(3)

Question 13.
(x +y)(x² – xy + y²) is equal to ……..
(a) (x + y)³
(b) (x – y)³
(c) x³ + y³
(d) x³ – y³
Solution:
(c) x³ + y³

Question 14.
(a + b – c)² is equal to ……..
(a) (a – b + c)²
(b) (-a – b + c)²
(c) (a + b + c)²
(d) (a – b – c)²
Solution:
(b) (-a – b + c)²
Hint:
(a + b – c)² = a² + b² + c² + 2ab – 2bc – 2ac
(- a – b + c)² = a² + b² + c² + 2ab – 2bc – 2ac
(OR)
(- a – b + c)² = (-1)² (a + b + c)² (taking – 1 as common)
= (a + b – c)²

Question 15.
In an expression ax² + bx + c the sum and product of the factors respectively ……..
(a) a, bc
(b) b, ac
(c) ac, b
(d) bc, a
Solution:
(b) b, ac

Question 16.
If (x + 5) and (x – 3) are the factors of ax² + bx + c, then values of a, b and c are ………
(a) 1, 2, 3
(b) 1, 2, 15
(c) 1, 2, -15
(d) 1, -2, 15
Solution:
(c) 1, 2, -15
Hint:
(x + 5) (x – 3) = x² + (5 – 3) x + (5) (-3)
= x² + 2x – 15
compare with ax² + bx + c
a = 1, b = 2 and c = -15

Question 17.
Cubic polynomial may have maximum of ……… linear factors.
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3

Question 18.
Degree of the constant polynomial is ……..
(a) 3
(b) 2
(c) 1
(d) 0
Solution:
(d) 0

Question 19.
Find the value of m from the equation 2x + 3y = m. If its one solution is x = 2 and y = -2.
(a) 2
(b) -2
(c) 10
(d) 0
Solution:
(b) – 2
Hint:
The equation is 2x + 3y = m
Substitute x – 2 and y = -2 we get
2(2) + 3(-2) = m ⇒ 4 – 6 = m ⇒ -2 = m

Question 20.
Which of the following is a linear equation?
(a) x + \(\frac{1}{2}\) = 2
(b) x (x – 1) = 2
(c) 3x + 5 = \(\frac{2}{3}\)
(d) x³ – x = 5
Solution:
(c) 3x + 5 = \(\frac{2}{3}\)

Question 21.
Which of the following is a solution of the equation 2x – y = 6?
(a) (2, 4)
(b) (4, 2)
(c) (3, -1)
(d) (0, 6)
Solution:
(b) (4, 2)
Hint:
2x – y = 6
Substitute x – 4 and y = 2 we get
2(4) – 2 = 6 ⇒ 8 – 2 = 6 ⇒ 6 = 6
∴ (4, 2) is the solution

Question 22.
If (2, 3) is a solution of linear equation 2x + 3y = k then, the value of k is ……..
(a) 12
(b) 6
(c) 0
(d) 13
Solution:
(d) 13
Hint:
The equation is 2x + 3y = k
Substitute x = 2 and y = 3 we get,
2(2) + 3(3) = k ⇒ 4 + 9 = k ⇒ 13 = k

Question 23.
Which condition does not satisfy the linear equation ax + by + c = 0 ……..
(a) a ≠ 0, b = 0
(b) a = 0, b ≠ 0
(c) a = 0, b = 0, c ≠ 0
(d) a ≠ 0, b ≠ 0
Solution:
(c) a = 0, b = 0, c ≠ 0

Question 24.
Which of the following is not a linear equation in two variable?
(a) ax + by + c = 0
(b) 0x + 0y + c = 0
(c) 0x + by + c = 0
(d) ax + 0y + c = 0
Solution:
(b) 0x + 0y + c = 0
Hint:
0x + 0y + c = 0
0 + 0 + c = 0 ⇒ c = 0
There is no variable.
∴ It is not a linear equation

Question 25.
The value of k for which the pair of linear equations 4x + 6y – 1 = 0 and 2x + ky – 1 = 0 represents parallel lines is ……..
(a) k = 3
(b) k = 2
(c) k = 4
(d) k = -3
Solution:
(a) k = 3
Hint:
Slope of 4x + 6y – 1 = 0 is
6y = -4x + 1 ⇒ y = \(\frac{-4}{6}\) x + \(\frac{1}{6}\)
Slope = \(\frac{-4}{6}\) = \(\frac{-2}{3}\)
Slope of 2x + ky – 7 = 0
ky = -2x + 7
y = \(\frac{-2}{k}\)x + \(\frac{7}{k}\)
Slope of a line = \(\frac{-2}{k}\)
Since the lines are parallel
\(\frac{-2}{3}\) = \(\frac{-2}{k}\)
-2k = – 6
k = \(\frac{6}{2}\)
= 3

Question 26.
A pair of linear equations has no solution then the graphical representation is ……..

Solution:

Hint:
Since there is no solution the two lines are parallel. (l11m)

Question 27.
If \(\frac{a_1}{a_2}\) ≠ \(\frac{b_1}{b_2}\) where a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 then the given pair of linear equation has …….. solution(s).
(a) no solution
(b) two solutions
(c) unique
(d) infinite
Solution:
(c) unique
Hint:
Since it has unique solution
\(\frac{a_1}{a_2}\) ≠ \(\frac{b_1}{b_2}\)

Question 28.
\(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\) where a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 then the given pair of linear equation has …….. solution(s).
(a) no solution
(b) two solutions
(c) infinite
(d) unique
Solution:
(a) no solution
Hint:
\(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\) the linear equation has no solution.

Question 29.
GCD of any two prime numbers is …….
(a) -1
(b) 0
(c) 1
(d) 2
Solution:
(c) 1

Question 30.
The GCD of x⁴ – y⁴ and x² – y² is ……..
(a) x⁴ – y⁴
(b) x² – y²
(c) (x + y)²
(d) (x + y)⁴
Solution:
(b) x² – y²
Hint:
x⁴ – y⁴ = (x²)² – (y²)²
= (x² + y²)(x² – y²)
x² – y² = (x² – y²)
G.C.D. = x² – y²

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.4 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.4

Question 1.
Find the coordinates of the point which divides the line segment joining the points A (4,-3) and B (9,7) in the ratio 3 : 2.
Solution:
A line divides internally in the ratio m : n

Question 2.
In what ratio does the point P(2, -5) divide the line segment joining A(-3, 5) and B(4, -9).
Solution:
A line divides internally in the ratio m : n

\(\frac{4m-3n}{m+n}\)
= 2
4m – 3n = 2m + 2n
4m – 2m = 3n + 2n
2m = 5n
\(\frac{m}{n}\) = \(\frac{5}{2}\)
m : n = 5 : 2
The ratio is 5 : 2.
and
\(\frac{-9m+5n}{m+n}\)
= -5
-9m + 5 n = -5(m + n)
-9m + 5 n = -5m – 5n
-9m + 5 m = -5n – 5n
-4m = -10
\(\frac{m}{n}\) = \(\frac{10}{4}\) ⇒ \(\frac{m}{n}\) = \(\frac{5}{2}\)
m : n = 5 : 2

Question 3.
Find the coordinates of a point P on the line segment joining A(1, 2) and B(6, 7) in such a way that AP = \(\frac{2}{5}\) AB.
Solution:
Let the point A (1, 2) and B (6, 7)
AP = \(\frac{2}{5}\) AB
\(\frac{AP}{PB}\) = \(\frac{2}{5}\)
∴ AP = 2; PB = 5 – 2 = 3
A line divides internally in the ratio m : n

Question 4.
Find the coordinates of the points of trisection of the line segment joining the points A (-5, 6) and B (4, -3).
Solution:
Let P and Q be the point of trisection
so that AP = PB = QB

(\(\frac{3}{3}\), \(\frac{0}{3}\)) = (1, 0)
The Point P is (-2, 3), The Point Q is (1, 0)

Question 5.
The line segment joining A(6, 3) and B(-1, -4) is doubled in length by adding half of AB to each end. Find the coordinates of the new end points.
Solution:

m : n = 3 : 1
A line divides externally in the ratio m : n

∴ BA’ divides in the ratio 2 : 1
A line divides internally in the ratio m : n the point is \(\frac{mx_{2}+nx_{1}}{m+n}, \frac{my_{2}+ny_{1}}{m+n}\)
Let the point A’ be (a, b)

\(\frac{2a-1}{3}\) = 6
2a – 1 = 18
2a = 19
a = \(\frac{19}{2}\)
and
\(\frac{2b-4}{3}\) = 3
2b – 4 = 9
2b = 13
b = \(\frac{13}{2}\)

The point A’ is (\(\frac{19}{2}\), \(\frac{13}{2}\))
To find B’
Let B’ be (a, b)
AB = 7√2
BB’ = \(\frac{1}{2}\) × 7√2 = \(\frac{7√2}{2}\)
\(\frac{AB}{BB’}\) = 7√2 ÷ \(\frac{7√2}{2}\) = \(\frac{7√2×2}{7√2}\) = 2
AB’ divides in the ratio 2 : 1
(-1, -4) = \(\frac{2a+6}{3}\), \(\frac{2b+3}{3}\)
\(\frac{2a+6}{3}\) = -1
2a + 6 = -3
2a = -3 – 6
2a = -9
a = –\(\frac{9}{2}\)
and
\(\frac{2b+3}{3}\) = -4
2b + 3 = -12
2b = -12 – 3
2b = -15
b = –\(\frac{15}{2}\) = -1
The point B’ is (-\(\frac{9}{2}\), –\(\frac{15}{2}\))

Question 6.
Using section formula, show that the points A (7, -5), B (9, -3) and C (13, 1) are collinear.
Solution:
If three points are collinear, then one of the points divide the line segment joining the other points in the ratio r : 1. If P is between A and B and \(\frac{AP}{PB}\) = r

The line divides in the ratio 1 : 2
A line divides internally in the ratio m : n
The point P = (\(\frac{mx_{2}+nx_{1}}{m+n}\), \(\frac{my_{2}+ny_{1}}{m+n}\))
m = 1, n = 2, x₁ = 7, x₂ = 13, y₁ = – 5, y₂ = 1
By the given equation,
The Point B = (\(\frac{13+14}{3}\), \(\frac{1-10}{3}\))
= (\(\frac{27}{3}\), \(\frac{-9}{3}\))
= (9, -3)
∴ The three points A, B and C are collinear.

Question 7.
A line segment AB is increased along its length by 25% by producing it to C on the side of B. If A and B have the coordinates (-2, -3) and (2, 1) respectively, then find the coordinates of C.
Solution:
Let the point C be (a, b)

The ratio is 4 : 1 (m : n)
A line divides internally in the ratio m : n

\(\frac{4a-2}{5}\) = 2
4a – 2 = 10
4a = 12
a = \(\frac{12}{4}\) = 3
and
\(\frac{4b-3}{5}\) = 1
4b – 3 = 5
4b = 8
b = \(\frac{8}{4}\) = 2
The co-ordinate of C is (3, 2)