Category: Class 9

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 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 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)

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.2 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 5.2

Question 1.
Find the distance between the following pairs of points.
(i) (1, 2) and (4, 3)
Solution:
Distance between the points (1, 2) and (4, 3)

(ii) (3, 4) and (-7, 2)
Solution:
Distance between the points (3,4) and (-7, 2)

(iii) (a, b) and (c, b)
Solution:
Distance between the two points (a, b) and (c, b)

= c – a units

(iv) (3,- 9) and (-2, 3)
Solution:
Distance between the two points (3, -9) and (-2, 3)

= 13 units

Question 2.
Determine whether the given set of points in each case are collinear or not.
(i) (7, -2), (5, 1), (3, 4)
Solution:
To prove that three points are collinear, sum of the distance between two pairs of points is equal to the third pair of points.

AB + BC = AC
\(\sqrt{13}\) + \(\sqrt{13}\) = 2\(\sqrt{13}\) ⇒ 2\(\sqrt{13}\) = 2\(\sqrt{13}\)
∴ The given three points are collinear.

(ii) (a, -2), (a, 3), (a, 0)
Solution:
A (a, -2) B (a, 3) C (a, 0)

√4
= 2
AC + BC = AB ⇒ 2 + 3 = 5
∴ The given three points are collinear.

Question 3.
Show that the following points taken in order to form an isosceles triangle.
(i) A (5, 4), B(2, 0), C (-2, 3)
Solution:

= 5√2
AB = BC = 5. (Two sides are equal)
∴ ABC is an isosceles triangle.

(ii) A (6, -1), B (-2, -4), C (2, 10)
Solution:

BC = AC = \(\sqrt{212}\) (TWO sides are equal)
∴ ABC is an isosceles triangle.

Question 4.
Show that the following points taken in order to form an equilateral triangle in each case.
(i) A(2, 2), B(-2, -2), C(-2√3, 2√3)
Solution:

AB = BC = AC (Three sides are equal)
∴ ABC is an equilateral triangle.

(ii) A(√3, 2), B (0, 1), C(0, 3)
Solution:

= √4
= 2
AB = BC = AC (Three sides are equal)
∴ ABC is an equilateral triangle.

Question 5.
Show that the following points taken in order to form the vertices of a parallelogram.
(i) A(-3, 1), B(-6, -7), C (3, -9) and D(6, -1)
Solution:

AB = CD = \(\sqrt{73}\) and BC = AD = \(\sqrt{85}\) (Opposite sides are equal)
∴ ABCD is a parallelogram.

(ii) A (-7, -3), B(5, 10), C(15, 8) and D(3, -5)
Solution:

AB = CD = \(\sqrt{313}\) and BC = AD = \(\sqrt{104}\) (Opposite sides are equal)
∴ ABCD is a parallelogram.

Question 6.
Verify that the following points taken in order to form the vertices of a rhombus.
(i) A(3, -2), B (7, 6),C (-1, 2) and D (-5, -6)
Solution:

AB = BC = CD = AD = \(\sqrt{80}\). All the four sides are equal.
∴ ABCD is a rhombus.

(ii) A (1, 1), B (2, 1),C (2, 2) and D (1, 2)
Solution:

AB = BC = CD = AD = 1. All the four sides are equal.
∴ ABCD is a rhombus.

Question 7.
A (-1, 1), B (1, 3) and C (3, a) are points and if AB = BC, then find ‘a’.
Solution:

4 + (a – 3)² = 8
(a – 3)² = 8 – 4
(a – 3)² = 4
a – 3 = √4
= ± 2
a – 3 = 2 (or) a – 3 = -2
a = 2 + 3 (or) a = 3 – 2
a = 5 (or) a = 1
∴ The value of a = 5 or a = 1.

Question 8.
The abscissa of a point A is equal to its ordinate, and its distance from the point B(1, 3) is 10 units, What are the coordinates of A?
Solution:
Let the point A be (a, a) B is (1, 3)
Distance AB = 10 (Given)
By distance formula \(\sqrt{(a – 1 )² + (a – 3)²}\) = 10
Simplifying 2a² – 8a + 10 = 100
a² – 4a – 45 = 0
(a – 9)(a + 5) = 0
⇒ a = – 5; A = (-5, -5)
a = 9; A = (9, 9)

Question 9.
The point (x, y) is equidistant from the points (3, 4) and (-5, 6). Find a relation between x and y.
Solution:
Let the point O be (x, y), A be (3, 4) and B be (-5, 6).

Distance = \(\sqrt{(x_{2} – x_{1})² + (y_{2} – y_{1})²}\)
Given ,OA = OB
\(\sqrt{(x – 3 )² + (y – 4)²}\) = \(\sqrt{(x + 5 )² + (y – 6)²}\)
Squaring on both sides
(x – 3)² + (y – 4)² = (x + 5)² + (y – 6)²
x² – 6x + 9 + y² – 8y + 16 = x² + 10x + 25 + y² – 12y + 36
x² + y² – 6x – 8y + 25 = x² + y² + 10x – 12y + 61
6x – 10x – 8y + 12y = 61 – 25 ⇒ -16x + 4y = 36
÷ 4 ⇒ -4x + y = 9
∴ The relation between x and y is y = 4x + 9

Question 10.
Let A(2,3) and B(2, -4) be two points. If P lies on the x-axis, such that AP = \(\frac{3}{7}\) AB, find the coordinates of P.
Solution:
Given points are A(2, 3) and B(2, -4)
The point P lies on the x-axis.
∴ The point P is (x, 0)

16x² – 64x + 208 = 9x² – 36x + 180
16x² – 9x² – 64x + 36x + 208 – 180 = 0
7x² – 28x + 28 = 0
x² – 4x + 4 = 0
(x – 2)² = 0
x – 2 = 0
x = 2
∴ The point P is (2, 0)

Question 11.
Show that the point (11, 2) is the centre of the circle passing through the points (1, 2), (3, -4) and (5, -6)
Solution:

\(\sqrt{100}\)
= 10
OA = OB = OC = 10
O is the centre of the circle passing through A, B and C.

Question 12.
The radius of a circle with centre at origin is 30 units. Write the coordinates of the points where the circle intersects the axes. Find the distance between any two such points.
Solution:

Radius of the circle = 30 units. The point O is (0, 0).
Let a intersect the x-axis and b intersect the y-axis.
∴ The point A is (a, 0) and B is (0, b)

Squaring on both sides
30² = a²
∴ a = 30
The point A is (30, 0)
OB = \(\sqrt{(0 – 0)² + (b – 0)²}\)
= \(\sqrt{0² + b²}\)
30 = \(\sqrt{b²}\)
Squaring on both sides
30² = b²
∴ b = 30
The point B is (0, 30)

= 30√2
∴ Distance between the two points = 30√2

Students can download Maths Chapter 3 Algebra Ex 3.14 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 3.14

Question 1.
The sum of a two digit number and the number formed by interchanging the digits is 110. If 10 is subtracted from the first number, the new number is 4 more than 5 times the sums of the digits of the first number. Find the first number.
Solution:
Let the ten’s digit be x and the unit digit be y.
The number is 10x + y
If the digits are interchanged
The new number is 10y + x
By the given first condition
10x + y + 10y + x = 110
11x + 11y = 110
x + y = 10 → (1) (Divided by 11)
Again by the given second condition
10x + y – 10 = 5(x + y ) + 4
10x + y – 10 = 5x + 5y + 4
5x – 4y = 14 → (2)
(1) × 5 ⇒ 5x + 5y = 50 → (3)
(2) × 1 ⇒ 5x – 4y = 14 → (2)
(3) – (2) ⇒ 9y = 36
y = 36/9
= 4
Substitute the value of y = 4 in (1)
x + y = 10
x + 4 = 10
x = 10 – 4
= 6
∴ The number is (10 × 6 + 4) = 64

Question 2.
The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes \(\frac{1}{2}\). Find the fraction.
Solution:
Let the numerator be “x” and the denominator be “y”
∴ The fraction is \(\frac{x}{y}\)
By the given first condition
x + y = 12 → (1)
Again by the second condition
\(\frac{x}{y+3}\) = \(\frac{1}{2}\)
2x = y + 3
2x – y = 3 → (2)
(1) + (2) ⇒ 3x = 15
x = \(\frac{15}{3}\) = 5
Substitute the value of x = 5 in (1)
5 + y = 12
y = 12 – 5
= 7
∴ The fraction is \(\frac{5}{7}\)

Question 3.
ABCD is a cyclic quadrilateral such that ∠A = (4y + 20)°, ∠B = (3y -5)°, ∠C = (4x)° and ∠D = (7x + 5)°. Find the four angles.
Solution:

ABCD is a cyclic quadrilateral ∠A + ∠C = 180°
(Sum of the opposite angles of a cyclic quadrilateral is 180°)
(4y + 20)° + (4x)° = 180°
4y + 20 + 4x = 180
4x + 4y = 180 – 20
4x + 4y = 160
x + y = 40 → (1) (divided by 4)
∠B + ∠D = 180° (Sum of the opposite angles of a cyclic quadrilateral)
(3y – 5)° + (7x + 5)° = 180°
3y – 5 + 7x + 5 = 180
7x + 3y = 180 → (2)
(1) × 3 ⇒ 3x + 3y = 120 → (3)
(3) – (2) ⇒ -4x = – 60
4x = 60
x = \(\frac{60}{4}\)
Substitute the value of x = 15 in (1)
15 + y = 40
y = 40 – 15 = 25
∠A = 4y + 20 = 4(25) + 20 = 100 + 20 = 120°
∴ ∠A = 120°
∠B = 3y – 5 = 3(25) – 5 = 75 – 5 = 70
∴ ∠B = 70°
∠C = 4x = 4(15) = 60
∴ ∠C = 60°
∠D = 7x + 5 = 7(15) + 5
∠D = 105 + 5 = 110°
∴ ∠A= 120°, ∠B = 70°, ∠C = 60° and ∠D = 110°

Question 4.
On selling a T.V. at 5% gain and a fridge at 10% gain, a shopkeeper gains Rs 2000. But if he sells the T.V. at 10% gain and the fridge at 5% loss, he gains Rs.1500 on the transaction. Find the actual price of the T.V. and the fridge.
Solution:
Let the cost price of the TV be Rs “x” and the cost price of the fridge be Rs “y”.
By the given condition

Multiply by 20
x + 2y = 40000 → (1)
Again by the given second condition

Multiply by 20
2x – y = 30000 → (2)
(2) × 2 ⇒ 4x – 2y = 60000 → (3)
(1) + (3) ⇒ 5x + 0 = 100000
x = \(\frac{100000}{5}\)
= 20000
Substitute the value of x = 20000 in (1)
20000 + 2y = 40000
2y = 40000 – 20000
= 20000
y = \(\frac{20000}{2}\)
= 10000
Cost price of a TV = Rs 20,000
Cost price of a fridge = Rs 10,000

Question 5.
Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5. Find the numbers.
Solution:
Let the two numbers be x and y.
By the given first condition
x : y = 5 : 6
6x = 5y (Product of the extreme is equal to the product of the means)
6x – 5y = 0 → (1)
Again by the given second condition
x – 8 : y – 8 = 4 : 5
5(x – 8) = 4(y – 8)
5x – 40 = 4y – 32
5x – 4y = – 32 + 40
5x – 4y = 8 → (2)
(1) × 4 ⇒ 24x – 20y = 0 → (3)
(2) × 5 ⇒ 25x – 20y = 40 → (4)
(3) – (4) ⇒ – x + 0 = -40
∴ x = 40
Substitute the value of x = 40 in (1)
6(40) – 5y = 0
240 – 5y = 0 ⇒ – 5y = -240
5y = 240
y = \(\frac{240}{5}\)
= 48
The two numbers are 40 and 48 [∴ The ratio of the number = 40 : 48 are 5 : 6]

Question 6.
4 Indians and 4 Chinese can do a piece of work in 3 days. While 2 Indian and 5 Chinese can finish it in 4 days. How long would it take for 1 Indian to do it? How long would it ‘ take for 1 Chinese to do it?
Solution:
Let the time taken by a Indian be “x”
Time taken by a Chinese be “y”
Work done by a Indian in one day = \(\frac{1}{x}\)
Work done by a Chinese in one day = \(\frac{1}{y}\)
By the given first condition
(4 Indian + 4 Chinese) finish the work in 3 days
\(\frac{4}{x}\) + \(\frac{4}{y}\) = \(\frac{1}{3}\) → (1)
Again by the given second condition
(2 Indian + 5 Chinese) finish the work in 4 days
\(\frac{2}{x}\) + \(\frac{5}{y}\) = \(\frac{1}{4}\) → (2)
Solve the equation (1) and (2)
Let \(\frac{1}{x}\) = a; \(\frac{1}{y}\) = b
4a + 4b = \(\frac{1}{3}\)
12a + 12b = 1 → (3) (Multiply by 3)
2a + 5b = \(\frac{1}{4}\)
8a + 20b = 1 → (4) (Multiply by 4)
(3) × (2) ⇒ 24a + 24b = 2 → (5)
(4) × (3) ⇒ 24a + 60b = 3 → (6)
(5) – (6) ⇒ -36b = -1
b = \(\frac{1}{36}\)
Substitute the value of b = \(\frac{1}{36}\) in (3)
12a + 12(\(\frac{1}{36}\)) = 1
12a + \(\frac{1}{3}\) = 1
36a + 1 = 3
36a = 2
a = \(\frac{2}{36}\) = \(\frac{1}{18}\)
But \(\frac{1}{x}\) = a ⇒ \(\frac{1}{x}\) = \(\frac{1}{18}\)
x = 18
\(\frac{1}{y}\) = b ⇒ \(\frac{1}{y}\) = \(\frac{1}{36}\)
y = 36
∴ Time taken by a 1 Indian is 18 days
Time taken by a 1 Chinese is 36 days

Students can download Maths Chapter 3 Algebra Ex 3.13 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 3.13

Question 1.
Solve by cross-multiplication method.
(i) 8x – 3y = 12; 5x = 2y + 7
Solution:
8x – 3y – 12 = 0 → (1)
5x – 2y – 7 = 0 → (2)
Use the coefficients for cross multiplication

\(\frac{x}{-3}\) = -1 ⇒ x = 3
\(\frac{y}{-4}\) = -1 ⇒ y = 4
∴ The value of x = 3 and y = 4

(ii) 6x + 7y – 11 = 0; 5x + 2y = 13
Solution:
6x + 7y – 11 = 0 → (1)
5x + 2y = 13 → (2)
Use the coefficient for cross multiplication

-23x = -69
∴ 23x = 69
x= \(\frac{69}{23}\)
= 3
\(\frac{y}{23}\) = \(\frac{1}{-23}\)
-23y = 23
23y = -23
y = –\(\frac{23}{23}\)
y = -1
∴ The value of x = 3 and y = -1

(iii) \(\frac{2}{x}\) + \(\frac{3}{y}\) =5; \(\frac{3}{x}\) – \(\frac{1}{y}\) + 9 = 0
Solution:
\(\frac{1}{x}\) = a; \(\frac{1}{y}\) = b
2a + 3b – 5 = 0 → (1)
3a – b + 9 = 0 → (2)
Using the coefficients for cross multiplication

-11b = -33
11b = 33
b = \(\frac{33}{11}\) = 3
But \(\frac{1}{x}\) = a ⇒ \(\frac{1}{x}\) = -2
-2x = 1 ⇒ 2x = -1
x = –\(\frac{1}{2}\)
but \(\frac{1}{y}\) = b
\(\frac{1}{y}\) = 3 ⇒ 3y = 1
y = \(\frac{1}{3}\)
∴ The value of x = –\(\frac{1}{2}\) and y = \(\frac{1}{3}\)

Question 2.
Akshaya has 2 rupee coins and 5 rupee coins in her purse. If in all she has 80 coins totalling Rs 220, how many coins of each kind does she have.
Solution:
Let the number of 2 rupee coins be “x” and the number of 5 rupee coins be “y”.
By the given first condition
x + y = 80 → (1)
Again by the given second condition
2x + 5y = 220 → (2)
x + y – 80 = 0 → (3)
2x + 5y – 220 = 0 → (4)
Using the coefficients for cross multiplication

\(\frac{x}{180}\) = \(\frac{1}{3}\)
3x = 180
x = \(\frac{180}{3}\)
= 60
But \(\frac{y}{60}\) = \(\frac{1}{3}\)
3y = 60
y = \(\frac{6}{30}\)
= 20
Number of 2 rupee coins = 60
Number of 5 rupee coins = 20

Question 3.
It takes 24 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 8 hours and the pipe of the smaller diameter is used for 18 hours. Only half of the pool is filled. How long would each pipe take to fill the swimming pool.
Solution:
Let the time taken by the larger diameter pipe be “x” hours and the time taken by the smaller diameter pipe be “y” hours.
By the given first condition
\(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{24}\) → (1)
Also
In 8 hours the large pipe fill \(\frac{8}{x}\)
In 18 hours the smaller pipe fill \(\frac{18}{y}\)
By the given second condition ( \(\frac{1}{2}\) of the tank)
\(\frac{8}{x}\) + \(\frac{18}{y}\) = \(\frac{1}{2}\) → (2)
Solve (1) and (2) we get
Let \(\frac{1}{x}\) = a; \(\frac{1}{y}\) = b
a + b = \(\frac{1}{24}\)
Multiply by 24
24a + 24b = 1
24a + 24b – 1 = 0 → (3)
8a + 18b = \(\frac{1}{2}\)
Multiply by 2
16a + 36b = 1
16a + 36b – 1 = 0 → (4)

x = 40
\(\frac{1}{y}\) = b ⇒ \(\frac{1}{y}\) = \(\frac{1}{60}\)
y = 60
To fill the remaining half of the pool.
Time taken by larger pipe = \(\frac{1}{2}\) × 40 = 20 hours
Time taken by smaller pipe = \(\frac{1}{2}\) × 60 = 30 hours

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.1 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 5 Coordinate Geometry Ex 5.1

Question 1.
Plot the following points in the coordinate system and identify the quadrants P(-7, 6), Q(7, -2), R(-6, -7), S(3, 5) and T(3, 9)
Solution:

(i) P(-7, 6) lies in the II quadrant because the x-coordinate is negative and y-coordinate is positive.
(ii) Q(7, -2) lies in the IV quadrant because the x-coordinate is positive and y-coordinate is negative.
(iii) R(-6, -7) lies in the III quadrant because the x-coordinate is negative and y-coordinate is negative.
(iv) S(3, 5) lies in the I quadrant because the x-coordinate is positive and y-coordinate is also positive.
(v) T(3, 9) lies in the I quadrant because the x-coordinate is positive and y-coordinate is also positive.

Question 2.
Write down the abscissa and ordinate of the following.
(i) P
(ii) Q
(iii) R
(iv) S

Solution:
(i) P is (-4, 4) [-4 is abscissa and 4 is ordinate]
(ii) Q is (3, 3) [3 is abscissa and 3 is ordinate]
(iii) R is (4, -2) [4 is abscissa and -2 is ordinate]
(iv) S is (-5, -3) [-5 is abscissa and -3 is ordinate]

Question 3.
Plot the following points in the coordinate plane and join them. What is your conclusion about the resulting figure?
(i) (-5, 3) (-1, 3) (0, 3) (5, 3)
Solution:

Straight line parallel to x-axis.

(ii) (0, -4) (0, -2) (0, 4) (0, 5)
Solution:

The line is on the y-axis.

Question 4.
Plot the following points in the coordinate plane. Join them in order. What type of geometrical shape is formed?
(i) (0, 0) (-4, 0) (-4, -4) (0, -4)
Solution:

The geometrical shape of the figure is square.

(ii) (-3, 3) (2, 3) (-6, -1) (5, -1)
Solution:

The shape of the geometrical figure is Trapezium.

Students can download Maths Chapter 3 Algebra Ex 3.5 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 3.5

Question 1.
Factorise the following expressions:
(i) 2a² + 4a²b + 8a²c
(ii) ab – ac – mb + mc
Solution:
(i) 2a² + 4a²b + 8a²c = 2a²(1 + 2b + 4c)
(ii) ab – ac – mb + mc = a(b – c) – m(b – c)
= (b – c) (a – m)

Question 2.
Factorise the following expressions:
(i) x² + 4x + 4
(ii) 3a² – 24ab + 48b²
(iii) x⁵ – 16x
(iv) m² + \(\frac{1}{m^2}\) – 23
(v) 6 – 216x²
(vi) a² + \(\frac{1}{a^2}\) – 18
Solution:
(i) x² + 4x + 4 = x² + 2 × x × 2 + 2² [a² + 2ab+ b² = (a + b)²]
= (x + 2)²

(ii) 3a² – 24ab + 48b² = 3[a² – 8ab + 16b²]
= 3[a² – 2 × a × 4b + (4b)²]
= 3(a- 4b)² [a² – 2ab + b² = (a – b)²]

(iii) x⁵ – 16x = x[x⁴ – 16] [a² – b² = (a + b) (a – b)]
= x[(x²)² – 4²]
= x(x² + 4) (x² – 4)
= x(x² + 4) (x² – 2²)
= x(x²+ 4) (x + 2) (x – 2)

(iv) m² + \(\frac{1}{m^2}\) – 23 = [Add + 2 and – 2 to make -23 as -25]
= m² + \(\frac{1}{m^2}\) + 2 – 2 – 23 = m² + \(\frac{1}{m^2}\) + 2 – 25
= m² + \(\frac{1}{m^2}\) + 2 × m × \(\frac{1}{m}\) – 5²

(v) 6 – 216x² = 6[1 – 36x²]
= 6[1 – (6x)²] [a² – b² = (a + b)(a – b)]
= 6(1 + 6x) (1 – 6x)

(vi) a² + \(\frac{1}{a^2}\) – 18 = a² + \(\frac{1}{a^2}\) – 2 + 2 – 18
(add -2 and +2 to make 18 as 16)
= a² + \(\frac{1}{a^2}\) – 2 × a × \(\frac{1}{a}\) – 16 [a² + b² – 2ab = (a-b)(a- b)²]

Question 3.
Factorise the following expressions:
(i) 4x² + 9y² + 25z² + 12xy + 30yz + 20xz
Solution:
[a² + b² + c² + 2ab + 2bc + 2ac = (a + b + c)²]
= (2x)² + (3y)² + (5z)² + 2(2x) (3y) + 2(3y) (5z) + 2(5z) (2x)
= (2x + 3y + 5z)²

(ii) 25x² + 4y² + 9z² – 20xy + 12yz – 30xz
Solution:
= (5x)² + (2y)² + (3z)² + 2(5x) (-2y) + 2(-2y) (- 3z) + 2(-3z) (5x)
= (5x – 2y – 3z)²

Question 4.
Factorise the following expressions:
(i) 8x³ + 125y³
(ii) 27x³ – 8y³
(iii) a⁶ – 64
Solution:
(i) 8x³ + 125y³ = (2x)³ + (5y)³ [a³ + b³ = (a + b)(a² – ab + b²)
= (2x + 5y) [(2x)² – (2x) (5y) + (5y)²]
= (2x + 5y) (4x² – 10xy + 25y²)

(ii) 27x³ – 8y³ = (3x)³ – (2y)³ [a³ – b³ = (a – b)(a² + ab + b²)]
= (3x – 2y) [(3x)² + (3x) (2y) + (2y)²]
= (3x – 2y) (9x² + 6xy + Ay²)

(iii) a⁶ – 64 = a⁶ – 2⁶
= (a²)³ – (22)³ [a² – b³ = (a- b) + (a² + ab + b²)]
= (a² – 22) [(a²)² + (a²) (2²) + (2²)²]
= (a + 2) (a – 2) (a⁴ + 4a² + 16)
= (a + 2) (a – 2) [(a²)² + 4² + 8a² – 4a²]
= (a + 2)(a- 2) [(a² + 4)² – (2a)²] {a² – b² = (a + b) (a – b)}
= (a + 2) (a – 2) [(a² + 4 + 2a) (a² + 4 – 2a)
= (a + 2) (a – 2) (a² + 2a + 4) (a² – 2a + 4)

Question 5.
Factorize the following
(i) x³ + 8y³ + 6xy – 1
(ii) l³ – 8m³ – 27n³ – 18lmn
Using the formula [a³ + b³ + c³ – 3abc] = (a + b + c) (a² + b² + c² – ab – bc – ac)
Solution:
(i) x³ + 8y² + 6xy – 1 = -(-x³ – 8y³ – 6xy + 1)
= – (-x³ – 8y³ + 1 – 6xy)
= -[(-x)³ + (-2y)³ + 1 – 3(x) (2y) (1)]
= -[-x – 2y + 1] [(-x)² + (-2y)² + 1² – (-x) (-2y) – (-2y) (1) – (1) (-x)]
= (x + 2y – 1)(x² + 4y² + 1 – 2xy + 2y + x)

(ii) l³ – 8m³ – 27n³ – 18lmn
= l³ + (-2m)³ + (-3n)² – 3(l) (-2m) (-3n)
= (l – 2m – 3n) [l² + (-2m)² + (-3n)² -1 (-2m)] – (-2m)(-3n) – (-3n)(l)
= (l – 2m – 3n) (l² + 4m² + 9n² + 2lm – 6mn + 3ln)