Category: Class 10

Students can download Maths Chapter 6 Trigonometry Ex 6.2 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 6 Trigonometry Ex 6.2

Question 1.
Find the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height 10 \(\sqrt { 3 }\) m.
Answer:
Height of the tower (AC) = 10 \(\sqrt { 3 }\) m
Distance between the base of the tower and point of observation (AB) = 30 m
Let the angle of elevation ∠ABC be θ
In the right ∆ ABC, tan θ = \(\frac { AC }{ AB } \)
= \(\frac{10 \sqrt{3}}{30}=\frac{\sqrt{3}}{3}\)

tan θ = \(\frac{1}{\sqrt{3}}\) = tan 30°
∴ Angle of inclination is 30°

Question 2.
A road is flanked on either side by continuous rows of houses of height 4\(\sqrt { 3 }\) m with no space in between them. A pedestrian is standing on the median of the road facing a row house. The angle of elevation from the pedestrian to the top of the house is 30° . Find the width of the road.
Answer:
Let the mid point of the road AB is “P” (PA = PB)
Height of the home = 4\(\sqrt { 3 }\) m
Let the distance between the pedestrian and the house be “x”
In the right ∆ APD, tan 30° = \(\frac { AD }{ AP } \)

\(\frac{1}{\sqrt{3}}=\frac{4 \sqrt{3}}{x}\)
x = 4 \(\sqrt { 3 }\) × \(\sqrt { 3 }\) = 12 m
∴ Width of the road = PA + PB
= 12 + 12
= 24 m

Question 3.
To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and if he is 5 m away from the wall, what is the height of the window? (\(\sqrt { 3 }\) = 1.732)
Answer:
Let the height of the window FE be “h” m
Let FC be “x” m
∴ EC = (h + x) m

In the right ∆ CDF, tan 45° = \(\frac { CE }{ CD } \)
1 = \(\frac { x }{ 5 } \) ⇒ x = 5
In the right ∆ CDE, tan 60° = \(\frac { CE }{ CD } \)
\(\sqrt { 3 }\) = \(\frac { x+h }{ 5 } \) ⇒ x + h = 5\(\sqrt { 3 }\)
5 + h = 5 \(\sqrt { 3 }\) (substitute the value of x)
h = 5 \(\sqrt { 3 }\) – 5 = 5 × 1.732 – 5 = 8. 66 – 5 = 3.66
∴ Height of the window = 3.66 m

Question 4.
A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 40° . Find the height of the pedestal.
(tan 40° = 0.8391, \(\sqrt { 3 }\) = 1.732)
Answer:
Height of the statue = 1.6 m
Let the height of the pedestal be “h”
AD = H + 1.6m
Let AB be x
In the right ∆ ABD, tan 60° = \(\frac { AD }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { h+1.6 }{ x } \)
x = \(\frac{h+1.6}{\sqrt{3}}\) ……..(1)

In the right ∆ ABC, tan 40° = \(\frac { AC }{ AB } \)
0.8391 = \(\frac { h }{ x } \)
x = \(\frac { h }{ 0.8391 } \)
Substitute the value of x in (1)
\(\frac{h}{0.8391}=\frac{h+1.6}{\sqrt{3}}\)
(h + 1.6) 0.8391 = \(\sqrt { 3 }\) h
0.8391 h + 1.34 = 1.732 h
1.34 = 1.732 h – 0.8391 h
1.34 = 0.89 h
h = \(\frac { 1.34 }{ 0.89 } \) = \(\frac { 134 }{ 89 } \) = 1.5 m
Height of the pedestal = 1.5 m

Question 5.
A Flag pole ‘h’ metres is on the top of the hemispherical dome of radius ‘r’ metres. A man is standing 7 m away from the dome. Seeing the top of the pole at an angle 45° and moving 5 m away from the dome and seeing the bottom of the pole at an angle 30° . Find (i) the height of the pole (ii) radius of the dome. (\(\sqrt { 3 }\) = 1.732)

Answer:
Height of the Flag pole (ED) = h m
AF and AD is the radius of the semi circle (r)
AC = (r + 7)
AB = (r + 7 + 5)
= (r + 12)
In the right ∆ ABD, tan 30° = \(\frac { AD }{ AB } \)

\(\frac{1}{\sqrt{3}}=\frac{r}{r+12}\)
\(\sqrt { 3 }\) r = r + 12
\(\sqrt { 3 }\) r – r = 12 ⇒ r (\(\sqrt { 3 }\) – 1) = 12
r[1.732 – 1] = 12 ⇒ 0.732 r = 12
r = \(\frac { 12 }{ 0.732 } \) ⇒ = 16.39 m
In the right ∆ ACE, tan 45° = \(\frac { AE }{ AC } \)
1 + \(\frac { r+h }{ r+7 } \)
r + 7 = r + h
∴ h = 7 m
Height of the pole (h) = 7 m
Radius of the dome (r) = 16.39 m

Question 6.
The top of a 15 m high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
Answer:
Let the height of the electric pole AD be “h” m
EC = 15 – h m
Let AB be “x”

In the right ∆ ABC, tan 60° = \(\frac { BC }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { 15 }{ x } \)
x = \(=\frac{15}{\sqrt{3}}=\frac{15 \times \sqrt{3}}{3}\)
= 5\(\sqrt { 3 }\)
In the right ∆ CDE, tan 30° = \(\frac { EC }{ DE } \)
\(\frac{1}{\sqrt{3}}=\frac{15-h}{x}\) ………….(1)
Substitute the value of x = 5 \(\sqrt { 3 }\) in (1)
\(\frac{1}{\sqrt{3}}=\frac{15-h}{5 \sqrt{3}} \Rightarrow \sqrt{3}(15-h)=5 \sqrt{3}\)
(15 – h) = \(\frac{5 \sqrt{3}}{\sqrt{3}}\) ⇒ 15 – h = 5
h = 15 – 5 = 10
∴ Height of the electric pole = 10 m

Question 7.
A vertical pole fixed to the ground is divided in the ratio 1:9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 25 m away from the base of the pole, what is the height of the pole?
Answer:
Let the first part of the pole be “x” and the second part be “9x”
∴ height of the pole (AC) = x + 9x = 10x
Given ∠CDB = ∠BDA
∴ BD is the angle bisector of ∠ADC
By angle bisector theorem

\(\frac { AB }{ BC } \) = \(\frac { AD }{ DC } \)
\(\frac { 9x }{ x } \) = \(\frac { AD }{ 25 } \) ⇒ AD = 9 × 25 = 225
In the right ∆ ACD
AD² = AC² + CD²
(225)² = (10x)² + 25²
50625 = 100x² + 625
∴ 100x² = 50625 – 625 = 50000
x² = \(\frac { 50000 }{ 100 } \) = 500
x = \(\sqrt { 500 }\) = \(\sqrt{5 \times 100}=10 \sqrt{5}\)
∴ AC = 10 × 10\(\sqrt { 5 }\) = 100 \(\sqrt { 5 }\) (AC = 10x)
∴ Height of the pole = 100 \(\sqrt { 5 }\) m

Question 8.
A traveler approaches a mountain on highway. He measures the angle of elevation to the peak at each milestone. At two consecutive milestones the angles measured are 4° and 8°. What is the height of the peak if the distance between consecutive milestones is 1 mile, (tan 4° = 0.0699, tan 8° = 0.1405)
Answer:
Let the height of the peak be “h” mile. Let AD be x mile.
∴ AB = (x + 1) mile.
In the right ∆ ADC, tan 8° = \(\frac { AC }{ AC } \)

0.1405 = \(\frac { h }{ x } \)
x = \(\frac { h }{ 0.1405 } \) ………..(1)
In ∆ ABC, tan 4° = \(\frac { AC }{ AB } \)
0.0699 = \(\frac { h }{ x+1 } \) ⇒ (x + 1) 0.0699 = h
0.0699x + 0.0699 = h
0.0699 x = h – 0.0699
x = \(\frac { h-0.0699 }{ 0.0699 } \) ………(2)
Equation (1) and (2) we get,
\(\frac { h-0.0699 }{ 0.0699 } \) = \(\frac { h }{ 0.1405 } \)
0.0699 h = 0.1405 (h – 0.0699)
0.0699 h = 0.1405 h – 0.0098
0.0098 = 0.1405 h – 0.0699 h
0.0098 = 0.0706 h
h = \(\frac { 0.0098 }{ 0.0706 } \) = \(\frac { 98 }{ 706 } \) = 0.1388
= 0.14 mile (approximately)
Height of the peak = 0.14 mile

Students can download Maths Chapter 6 Trigonometry Ex 6.4 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 6 Trigonometry Ex 6.4

Question 1.
From the top of a tree of height 13 m the angle of elevation and depression of the top and bottom of another tree are 45° and 30° respectively. Find the height of the second tree, (\(\sqrt { 3 }\) = 1.732)
Answer:
Let the height of the second tree be “h”
ED = (h – 13) m
Let AB = x m
In the right ∆ ABC, tan 30° = \(\frac { BC }{ AB } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { 13 }{ x } \)
x = 13 \(\sqrt { 3 }\) ……..(1)

In the right ∆ CED, tan 45° = \(\frac { DE }{ EC } \)
1 = \(\frac { h-13 }{ x } \)
x = h – 13 ……..(2)
From (1) and (2) we get
h – 13 = 13 \(\sqrt { 3 }\) ⇒ h = 13 \(\sqrt { 3 }\) + 13
= 13 × 1.732 + 13
= 22.52 + 13 = 35.52 m
∴ Height of the second tree = 35.52 m

Question 2.
A man is standing on the deck of a ship, which is 40 m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30° . Calculate the distance of the hill from the ship and the height of the hill. (\(\sqrt { 3 }\) = 1.732)
Answer:
Let the height of the hill BE be “h” m and the distance of the hill from the ship be “x” m
In the right ∆ ABD
tan 30° = \(\frac { AD }{ DB } \)
\(\frac{1}{\sqrt{3}}=\frac{40}{x}\)
x = 40 \(\sqrt { 3 }\) ……..(1)
In the right ∆ CDE
tan 60° = \(\frac { CE }{ DC } \)
\(\sqrt { 3 }\) = \(\frac { h-40 }{ x } \)
x = \(\frac{h-40}{\sqrt{3}}\) ……..(2)

From (1) and (2) we get
\(\frac{h-40}{\sqrt{3}}\) = 40\(\sqrt { 3 }\)
h – 40 = 40 × 3
h = 120 + 40 = 160 m
Height of the hill = 160 m
Distance of the hill from the ship = 40 × \(\sqrt { 3 }\) = 40 × 1.732 = 69.28 m

Question 3.
If the angle of elevation of a cloud from a point ‘h’ metres above a lake is θ1 and the angle of depression of its reflection in the lake is θ2. Prove that the height that the cloud is located from the ground is \(\frac{h\left(\tan \theta_{1}+\tan \theta_{2}\right)}{\tan \theta_{2}-\tan \theta_{1}}\)
Answer:
Let P be the cloud and Q be its reflection.
Let A be the point of observation such that AB = h
Let the height of the cloud be x. (PS = x)
PR = x – h and QR = x + h

Let AR = y
In the right ∆ ARP, tan θ₁ = \(\frac { PR }{ AR } \)
tan θ₁ = \(\frac { x-h }{ y } \) ………(1)
In the ∆ AQR,
tan θ₂ = \(\frac { QR }{ AR } \)
tan θ₂ = \(\frac { x+h }{ y } \) ……….(2)
Add (1) and (2)

Question 4.
The angle of elevation of the top of a cell phone tower from the foot of a high apartment is 60° and the angle of depression of the foot of the tower from the top of the apartment is 30° . If the height of the apartment is 50 m, find the height of the cell phone tower. According to radiations control norms, the minimum height of a cell phone tower should be 120 m. State if the height of the above mentioned cell phone tower meets the radiation norms.
Answer:
Let the height of the cell phone tower be “h” m
AD is the height of the apartment; AD = 50 m
Let AB be “x”
In the right triangle ABC
tan 60° = \(\frac { BC }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { h }{ x } \)

x = \(\frac{h}{\sqrt{3}}\) …….(1)
In the right triangle ABD, tan 30° = \(\frac { AD }{ AB } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { 50 }{ x } \)
x = 50 \(\sqrt { 3 }\) ……(2)
From (1) and (2) We get
\(\frac{h}{\sqrt{3}}\) = 50 \(\sqrt { 3 }\)
h = 50\(\sqrt { 3 }\) × \(\sqrt { 3 }\) = 50 × 3 = 150
Height of the cell phone tower is 150 m.
Yes, the cell phone tower meets the radiation norms.

Question 5.
The angles of elevation and depression of the top and bottom of a lamp post from the top of a 66 m high apartment are 60° and 30° respectively. Find
(i) The height of the lamp post.
(ii) The difference between height of the lamp post and the apartment.
(iii) The distance between the lamp post and the apartment. (\(\sqrt { 3 }\) = 1.732)
Answer:
(i) Let the height of the lamp post AE be “h” m
DE = h – 66
Let AB be “x”
In the right ∆ ABC, tan 30° = \(\frac { BC }{ AB } \)
\(\frac{1}{\sqrt{3}}=\frac{66}{x}\)
x = 66 \(\sqrt { 3 }\) ……(1)

In the right ∆ CDE, tan 60° = \(\frac { DE }{ DC } \)
\(\sqrt { 3 }\) = \(\frac { h-66 }{ x } \) ⇒ \(\sqrt { 3 }\) x = h – 66
x = \(\frac{h-66}{\sqrt{3}}\) ………….(2)
From (1) and (2) we get
\(\frac{h-66}{\sqrt{3}}\) = 66 \(\sqrt { 3 }\)
h – 66 = 66 \(\sqrt { 3 }\) × \(\sqrt { 3 }\) = 66 × 3
h – 66 = 198 ⇒ h = 198 + 66
h = 264 m
(i) the height of the lamp post = 264 m
(ii) Difference of the height of lamp post and apartment = 264 – 66
= 198 m
(ii) Distance between the lamp post and the apartment = 66 \(\sqrt { 3 }\) m
= 66 × 1.732 = 114.31 m

Question 6.
Three villagers A, B and C can see each other across a valley. The horizontal distance between A and B is 8 km and the horizontal distance between B and C is 12 km. The angle of depression of B from A is 20° and the angle of elevation of C from B is 30°. Calculate:
(i) the vertical height between A and B.
(ii) the vertical height between B and C. (tan 20° = 0 .3640, \(\sqrt { 3 }\) = 1. 732)
Answer:
Let AD is the vertical height between A and B
In the right ∆ ABD

tan 20° = \(\frac { AD }{ BD } \)
0.3640 = \(\frac { AD }{ 8 } \)
AD = 0.3640 × 8 = 2.912 km
∴ AD = 2.91 km
CE is the vertical height between C and B
In the right ∆ BCE, tan 30° = \(\frac { CE }{ BE } \)
\(\frac{1}{\sqrt{3}}=\frac{C E}{12} \Rightarrow \sqrt{3} C E=12\)
CE = \(\frac{12}{\sqrt{3}}=\frac{12 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{12 \times \sqrt{3}}{3}\)
= 4 \(\sqrt { 3 }\) = 4 × 1.732 = 6.928
= 6.93 km
(i) The vertical height between A and B = 2.91 km
(ii) The vertical height between B and C = 6.93 km

Students can download Maths Chapter 6 Trigonometry Ex 6.1 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 6 Trigonometry Ex 6.1

Question 1.
Prove the following identities.
(i) cot θ + tan θ = sec θ cosec θ
(ii) tan⁴ θ + tan² θ = sec⁴ θ – sec² θ
Answer:
(i) L. H. S = cot θ + tan θ
= \(\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}\)
= \(\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta \cos \theta}\)
[cos² θ + sin² θ = 1]
= \(\frac{1}{\sin \theta \cos \theta}\)
= sec θ . cosec θ = R. H. S
∴ cot θ + tan θ = sec θ cosec θ

(ii) tan⁴ θ + tan² θ = sec⁴ θ – sec² θ
L.H.S = tan⁴ θ + tan² θ
= tan² θ (tan² θ + 1)
= tan² θ sec² θ
R.H.S = sec⁴ θ – sec² θ
= sec² θ (sec² θ – 1)
= sec² θ tan² θ
L.H.S = R.H.S
∴ tan⁴ θ + tan² θ = sec⁴ θ – sec² θ

Question 2.
Prove the following identities.
(i) \(\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}\) = tan² θ
(ii) \(\frac{\cos \theta}{1+\sin \theta}\) = sec θ – tan θ
Answer:
(i) \(\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}\) = tan² θ

(ii) \(\frac{\cos \theta}{1+\sin \theta}\) = sec θ – tan θ

Aliter:
L.H.S = \(\frac{\cos \theta}{1-\sin \theta}\)
[conjugate (1 – sin θ)]

Question 3.
Prove the following identities.

Solution:

Question 4.
Prove the following identities.
(i) sec⁶ θ = tan⁶ θ + 3 tan² θ sec² θ + 1
(ii) (sin θ + sec θ)² + (cos θ + cosec θ)² = 1 + (sec θ + cosec θ)²
Answer:
(i) sec⁶ θ = tan⁶ θ + 3 tan² θ sec² θ + 1
L.H.S = sec⁶ θ
= (sec² θ)³ = (1 + tan² θ)³
= 1 + (tan² θ)³ + 3 (1) (tan² θ) (1 + tan² θ) [(a + b)³ = a³ + b³ + 3 ab (a + b)]
= 1 + tan⁶ θ + 3 tan² θ(1 + tan² θ)
= 1 + tan⁶ θ + 3 tan² θ (sec² θ)
= 1 + tan⁶ θ + 3 tan² θ sec² θ
= tan⁶ θ + 3 tan² θ sec² θ + 1
L.H.S = R.H.S

(ii) (sin θ + sec θ)² + (cos θ + cosec θ)² = 1 + (sec θ + cosec θ)²
L.H.S = (sin θ + sec θ)² + (cos θ + cosec θ)²]
= [sin² θ + sec² θ + 2 sin θ sec θ + cos² θ + cosec² θ + 2 cos θ cosec θ]
= (sin² θ + cos² θ) + (sec² θ + cosec² θ) + 2 (sin θ sec θ + cos θ cosec θ)

= 1 + sec² θ + cosec² θ + 2 sec θ cosec θ
= 1 + (sec θ + cosec θ)²
L.H.S = R.H.S
∴ (sin θ + sec θ)² + (cos θ + cosec θ)² = 1 + (sec θ + cosec θ)²

Question 5.
Prove the following identities.
(i) sec⁴ θ (1 – sin⁴ θ) – 2 tan² θ = 1
(ii) \(\frac{\cot \theta-\cos \theta}{\cot \theta+\cos \theta}=\frac{\csc \theta-1}{\csc \theta+1}\)
Answer:
(i) L.H.S = sec⁴ θ (1 – sin⁴ θ) – 2 tan² θ

L.H.S = R.H.S
∴ sec⁴ θ (1 – sin⁴ θ) – 2 tan² θ = 1

(ii) \(\frac{\cot \theta-\cos \theta}{\cot \theta+\cos \theta}=\frac{\csc \theta-1}{\csc \theta+1}\)

Question 6.
Prove the following identities.


Answer:

Question 7.
(i) If sin θ + cos θ = \(\sqrt { 3 }\), then prove that tan θ + cot θ = 1.
(ii) If \(\sqrt { 3 }\) sin θ – cos θ = θ, then show that tan 3θ = \(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\)
Answer:
sin θ + cos θ = \(\sqrt { 3 }\) (squaring on both sides)
(sin θ + cos θ)² = (\(\sqrt { 3 }\))²
sin² θ + cos² θ + 2 sin θ cos θ = 3
1 + 2 sin θ cos θ = 3
2 sin θ cos θ = 3 – 1
2 sin θ cos θ = 2
∴ sin θ cos θ = 1
L.H.S = tan θ + cot θ

L.H.S = R.H.S ⇒ tan θ + cot θ = 1

(ii) If \(\sqrt { 3 }\) sin θ – cos θ = 0
To prove tan 3θ = \(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\)
\(\sqrt { 3 }\) sin θ – cos θ = 0
\(\sqrt { 3 }\) sin θ = cos θ
\(\frac{\sin \theta}{\cos \theta}=\frac{1}{\sqrt{3}}\)
tan θ = tan 30°
θ = 30°
L.H.S = tan 3θ°
= tan3 (30°)
= tan 90°
= undefined (∝)

∴ tan 3θ = \(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\)

Question 8.
(i) If \(\frac{\cos \alpha}{\cos \beta}=m\) and \(\frac{\cos \alpha}{\cos \beta}=n\) then prove that (m2 + n2) cos2 β = n2
(ii) If cot θ + tan θ = x and sec θ – sec θ – cos θ = y, then prove that (x2y)2/3 – (xy2)2/3 = 1
Answer:
(i) L.H.S = (m² + n²) cos² β

L.H.S = R.H.S ⇒ ∴ (m² + n²) cos² β = n²

(ii) Given cot θ + tan θ = x sec θ – cos θ = y
x = cot θ + tan θ
x = \(\frac{1}{\tan \theta}\) + tan θ
= \(\frac{1+\tan ^{2} \theta}{\tan \theta}\) = \(\frac{\sec ^{2} \theta}{\tan \theta}\)

y = sec θ – cos θ
= \(\frac{1}{\cos \theta}-\cos \theta=\frac{1-\cos ^{2} \theta}{\cos \theta}\)
y = \(\frac{\sin ^{2} \theta}{\cos \theta}\)

Question 9.
(i) If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q (p² – 1) = 2 p
(ii) If sin θ (1 + sin² θ) = cos² θ, then prove that cos⁶ θ – 4 cos⁴ θ + 8 cos² θ = 4
Answer:
(i) p = sin θ + cos θ
p² = (sin θ + cos θ)²
= sin² θ + cos² θ + 2 sin θ cos θ = 1 + 2 sin θ cos θ
q = sec θ + cosec θ
= \(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}=\frac{\sin \theta+\cos \theta}{\sin \theta \cos \theta}\)
L.H.S = q(p² – 1)

(ii) sin θ (1 + sin² θ) = cos² θ
sin θ (1 + 1 – cos² θ) = cos² θ
sin θ (2 – cos² θ) = cos² θ
Squaring on both sides,
sin² θ (2 – cos² θ)² = cos⁴ θ
(1 – cos² θ) (4 + cos⁴ θ – 4 cos² θ) = cos⁴ θ
4 cos⁴ θ – 4 cos² θ – cos⁶ θ + 4 cos⁴ θ = cos⁴ θ
4 + 5 cos⁴ θ – 8 cos² θ – cos⁶ θ = cos⁴ θ
– cos⁶ θ + 5 cos⁴ θ – cos⁴ θ – 8 cos² θ = -4
– cos⁶ θ + 4 cos⁴ θ – 8 cos² θ = -4
cos⁶ θ – 4 cos⁴ θ + 8 cos² θ = 4
Hence it is proved

Question 10.
If \(\frac{\cos \theta}{1+\sin \theta}\) = \(\frac { 1 }{ a } \), then prove that \(\frac{a^{2}-1}{a^{2}+1}\) = sin θ
Answer:

Students can download Maths Chapter 6 Trigonometry Ex 6.3 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 6 Trigonometry Ex 6.3

Question 1.
From the top of a rock 50 \(\sqrt { 3 }\) m high, the angle of depression of a car on the ground is observed to be 30°. Find the distance of the car from the rock.
Answer:
Let the distance of the car from the rock is “x” m

In the right ∆ ABC, tan 30° = \(\frac { AB }{ BC } \)
\(\frac{1}{\sqrt{3}}=\frac{50 \sqrt{3}}{x}\)
x = 50 \(\sqrt { 3 }\) × \(\sqrt { 3 }\) = 50 × 3
= 150 m
∴ Distance of the car from the rock = 150 m

Question 2.
The horizontal distance between two buildings is 70 m. The angle of depression of the top of the first building when seen from the top of the second building is 45°. If the height of the second building is 120 m, find the height of the first building.
Answer:
Let the height of the first building AD be “x” m
∴ EC = 120 – x
In the right ∆ CDE,
tan 45° = \(\frac { CE }{ CD } \)

1 = \(\frac { 120-x }{ 70 } \) ⇒ 70 = 120 – x
x = 50 cm
∴ The height of the first building is 50 m

Question 3.
From the top of the tower 60 m high the anles of depression the top and bottom of a vertical lamp post are observed be 38° and 60° respectively
Find the height of the lamp post. (tan 38° = 0.7813,\(\sqrt { 3 }\) = 1.732)
Answer:
Let the height of the lamp post be “h”
The height of the tower (BC) = 60 m
∴ EC = 60 – h
Let AB be x
In the right ∆ ABC,

tan 60° = \(\frac { BC }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { 60 }{ x } \)
x = \(\frac{60}{\sqrt{3}}\) ……..(1)
In the right ∆ DEC, tan 38° = \(\frac { EC }{ DE } \)
0.7813 = \(\frac { 60-h }{ x } \)
x = \(\frac { 60-h }{ 0.7813 } \) …….(2)
From (1) and (2) we get
\(\frac{60}{\sqrt{3}}\) = \(\frac { 60-h }{ 0.7813 } \)
60 × 0.7813 = 60 \(\sqrt { 3 }\) – \(\sqrt { 3 }\) h
\(\sqrt { 3 }\) h = 60 \(\sqrt { 3 }\) – 46.88
= 60 × 1.732 – 46.88
= 103.92 – 46.88
1.732 h = 57.04 ⇒ h = \(\frac { 57.04 }{ 1.732 } \)
h = \(\frac { 570440 }{ 1732 } \) = 32.93 m
∴ Height of the lamp post = 32.93 m

Question 4.
An aeroplane at an altitude of 1800 m finds that two boats are sailing towards it in the same direction. The angles of depression of the boats as observed from the aeroplane are
60° and 30° respectively. Find the distance between the two boats. (\(\sqrt { 3 }\) = 1.732)
Answer:
C and D are the position of the two boats.
Let the distance between the two boats be “x”
Let BC = y
∴ BD = (x + y)

In the right ∆ ABC, tan 30° = \(\frac { AB }{ BD } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { 1800 }{ x+y } \)
x + y = 1800 \(\sqrt { 3 }\)
y = 1800 \(\sqrt { 3 }\) – x ……(1)
In the right ∆ ABC, tan 60° = \(\frac { AB }{ BC } \)
\(\sqrt { 3 }\) = \(\frac { 1800 }{ y } \)
y = \(\frac{1800}{\sqrt{3}}\) ……….(2)
From (1) and (2) we get
\(\frac{1800}{\sqrt{3}}\) = 1800 \(\sqrt { 3 }\) – x
1800 = 1800 × 3 – \(\sqrt { 3 }\)x
\(\sqrt { 3 }\)x = 5400 – 1800
x = \(\frac{3600}{\sqrt{3}}=\frac{3600 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{3600 \times \sqrt{3}}{3}\)
= 1200 × 1.732 = 2078.4 m
Distance between the two boats = 2078.4 m

Question 5.
From the top of a lighthouse, the angle of depression of two ships on the opposite sides of it are observed to be 30° and 60°. If the height of the lighthouse is h meters and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is \(\frac{4 h}{\sqrt{3}}\) m.
Answer:
A and C be the position of two ships.
Let AB be x and BC be y. Distance between the two ships is x + y

In the right ∆ ABD, tan 60° = \(\frac { BD }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { h }{ x } \)
x = \(\frac{h}{\sqrt{3}}\) ……(1)
In the right ∆ BCD,
tan 30° = \(\frac { BD }{ BC } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h }{ y } \)
y = \(\sqrt { 3 }\) h
Distance between the two ships (x + y) = \(\frac{h}{\sqrt{3}}+\sqrt{3} h\)
= \(\frac{h+3 h}{\sqrt{3}}=\frac{4 h}{\sqrt{3}}\)
Hence it is verified

Question 6.
A lift in a building of height 90 feet with transparent glass walls is descending from the top of the building. At the top of the building, the angle of depression to a fountain in the garden is 60°. Two minutes later, the angle of depression reduces to 30°. If the fountain is 30 \(\sqrt { 3 }\) feet from the entrance of the lift, find the speed of the lift which is descending.
Answer:
Let the speed of the lift is “x” feet / minute
Distance AB = 2 x feet (speed × time)
BC = (90 – 2x)
In the right ∆ BCD,

tan 30° = \(\frac { BC }{ DC } \)
\(\frac{1}{\sqrt{3}}=\frac{90-2 x}{30 \sqrt{3}}\)
\(\sqrt { 3 }\) (90 – 2x) = 30\(\sqrt { 3 }\)
(90 – 2x) = \(\frac{30 \sqrt{3}}{\sqrt{3}}\) ⇒ (90 – 2x) = 30
2x = 60
x = \(\frac { 60 }{ 2 } \) = 30
x = 30 feet/minute
Speed of the lift = 30 feet / minute (or) [ \(\frac { 30 }{ 60 } \) second) 0.5 feet / second

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.5 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 5 Coordinate Geometry Ex 5.5

Multiple Choice Questions

Question 1.
The area of triangle formed by the points (-5, 0), (0, – 5) and (5, 0) is …………..
(1) 0 sq.units
(2) 25 sq.units
(3) 5 sq.units
(4) none of these
Answer:
(2) 25 sq.units Hint.
Hint:
Area of the ∆

Question 2.
A man walks near a wall, such that the distance between him and the wall is 10 units. Consider the wall to be the Y axis. The path travelled by the man is …………
(1) x = 10
(2) y = 10
(3) x = 0
(4) y = 0
Answer:
(1) x = 10
Hint:

Question 3.
The straight line given by the equation x = 11 is …………….
(1) parallel to X axis
(2) parallel to Y axis
(3) passing through the origin
(4) passing through the point (0,11)
Answer:
(2) parallel to Y axis

Question 4.
If (5,7), (3,p) and (6,6) are collinear, then the value of p is ……………
(1) 3
(2) 6
(3) 9
(4) 12
Answer:
(3) 9
Hint:
Since the three points are collinear. Area of a triangle is 0

5p + 18 + 42 – (21 + 6p + 30) = 0
5p + 60 – (51 + 6p) = 0
5p + 60 – 51 – 6p = 0
-p + 9 = 0
-p = -9
p = 9

Question 5.
The point of intersection of 3x – y = 4 and x + 7 = 8 is ……………
(1) (5,3)
(2) (2,4)
(3) (3,5)
(4) (4, 4)
Answer:
(3) (3, 5)

Substitute the value of x = 3 in (2)
3 + 7 = 8
y = 8 – 3 = 5
The point of intersection is (3, 5)

Question 6.
The slope of the line joining (12, 3), (4, a) is \(\frac { 1 }{ 8 } \). The value of ‘a’ is …………….
(1) 1
(2) 4
(3) -5
(4) 2
Answer:
(4) 2
Hint:
Slope of a line = \(\frac { 1 }{ 8 } \)

Question 7.
The slope of the line which is perpendicular to a line joining the points (0, 0) and (- 8, 8) is ………..
(1) -1
(2) 1
(3) \(\frac { 1 }{ 3 } \)
(4) -8
Answer:
(2) 1
Hint:
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { 8-0 }{ -8-0 } \) = \(\frac { 8 }{ -8 } \) = -1
Slope of the Perpendicular = 1

Question 8.
If slope of the line PQ is \(\frac{1}{\sqrt{3}}\) then slope of the perpendicular bisector of PQ is …………..
(1) \(\sqrt { 3 }\)
(2) –\(\sqrt { 3 }\)
(3) \(\frac{1}{\sqrt{3}}\)
(4) 0
Answer:
(2) –\(\sqrt { 3 }\)
Hint:
Slope of a line = \(\frac{1}{\sqrt{3}}\)
Slope of the ⊥r bisector = –\(\sqrt { 3 }\)

Question 9.
If A is a point on the Y axis whose ordinate is 8 and B is a point on the X axis whose abscissae is 5 then the equation of the line AB is ……………
(1) 8x + 5y = 40
(2) 8x – 5y = 40
(3) x = 8
(4) y = 5
Answer:
(1) 8x + 5y = 40
Hint:
Let the point A be (0, 8) and B (5, 0)

Question 10.
The equation of a line passing through the origin and perpendicular to the line lx -3y + 4 = 0 is
(1) 7x – 3y + 4 = 0
(2) 3x – 7y + 4 = 0
(3) 3x + 7y = 0
(4) 7x – 3y = 0
Answer:
(3) 3x + 7y = 0
Hint:
Slope of the line of 7x – 3y + 4 = 0
= \(\frac { -7 }{ -3 } \) = \(\frac { 7 }{ 3 } \)
Slope of its ⊥^r = \(\frac { -3 }{ 7 } \)
The line passes through (0,0)
Equation of a line is
y – y₁ = m(x – x₁)
y – 0 = \(\frac { -3 }{ 7 } \) (x – 0)
y = \(\frac { -3 }{ 7 } \) x ⇒ 7y = -3x
3x + 7y = 0

Question 11.
Consider four straight lines
(i) l₁ : 3y = 4x + 5
(ii) l₂ : 4y = 3x – 1
(iii) l₃ : 4y + 3x = 7
(iv) l₄ : 4x + 3y = 2
Which of the following statement is true?
(1) l₁ and l₂ are perpendicular
(2) l₂ and l₄ are parallel
(3) l₂ and l₄ are perpendicular
(4) l₂ and l₃ are parallel
Answer:
(3) l₂ and l₄ are perpendicular
Hint:
Slope of l₁ = \(\frac { 4 }{ 3 } \); Slope of l₂ = \(\frac { 3 }{ 4 } \)
Slope of l₃ = – \(\frac { 3 }{ 4 } \); Slope of l₄ = –\(\frac { 4 }{ 3 } \)
(1) l₁ × l₂ = \(\frac { 4 }{ 3 } \) × \(\frac { 3 }{ 4 } \) = 1 …….False
(2) l₁ = \(\frac { 4 }{ 3 } \); l₄ = – \(\frac { 4 }{ 3 } \) not parallel ………False
(3) l₂ × l₄ = \(\frac { 3 }{ 4 } \) × – \(\frac { 4 }{ 3 } \) = -1 …….True
(4) l₂ = \(\frac { 3 }{ 4 } \); l₃ = – \(\frac { 3 }{ 4 } \) not parallel ………False

Question 12.
A straight line has equation 87 = 4x + 21. Which of the following is true …………………….
(1) The slope is 0.5 and the y intercept is 2.6
(2) The slope is 5 and the y intercept is 1.6
(3) The slope is 0.5 and they intercept is 1.6
(4) The slope is 5 and the y intercept is 2.6
Answer:
(1) The slope is 0.5 and they intercept is 2.6
Hint:
8y = 4x + 21
y = \(\frac { 4 }{ 8 } \) x + \(\frac { 21 }{ 8 } \)
= \(\frac { 1 }{ 2 } \) x + \(\frac { 21 }{ 8 } \)
\(\frac { 1 }{ 2 } \) = 0.5
\(\frac { 21 }{ 8 } \) = 2.625
Slope = \(\frac { 1 }{ 2 } \) = 0.5
y intercept = \(\frac { 21 }{ 8 } \) = 2.6

Question 13.
When proving that a quadrilateral is a trapezium, it is necessary to show
(1) Two sides are parallel.
(2) Two parallel and two non-parallel sides.
(3) Opposite sides are parallel.
(4) All sides are of equal length.
Solution:
(2) Two parallel and two non-parallel sides.

Question 14.
When proving that a quadrilateral is a parallelogram by using slopes you must find …………………
(1) The slopes of two sides
(2) The slopes of two pair of opposite sides
(3) The lengths of all sides
(4) Both the lengths and slopes of two sides
Answer:
(2) The slopes of two pair of opposite sides

Question 15.
(2,1) is the point of intersection of two lines.
(1) x – y – 3 = 0; 3x – y – 7 = 0
(2) x + y = 3; 3x + y = 7
(3) 3x + y = 3; x + y = 7
(4) x + 3y – 3 = 0; x – y – 7 = 0
Solution:
(2) x + y = 3; 3x + y = 7

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.4 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 5 Coordinate Geometry Ex 5.4

Question 1.
Find the slope of the following straight lines.
(i) 5y – 3 = 0
(ii) 7x – \(\frac { 3 }{ 17 } \) = 0
Solution:
(i) 5y – 3 = 0
5y = 3 ⇒ y = \(\frac { 3 }{ 5 } \)
Slope = 0

(ii) 7x – \(\frac { 3 }{ 17 } \) = 0 (Comparing with y = mx + c)
7x = \(\frac { 3 }{ 17 } \)
Slope is undefined

Question 2.
Find the slope of the line which is
(i) parallel to y = 0.7x – 11
(ii) perpendicular to the line x = -11
Solution:
(i) y = 0.7x – 11
Slope = 0.7 (Comparing with y = mx + c)
(ii) Perpendicular to the line x = – 11
Slope is undefined (Since the line is intersecting the X-axis)

Question 3.
Check whether the given lines are parallel or perpendicular
(i) \(\frac { x }{ 3 } \) + \(\frac { y }{ 4 } \) + \(\frac { 1 }{ 7 } \) = 0 and \(\frac { 2x }{ 3 } \) + \(\frac { y }{ 2 } \) + \(\frac { 1 }{ 10 } \) = 0
(ii) 5x + 23y + 14 = 0 and 23x – 5x + 9 = 0
Solution:
(i) \(\frac { x }{ 3 } \) + \(\frac { y }{ 4 } \) + \(\frac { 1 }{ 7 } \) = 0 ; \(\frac { 2x }{ 3 } \) + \(\frac { y }{ 2 } \) + \(\frac { 1 }{ 10 } \) = 0
Slope of the line (m₁) = \(\frac { -a }{ b } \)
= – \(\frac { 1 }{ 3 } \) ÷ \(\frac { 1 }{ 4 } \) = –\(\frac { 1 }{ 3 } \) × \(\frac { 4 }{ 1 } \) = – \(\frac { 4 }{ 3 } \)
Slope of the line (m₂) = – \(\frac { 2 }{ 3 } \) ÷ \(\frac { 1 }{ 2 } \) = –\(\frac { 2 }{ 3 } \) × \(\frac { 2 }{ 1 } \) = – \(\frac { 4 }{ 3 } \)
m₁ = m₂ = – \(\frac { 4 }{ 3 } \)
∴ The two lines are parallel.

(ii) 5x + 23y + 14 = 0 and 23x – 5x + 9 = 0
Slope of the line (m₁) = \(\frac { -5 }{ 23 } \)
Slope of the line (m₂) = \(\frac { -23 }{ -5 } \) = \(\frac { 23 }{ 5 } \)
m₁ × m₂ = \(\frac { -5 }{ 23 } \) × \(\frac { 23 }{ 5 } \) = -1
∴ The two lines are perpendicular

Question 4.
If the straight lines 12y = -(p + 3)x + 12, 12x – 7y = 16 are perpendicular then find ‘p’
Solution:
Slope of the first line 12y = -(p + 3)x +12
y = \(-\frac{(p+3) x}{12}+1\) (Comparing with y = mx + c)
Slope of the second line (m₁) = \(\frac { -(p+3) }{ 12 } \)
Slope of the second line 12x – 7y = 16
(m₂) = \(\frac { -a }{ b } \) = \(\frac { -12 }{ -7 } \) = \(\frac { 12 }{ 7 } \)
Since the two lines are perpendicular
m₁ × m₂ = -1
\(\frac { -(p+3) }{ 12 } \) × \(\frac { 12 }{ 7 } \) = -1 ⇒ \(\frac { -(p+3) }{ 7 } \) = -1
-(p + 3) = -7
– p – 3 = -7 ⇒ -p = -7 + 3
-p = -4 ⇒ p = 4
The value of p = 4

Question 5.
Find the equation of a straight line passing through the point P(-5,2) and parallel to the line joining the points Q(3, -2) and R(-5,4).
Solution:
Slope of the line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of the line QR = \(\frac { 4+2 }{ -5-3 } \) = \(\frac { 6 }{ -8 } \) = \(\frac { 3 }{ -4 } \) ⇒ – \(\frac { 3 }{ 4 } \)
Slope of its parallel = – \(\frac { 3 }{ 4 } \)
The given point is p(-5, 2)
Equation of the line is y – y₁ = m(x – x₁)
y – 2 = – \(\frac { 3 }{ 4 } \) (x + 5)
4y – 8 = -3x – 15
3x + 4y – 8 + 15 = 0
3x + 4y + 7 = 0
The equation of the line is 3x + 4y + 7 = 0

Question 6.
Find the equation of a line passing through (6, -2) and perpendicular to the line joining the points (6, 7) and (2, -3).
Solution:
Let the vertices A (6, 7), B (2, -3), D (6, -2)

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { -3-7 }{ 2-6 } \) = \(\frac { -10 }{ -4 } \) = \(\frac { 5 }{ 2 } \)
Slope of its perpendicular (CD) = – \(\frac { 2 }{ 5 } \)
Equation of the line CD is y – y₁ = m(x – x₁)
y + 2 = –\(\frac { 2 }{ 5 } \) (x – 6)
5(y + 2) = -2 (x – 6)
5y + 10 = -2x + 12
2x + 5y + 10 – 12 = 0
2x + 5y – 2 = 0
The equation of the line is 2x + 5y – 2 = 0

Question 7.
A(-3,0) B(10, -2) and C(12,3) are the vertices of ∆ABC. Find the equation of the altitude through A and B.
Solution:
To find the equation of the altitude from A.
The vertices of ∆ABC are A(-3, 0), B(10, -2) and C(12, 3)

Slope of BC = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { 3+2 }{ 12-10 } \) = \(\frac { 5 }{ 2 } \)
Slope of the altitude AD is – \(\frac { 2 }{ 5 } \)
Equation of the altitude AD is
y – y₁ = m (x – x₁)
y – 0 = – \(\frac { 2 }{ 5 } \) (x + 3)
5y = -2x -6
2x + 5y + 6 = 0
Equation of the altitude AD is 2x + 5y + 6 = 0
Equation of the altitude from B

Slope of AC = \(\frac { 3-0 }{ 12+3 } \) = \(\frac { 3 }{ 15 } \) = \(\frac { 1 }{ 5 } \)
Slope of the altitude AD is -5
Equation of the altitude BD is y – y₁= m (x – x₁)
7 + 2 = -5 (x – 10)
y + 2 = -5x + 50
5x + 7 + 2 – 50 = 0 ⇒ 5x + 7 – 48 = 0
Equation of the altitude from B is 5x + y – 48 = 0

Question 8.
Find the equation of the perpendicular bisector of the line joining the points A(-4,2) and B(6, -4).
Solution:
“C” is the mid point of AB also CD ⊥ AB.

Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { -4-2 }{ 6+4 } \) = \(\frac { -6 }{ 10 } \) = – \(\frac { 3 }{ 5 } \)
Slope of the ⊥^r AB is \(\frac { 5 }{ 3 } \)
Mid point of AB = (\(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\))
= (\(\frac { -4+6 }{ 2 } \),\(\frac { 2-4 }{ 2 } \)) = (\(\frac { 2 }{ 2 } \),\(\frac { -2 }{ 2 } \)) = (1,-1)
Equation of the perpendicular bisector of CD is
y – y₁ = m(x – x₁)
y + 1 = \(\frac { 5 }{ 3 } \) (x – 1)
5(x – 1) = 3(y + 1)
5x – 5 = 3y + 3
5x – 3y – 5 – 3 = 0
5x – 3y – 8 = 0
Equation of the perpendicular bisector is 5x – 3y – 8 = 0

Question 9.
Find the equation of a straight line through the intersection of lines 7x + 3y = 10, 5x – 4y = 1 and parallel to the line 13x + 5y + 12 = 0.
Solution:
Given lines are.

x = \(\frac { 43 }{ 43 } \) = 1
Substitute the value of x = 1 in (1)
7(1) + 3y = 10 ⇒ 3y = 10 – 7
y = \(\frac { 3 }{ 3 } \) = 1
The point of intersection is (1,1)
Equation of the line parallel to 13x + 5y + 12 = 0 is 13x + 5y + k = 0
This line passes through (1,1)
13 (1) + 5 (1) + k = 0
13 + 5 + k = 0 ⇒ 18 + k = 0
k = -18
∴ The equation of the line is 13x + 5y – 18 = 0

Question 10.
Find the equation of a straight line through the intersection of lines 5x – 6y = 2, 3x + 2y = 10 and perpendicular to the line 4x – 7y + 13 = 0.
Solution:
Given lines are.

Substitute the value of x = \(\frac { 16 }{ 7 } \) in (2)
3 × \(\frac { 16 }{ 7 } \) + 2y = 10 ⇒ 2y = 10 – \(\frac { 48 }{ 7 } \)
2y = \(\frac { 70-48 }{ 7 } \) ⇒ 2y = \(\frac { 22 }{ 7 } \)
y = \(\frac{22}{2 \times 7}\) = \(\frac { 11 }{ 7 } \)
The point of intersect is (\(\frac { 16 }{ 7 } \),\(\frac { 11 }{ 7 } \))
Equation of the line perpendicular to 4x – 7y + 13 = 0 is 7x + 4y + k = 0
This line passes through (\(\frac { 16 }{ 7 } \),\(\frac { 11 }{ 7 } \))
7 (\(\frac { 16 }{ 7 } \)) + 4 (\(\frac { 11 }{ 7 } \)) + k = 0 ⇒ 16 + \(\frac { 44 }{ 7 } \) + k = 0
\(\frac { 112+44 }{ 7 } \) + k = 0 ⇒ \(\frac { 156 }{ 7 } \) + k = 0
k = – \(\frac { 156 }{ 7 } \)
Equation of the line is 7x + 4y – \(\frac { 156 }{ 7 } \) = 0
49x + 28y – 156 = 0

Question 11.
Find the equation of a straight line joining the point of intersection of 3x + y + 2 = 0 and x – 2y -4 = 0 to the point of intersection of 7x – 3y = -12 and 2y = x + 3.
Solution:
The given lines are.


Substitute the value of x = 0 in (1)
3 (0) + y = -2
y = -2
The point of intersection is (0, -2).
The given equation is

Substitute the value of y = \(\frac { 9 }{ 11 } \) in (6)
– x + 2 (\(\frac { 9 }{ 11 } \)) = 3 ⇒ -x + \(\frac { 18 }{ 11 } \) = 3
-x = 3 – \(\frac { 18 }{ 11 } \) = \(\frac { 33-18 }{ 11 } \) = \(\frac { 15 }{ 11 } \)
x = – \(\frac { 15 }{ 11 } \)
The point of intersection is (-\(\frac { 15 }{ 11 } \),\(\frac { 9 }{ 11 } \))
Equation of the line joining the points (0, -2) and (-\(\frac { 15 }{ 11 } \),\(\frac { 9 }{ 11 } \)) is


31 × (- 11x) = 11 × 15 (y + 2) = 165 (y + 2)
– 341 x = 165 y + 330
– 341 x – 165 y – 330 = 0
341 x + 165 y + 330 = 0
(÷ by 11) ⇒ 31 x + 15 y + 30 = 0
The required equation is 31 x + 15 y + 30 = 0

Question 12.
Find the equation of a straight line through the point of intersection of the lines 8JC + 3j> = 18, 4JC + 5y = 9 and bisecting the line segment joining the points (5, -4) and (-7,6).
Solution:
Given lines are.
8x + 3y = 18 …..(1)
4x + 5y = 9 …..(2)

x = \(\frac { 63 }{ 28 } \) = \(\frac { 9 }{ 4 } \)
Substitute the value of x = \(\frac { 9 }{ 4 } \) in (2)
4 (\(\frac { 9 }{ 4 } \)) + 5y = 9
9 + 5y = 9 ⇒ 5y = 9 – 9
5y = 0 ⇒ y = 0
The point of intersection is (\(\frac { 9 }{ 4 } \),0)
Mid point of the points (5, -4) and (-7, 6)

Equation of the line joining the points (\(\frac { 9 }{ 4 } \),0) and (-1,1)

-13y = 4x – 9
-4x – 13y + 9 = 0 ⇒ 4x + 13y – 9 = 0
The equation of the line is 4x + 13y – 9 = 0

Students can download Maths Chapter 5 Coordinate Geometry Unit Exercise 5 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 5 Coordinate Geometry Unit Exercise 5

Question 1.
PQRS is a rectangle formed by joining the points P(- 1, – 1), Q(- 1, 4) , R(5, 4) and S (5, – 1). A, B, C and D are the mid-points of PQ, QR, RS and SP respectively. Is the quadrilateral ABCD a square, a rectangle or a rhombus? Justify your answer.
Answer:

Mid point of a line = (\(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\))
Mid point of PQ (A) = (\(\frac { -1-1 }{ 2 } \),\(\frac { -1+4 }{ 2 } \))
= (\(\frac { -2 }{ 2 } \),\(\frac { 3 }{ 2 } \)) = (-1,\(\frac { 3 }{ 2 } \))
Mid point of QR (B) = (\(\frac { -1+5 }{ 2 } \),\(\frac { 4+4 }{ 2 } \)) = (\(\frac { 4 }{ 2 } \),\(\frac { 8 }{ 2 } \)) = (2,4)
Mid point of RS (C) = (\(\frac { 5+5 }{ 2 } \),\(\frac { 4-1 }{ 2 } \)) = (\(\frac { 10 }{ 2 } \),\(\frac { 3 }{ 2 } \)) = (5,\(\frac { 3 }{ 2 } \))
Mid point of PS (D) = (\(\frac { 5-1 }{ 2 } \),\(\frac { -1-1 }{ 2 } \)) = (\(\frac { 4 }{ 2 } \),\(\frac { -2 }{ 2 } \)) = (2,-1)


img 355
AB = BC = CD = AD = \(\sqrt{\frac{61}{4}}\)
Since all the four sides are equal,
∴ ABCD is a rhombus.

Question 2.
The area of a triangle is 5 sq. units. Two of its vertices are (2,1) and (3, -2). The third vertex is (x, y) where y = x + 3 . Find the coordinates of the third vertex.
Answer:
Let the vertices A(2,1), B(3, – 2) and C(x, y)
Area of a triangle = 5 sq. unit

\(\frac { 1 }{ 2 } \) [x₁y₂ + x₂y₃ + x₃y₁ – (x₂y₁ + x₃y₂ + x₁y₃)] = 5
\(\frac { 1 }{ 2 } \) [-4 + 3y + x – (3 – 2x + 2y)] = 5
-4 + 3y + x – 3 + 2x – 2y = 10
3x + y – 7 = 10
3x + y = 17 ……(1)
Given y = x + 3
Substitute the value ofy = x + 3 in (1)
3x + x + 3 = 17
4x = 17 – 3
4x = 14
x = \(\frac { 14 }{ 4 } \) = \(\frac { 7 }{ 2 } \)
Substitute the value of x in y = x + 3
y = \(\frac { 7 }{ 2 } \) + 3 ⇒ y = \(\frac { 7+6 }{ 2 } \) = \(\frac { 13 }{ 2 } \)
∴ The coordinates of the third vertex is (\(\frac { 7 }{ 2 } \),\(\frac { 13 }{ 2 } \))

Question 3.
Find the area of a triangle formed by the lines 3x + y – 2 = 0, 5x + 2y – 3 = 0 and 2x – y – 3 = 0
Answer:
3x + y = 2 ……..(1)
5x + 2y = 3 ………(2)
2x – y = 3 ……….(3)
Solve (1) and (2) to get the vertices B


Substitute the value of x = 1 in (1)
3(1) + y = 2
y = 2 – 3 = – 1
The point B is (1, – 1)
Solve (2) and (3) to get the vertices C

Substitute the value of x = 1 in (3)
2(1) – y = 3 ⇒ -y = 3 – 2
– y = 1 ⇒ y = – 1
The point C is (1, – 1)
Solve (1) and (3) to get the vertices A

Substitute the value of x = 1 in (1)
3(1) + y = 2
y = 2 – 3 = -1
The point A is (1, – 1)
The points A (1, – 1), B (1, -1), C(1, -1)
Area of ∆ABC = \(\frac { 1 }{ 2 } \) [x₁y₂ + x₂y₃ + x₃y₁ – (x₂y₁ + x₃y₂ + x₁y₃)]

Area of the triangle = 0 sq. units.
Note: All the three vertices are equal, all the point lies in a same points.

Question 4.
If vertices of a quadrilateral are at A(- 5, 7), B(- 4, k), C(- 1, – 6) and D(4, 5) and its area is 72 sq.units. Find the value of k.
Answer:

Area of the quadrilateral ABCD = 72 sq. units.
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)] = 72

-5k + 24 – 5 + 28 – (- 28 – K – 24 – 25) = 144
– 5k + 47 – k – 77 = 144
– 5k + 47 + k + 77 = 144
– 4k + 124 = 144
-4k = 144 – 124
– 4k = 20
k = -5
The value of k = – 5

Question 5.
Without using distance formula, show that the points (-2,-1), (4,0), (3,3) and (-3,2) are vertices of a parallelogram.
Answer:
The vertices A(-2, -1), B(4, 0), C(3, 3) and D(- 3, 2)

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { 0+1 }{ 4+2 } \) = \(\frac { 1 }{ 6 } \)
Slope of BC = \(\frac { 3-0 }{ 3-4 } \) = \(\frac { 3 }{ -1 } \) = -3
Slope of CD = \(\frac { 2-3 }{ -3-3 } \) = \(\frac { -1 }{ -6 } \) = \(\frac { 1 }{ 6 } \)
Slope of AD = \(\frac { 2+1 }{ -3+2 } \) = \(\frac { 3 }{ -1 } \) = -3
Slope of AB = Slope of CD = \(\frac { 1 }{ 6 } \)
∴ AB || CD ……(1)
Slope of BC = Slope of AD = -3
∴ BC || AD …..(2)
From (1) and (2) we get ABCD is a parallelogram.

Question 6.
Find the equations of the lines, whose sum and product of intercepts are 1 and – 6 respectively.
Answer:
Let the “x” intercept be “a”
y intercept = 1 – a (sum of the intercept is 1)
Product of the intercept = – 6
a (1 – a) = – 6 ⇒ a – a² = – 6
– a² + a + 6 = 0 ⇒ a² – a – 6 = 0
(a – 3) (a + 2) = 0 ⇒ a – 3 = 0 (or) a + 2 = 0
a = 3 (or) a = -2
When a = 3
x – intercept = 3
y – intercept = 1 – 3 = – 2
Equation of a line is
\(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1
\(\frac { x }{ 3 } \) + \(\frac { y }{ -2 } \) = 1
\(\frac { x }{ 3 } \) – \(\frac { y }{ 2 } \) = 1
2x – 3y = 6
2x – 3y – 6 = 0

When a =-2
x – intercept = -2
y – intercept = 1 – (- 2) = 1 + 2 = 3
Equation of a line is
\(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1
\(\frac { x }{ -2 } \) + \(\frac { y }{ 3 } \) = 1
– \(\frac { x }{ 2 } \) + \(\frac { y }{ 3 } \) = 1
– 3x + 2y = 6
3x – 2y + 6 = 0

Question 7.
The owner of a milk store finds that, he can sell 980 litres of milk each week at ₹ 14/litre and 1220 litres of milk each week at ₹ 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at ₹ 17/litre?
Answer:
Let the selling price of a milk be “x”
Let the demand be “y”
We have to find the linear equation connecting them
Two points on the line are (14, 980) and (16,1220)
Slope of the line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { 1220-980 }{ 16-14 } \) = \(\frac { 240 }{ 2 } \) = 120
Equation of the line is y – y₁ = m (x – x₁)
y – 980 = 120 (x – 14) ⇒ y – 980 = 120 x – 1680
-120 x + y = -1680 + 980 ⇒ -120 x + y = -700 ⇒ 120 x – y = 700
Given the value of x = 17
120(17) – y = 700
-y = 700 – 2040 ⇒ – y = – 1340
y = 1340
The demand is 1340 liters

Question 8.
Find the image of the point (3,8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.
Answer:
Let the image of P(3, 8) and P’ (a, b)
Let the point of intersection be O

Slope of x + 3y = 7 is – \(\frac { 1 }{ 3 } \)
Slope of PP’ = 3 (perpendicular)
Equation of PP’ is
y – y₁ = m(x – x₁)
y – 8 = 3 (x – 3)
y – 8 = 3x – 9
-8 + 9 = 3x – y
∴ 3x – y = 1 ………(1)
The two line meet at 0

Substitute the value of x = 1 in (1)
3 – y = 1
3 – 1 = y
2 = y
The point O is (1,2)
Mid point of pp’ = (\(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\))
(1,2) = (\(\frac { 3+a }{ 2 } \),\(\frac { 8+b }{ 2 } \))
∴ \(\frac { 3+a }{ 2 } \) = 1 ⇒ 3 + a = 2
a = 2 – 3 = -1
\(\frac { 8+b }{ 2 } \) = 2
8 + b = 4
b = 4 – 8 = – 4
The point P’ is (-1, -4)

Question 9.
Find the equation of a line passing through the point of intersection of the lines 4x + 7y – 3 = O and 2x – 3y + 1 = 0 that has equal intercepts on the axes.
Answer:
Given lines

Substitute the value of y = \(\frac { 5 }{ 13 } \) in (2)
2x – 3 × \(\frac { 5 }{ 13 } \) = -1
2x – \(\frac { 15 }{ 13 } \) = -1
26x – 15 = -13
26x = -13 + 15
26x = 2
x = \(\frac { 2 }{ 26 } \) = \(\frac { 1 }{ 13 } \)
The point of intersection is (\(\frac { 1 }{ 13 } \),\(\frac { 5 }{ 13 } \))

Let the x – intercept and y intercept be “a”
Equation of a line is
\(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1
\(\frac { x }{ a } \) + \(\frac { y }{ a } \) = 1 (equal intercepts)
It passes through (\(\frac { 1 }{ 13 } \),\(\frac { 5 }{ 13 } \))
\(\frac { 1 }{ 13a } \) + \(\frac { 5 }{ 13a } \) = 1
\(\frac { 1+5 }{ 13a } \) = 1
13a = 6
a = \(\frac { 6 }{ 13 } \)
The equation of the line is

Question 10.
A person standing at a junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 seek to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find the equation of the path that he should follow.
Answer:
Two straight path will intersect at one point.
Solving this equations

2x – 3y + 4 = 0

Substitute the value of x = \(\frac { -1 }{ 17 } \) in (2)

The point of intersection is (-\(\frac { 1 }{ 17 } \),\(\frac { 22 }{ 17 } \))
Any equation perpendicular to 6x – 7y + 8 = 0 is 7x + 6y + k = 0
It passes through (-\(\frac { 1 }{ 17 } \),\(\frac { 22 }{ 17 } \))
7(-\(\frac { 1 }{ 17 } \)) + 6 (\(\frac { 22 }{ 17 } \)) + k = 0
Multiply by 17
-7 + 6 (22) + 17k = 0
-7 + 132 + 17k = 0
17k = -125 ⇒ k = – \(\frac { 125 }{ 17 } \)
The equation of a line is 7x + 6y – \(\frac { 125 }{ 17 } \) = 0
119x + 102y – 125 = 0
∴ Equation of the path is 119x + 102y – 125 = 0

Students can download Maths Chapter 5 Coordinate Geometry Additional 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 5 Coordinate Geometry Additional Questions

I. Multiple Choice Questions

Question 1.
If the three points (-3, 7), (a, 1), (-3, 2) are collinear then the value of “a” is
(1) 0
(2) -1
(3) -3
(4) 1
Answer:
(3) -3
Hint:
Since the three points are collinear
Area of a ∆ = 0

-3 + 2a – 21 – (7a – 3 – 6) = 0 ⇒ 2a – 24 – 7a + 9 = 0
– 5a – 15 = 0 ⇒ – 5(a + 3) = 0
a + 3 = 0 ⇒ a = -3

Question 2.
If A (5, 5), B (-5, 1), C (10, 7) lie in a straight line, then the area of ∆ ABC is …………….
(1) \(\frac { 13 }{ 2 } \) sq.units
(2) 9 sq.units
(3) 25 sq.units
(4) 0
Answer:
(4) 0
Hint:
Area of the ∆^le

Question 3.
In a rectangle ABCD, area of ∆ ABC is \(\frac { 31 }{ 2 } \) sq. units. Then the area of rectangle is ……………
(1) 62 sq. units
(2) 31 sq. units
(3) 60 sq. units
(4) 30 sq. units
Answer:
(2) 31 sq. units
Hint:
In a rectangle area of ∆ ABC and area of ∆ ACD are equal.
Area of rectangle ABCD = 2 × \(\frac { 31 }{ 2 } \) = 31 sq.units

Question 4.
If the points (k, 2k), (3k, 3k) and (3,1) are collinear, then k is ……………..
(1) \(\frac { 1 }{ 3 } \)
(2) – \(\frac { 1 }{ 3 } \)
(3) \(\frac { 2 }{ 3 } \)
(4) – \(\frac { 2 }{ 3 } \)
Answer:
(2) – \(\frac { 1 }{ 3 } \)
Hint:
Since the three points are collinear. Area of a ∆ = 0

3k² + 3k + 6k – (6k² + 9k + k) = 0 ⇒ 3k² + 9k – 6k² – 10k = 0
-3 k² – k = 0 ⇒ -k(3k + 1) = 0
3k + 1 = 0 ⇒ 3 k = -1 ⇒ k = – \(\frac { 1 }{ 3 } \)

Question 5.
If the area of the triangle formed by the points (x, 2x), (-2, 6) and (3, 1) is 5 square units then x = ………….
(1) 2
(2) \(\frac { 3 }{ 5 } \)
(3) 3
(4) 5
Answer:
(1) 2
Hint:
Area of the triangle = 5 sq. units

6x – 2 + 6x – (-4x + 18 + x) = 10 ⇒ 12x – 2 – (-3x + 18) = 10
12x – 2 + 3x – 18 = 10
15x – 20 = 10 ⇒ 15x = 10 + 20 = 30
x = \(\frac { 30 }{ 15 } \) = 2

Question 6.
The slope of a line parallel to y-axis is equal to …………..
(1) 0
(2) -1
(3) 1
(4) not defined
Answer:
(4) not defined

Question 7.
In a rectangle PQRS, the slope of PQ = \(\frac { 5 }{ 6 } \) then the slope of RS is ………..
(1) \(\frac { -5 }{ 6 } \)
(2) \(\frac { 6 }{ 5 } \)
(3) \(\frac { -6 }{ 5 } \)
(4) \(\frac { 5 }{ 6 } \)
Answer:
\(\frac { 5 }{ 6 } \)
Hint:
In a rectangle opposite sides are parallel.
∴ Slope of the line RS is \(\frac { 5 }{ 6 } \).

Question 8.
The y – intercept of the line y = 2x is ………
(1) 1
(2) 2
(3) \(\frac { 1 }{ 2 } \)
(4) 0
Answer:
(4) 0

Question 9.
The straight line given by the equation y = 5 is …………..
(1) Parallel to x – axis
(2) Parallel to y – axis
(3) Passes through the origin
(4) None of these
Answer:
(1) Parallel to x – axis

Question 10.
The x – intercept of the line 2x – 3y + 5 = 0 is ………….
(1) \(\frac { 5 }{ 2 } \)
(2) \(\frac { -5 }{ 2 } \)
(3) \(\frac { 2 }{ 5 } \)
(4) \(\frac { -2 }{ 5 } \)
Answer:
(2) \(\frac { -5 }{ 2 } \)
Hint:
2x – 3y + 5 = 0 ⇒ 2x – 3y = – 5

Question 11.
The lines 3x – 5y + 1 = 0 and 5x + ky + 2 = 0 are perpendicular if the value of k is ………..
(1) -5
(2) 3
(3) -3
(4) 5
Answer:
(2) 3
Hint:
Slope of the first line (m₁) = \(\frac { -3 }{ -5 } \) = \(\frac { 3 }{ 5 } \)
Slope of the second line (m₂) = \(\frac { -5 }{ k } \)
Since the two lines are perpendicular.
m₁ × m₂ = -1
\(\frac { 3 }{ 5 } \) × \(\frac { -5 }{ k } \) = -1 ⇒ \(\frac { -3 }{ k } \) = -1
-k = -3 ⇒ The value of k = 3

Question 12.
If x – y = 3 and x + 2y = 6 are the diameters of a circle then the centre is at the point ………..
(1) (0, 0)
(2) (1, 2)
(3) (1, -1)
(4) (4, 1)
Answer:
(4) (4, 1)
Hint:
Centre of the circle is the intersection of the two diameters.

Centre of the circle is (4, 1)

Question 13.
The line 4x + 3y – 12 = 0 meets the x-axis at the point ……….
(1) (4, 0)
(2) (3, 0)
(3) (-3, 0)
Answer:
(2) (3,0)
Hint:
4x + 3y – 12 = 0 meet the x-axis the value of y = 0
4x- 12 = 0 ⇒ 4x = 12
x = \(\frac { 12 }{ 4 } \) = 3 ⇒ The point is (3, 0)

Question 14.
The equation of a straight line passing through the point (2, -7) and parallel to x-axis is ……………….
(1) x = 2
(2) x = -7
(3) y = -7
(4) y = 2
Answer:
(3) y = -7
Hint:
Equation of a line parallel to x-axis is y = -7

Question 15.
The equation of a straight line having slope 3 and y intercept – 4 is ………………
(1) 3x – y – 4 = 0
(2) 3x + y – 4 = 0
(3) 3x – y + 4 = 0
(4) 3x – y + 4 = 0
Answer:
(1) 3x – y – 4 = 0
Hint. The equation of a line is y = mx + c
y = 3 (x) + (-4) ⇒ y = 3x – 4
3x – y – 4 = 0

II. Answer the following questions:

Question 1.
If the points (3, – 4) (1, 6) and (- 2, 3) are the vertices of a triangle, find its area.
Answer:
Let the vertices A (3, – 4), B (1, 6) and C (- 2, 3)

Area of ∆ ABC = \(\frac { 1 }{ 2 } \) [x₁y₂ + x₂y₃ + x₃y₁, – (x₂y₁ + x₃y₂ + x₁y₃)]

Area of a ∆ = 18 sq. units

Question 2.
If the area of the triangle formed by the points (1,2) (2,3) and (a, 4) is 8 sq. units, find a.
Answer:
Area of a triangle = 8 sq. units.

\(\frac { 1 }{ 2 } \) [x₁y₂ + x₂y₃ + x₃y₁ – (x₂y₁ + x₃y₂ + x₁y₃)] = 8.
\(\frac { 1 }{ 2 } \) [3 + 8 + 2a – (4 + 3a + 4)] = 8
11 + 2a – 8 – 3a= 16 ⇒ – a + 3 = 16
– a = 16 – 3 ⇒ a = -13
The value of a = -13

Question 3.
If the points A (2, 5), B (4, 6) and C (8, a) are collinear find the value of “a” using slope concept.
Answer:
Since the three points are collineal
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = Slope of BC
\(\frac { 6-5 }{ 4-2 } \) = \(\frac { a-6 }{ 8-4 } \) ⇒ \(\frac { 1 }{ 2 } \) = \(\frac { a-6 }{ 4 } \) ⇒ 2a – 12 = 4 ⇒ 2a = 16
a = \(\frac { 16 }{ 2 } \) = 8 ⇒ The value of a = 8

Question 4.
If the points (x,y) is collinear with the points (a, 0) and (0, b) then prove that \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1
Answer:
Let A (x, y), B (a, 0), C(0, b)
Since the three points are collinear
Slope of AB = Slope of BC
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
\(\frac { 0-y }{ a-x } \) = \(\frac { b-0 }{ 0-a } \)
\(\frac { -y }{ a-x } \) = \(\frac { b }{ -a } \)
ay = b (a – x)
ay = ba – bx
ay + bx = ab
Divided by ab
\(\frac { ay }{ ab } \) + \(\frac { bx }{ ab } \) = \(\frac { ab }{ ab } \)
\(\frac { y }{ b } \) + \(\frac { x }{ a } \) = 1 ⇒ \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1

Question 5.
A straight line passes through (1, 2) and has the equation y – 2x – k = 0. Find k.
Answer:
The given line is y – 2x – k = 0
It passes through (1,2)
(2) -2 (1) -k = 0 ⇒ 2 – 2 – k = 0
0 – k = 0 ⇒ k = 0
The value of k = 0

Question 6.
If a line passes through the mid point of AB where A is (3, 0) and B is (5, 4) and makes an angle 60° with x – axis find its equation.
Answer:

Slope of a line = tan 60°
= \(\sqrt { 3 }\)
Equation of a line is y – y₁ = m (x – x₁)
y – 2 = \(\sqrt { 3 }\) (x – 4)
y – 2 = \(\sqrt { 3 }\) x – 4 \(\sqrt { 3 }\)
\(\sqrt { 3x }\) – y + 2 – 4\(\sqrt { 3 }\) = 0

Question 7.
Find the equation of the line through (3, 2) and perpendicular to the line joining (4, 5) and (1,2)
Answer:
Slope of a line = \(\frac { 2-5 }{ 1-4 } \) ⇒ \(\frac { -3 }{ -3 } \) = 1
Slope of the line perpendicular to it is – 1
Equation of the line joining -1 and (3, 2) is
y – y₁ = m (x – x₁) ⇒ y – 2 = -1(x – 3)
y – 2 = -x + 3 ⇒ x + y – 5 = 0

Question 8.
P and Q trisect the line segment joining the points (2, 1) and (5, – 8). If the point P lies on 2x – y + k = 0, then find the value of k.
Answer:

A line divides internally in the ratio 1 : 2
A line divide internally in the ratio l : m

The point P = (\(\frac { 5+4 }{ 3 } \),\(\frac { -8+2 }{ 3 } \))
= (\(\frac { 9 }{ 3 } \),\(\frac { -6 }{ 3 } \)) = (3, -2)
The given line 2x – y + k = 0 passes through the point (3,-2)
2 (3) – (- 2) + k = 0
6 + 2 + k = 0
8 + k = 0
k = – 8
The value of k = – 8

Question 9.
The line 4x + 3y – 12 = 0 intersect the X, Y – axis at A and B respectively. Fine the area of ∆AOB.
Answer:
The equation of the line AB is 4x + 3y – 12 = 0

4x + 3y = 12
\(\frac { 4x }{ 12 } \) + \(\frac { 3y }{ 12 } \) = 1 ⇒ \(\frac { x }{ 3 } \) + \(\frac { y }{ 4 } \) = 1
The point A is (3, 0) (it intersect the X – axis)
and B is (0, 4) (it intersect the Y – axis)
Area of ∆ AOB = \(\frac { 1 }{ 2 } \) [x₁y₂ + x₂y₃ + x₃y₁ (x₂y₁ + x₃y₂ + x₁y₃)]

Question 10.
Find the equation of the line passing through (4, 5) and making equal intercept in the axes.
Answer:
Let the equal intercept on the axes be a, a.
Equation of the line is \(\frac { x }{ a } \) + \(\frac { y }{ a } \) = 1 (Given equal intercepts)
The line passes through (4, 5)
\(\frac { 4 }{ a } \) + \(\frac { 5 }{ a } \) = 1 ⇒ \(\frac { 9 }{ a } \) = 1 ⇒ a = 9
The equation of the line is \(\frac { x }{ 9 } \) + \(\frac { y }{ 9 } \) = 1
Multiply by 9
x + y – 9 = 0

Question 11.
Find the equation of the line passing through (2, – 1) and whose intercepts on the axes are equal in magnitude but opposite in sign.
Answer:
Let the x – intercept be “a” and y intercept be = “-a”
The equation of the line is
\(\frac { x }{ a } \) + \(\frac { y }{ -a } \) = 1 (y – intercept is – a)
\(\frac { x }{ a } \) – \(\frac { y }{ a } \) = 1
It passes through (2, -1)
\(\frac { 2 }{ a } \) – \(\frac { (-1) }{ a } \) = 1
\(\frac { 2 }{ a } \) + \(\frac { 1 }{ a } \) = 1 ⇒ \(\frac { 3 }{ a } \) = 1
a = 3
The equation of the line is
\(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1
\(\frac { x }{ 3 } \) + \(\frac { y }{ -3 } \) = 1 ⇒ \(\frac { x }{ 3 } \) – \(\frac { y }{ 3 } \) = 1
x – y = 3
The equation is x – y – 3 = 0

Question 12.
The straight line cuts the coordinate axes at A and B. If the mid point of AB is (3,2) then find the equation of AB.
Answer:
Let the point A be (a, 0) and B be (0, b)


The point A (6, 0) and B (0, 4)
Equation of the line AB is

III. Answer the following questions

Question 1.
If the coordinates of two points A and B are (3, 4) and (5, – 2) respectively. Find the ‘ coordinates of any point “c”, if AC = BC and Area of triangle ABC = 10 sq. units.
Answer:
Let the coordinates C be (a, 6) then AC = BC
AC² = BC²
(a – 3)² + (b – 4)² = (a – 5)² + (b + 2)²
a² + 9 – 6a + b² + 16 – 8b = a² + 25 – 10a + b² + 4 – 4b
a² + b² + 25 – 6a – 86 = a² + b² + 29 – 10a + 4b

25 – 6a – 8b = 29 – 10a + 46
4a – 12b = 4 ⇒ a – 3b = 1 ………… (1)
Area of ∆ ABC = 10 sq. units
\(\frac { 1 }{ 2 } \) [x₁y₂ + x₂y₃ + x₃y₁ – (x₂y₁ + x₃y₂ + x₁y₃)] = 10

-6 + 5b + 4a – (20 – 2a + 3b) = 20
-6 + 5b + 4a – 20 + 2a – 3b = 20
6a + 2b – 26 = 20 ⇒ 6a + 2b = 46

Substitute the value of a = 7 in (2)
3 (7) + b = 23 ⇒ b = 23 – 21 = 2
The coordinate C is (7, 2)

Question 2.
The four vertices of a Quadrilateral are (1,2) (- 5,6) (7, – 4) and (k, – 2) taken in order. If the area of the Quadrilateral is 9 sq. units, find the value of k.
Answer:
Let A (1, 2) B (- 5, 6) C (7, – 4) and D (k, – 2)
Area of the
Quadrilateral ABCD = \(\frac { 1 }{ 2 } \)[(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)]

Area of the Quadrilateral ABCD = 3k – 9
Given area of a Quadrilateral is 9 sq. units.
3k – 9 = 9 ⇒ 3k = 18 ⇒ k = \(\frac { 18 }{ 3 } \) = 6
The value of k = 6

Question 3.
Find the area of a triangles whose three sides are having the equations x + y = 2, x – y = 0 and x + 2y – 6 = 0.
Answer:
Find the three vertices of the triangles by solving their equation.
To find vertices A


Substitute the value of y = 4 in (1)
x + 4 = 2 ⇒ x = 2 – 4 = -2
The vertices A is (- 2, 4)
To find vertices B

Substitute the value of x = 1 in (1)
1 + y = 2 ⇒ y = 2 – 1 = 1
The vertices B is (1, 1)
To find vertices C

y = \(\frac { 6 }{ 3 } \) = 2
Substitute the value y = 2 in (3)
x – 2 = 0 ⇒ x = 2
The vertices C is (2, 2)

Area of the ∆ BC = 3 sq. units

Question 4.
Verify the Median of a triangle divides into two triangles of equal areas whose vertices are A (4, – 6), B (3, – 2) and C (5, 2)
Answer:

Let D be the mid point of AC .
Mid point of AC = (\(\frac { 5+4 }{ 2 } \),\(\frac { 2-6 }{ 2 } \)) = (\(\frac { 9 }{ 2 } \),-2)
Area of the triangle = \(\frac { 1 }{ 2 } \) [x₁y₂ + x₂y₃ + x₃y₁ – (x₂y₁ + x₃y₂ + x₁y₃)]

Area of ∆ ADB = Area of ∆ BDC
A median divides the triangle of equal areas.

Question 5.
Find the area of the ∆ ABC with A (1, – 4) and the mid points of sides through A being (2,-1) and (0,-1)
Answer:
Let the coordinates of B and C are (a, b) and (c, d) respectively.
Sides through A are AB and AC

Mid point of AB = (\(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\))
(2, -1) = (\(\frac { 1+a }{ 2 } \),\(\frac { -4+b }{ 2 } \))
\(\frac { 1+a }{ 2 } \) = 2
1 + a = 4
a = 4 – 1
= 3
The point B is (3,2)
\(\frac { -4+b }{ 2 } \) = -1
-4 + b = -2
b = -2 + 4
= 2
Mid point of AC = (\(\frac { 1+c }{ 2 } \),\(\frac { -4+d }{ 2 } \))
(0,-1) = (\(\frac { 1+c }{ 2 } \),\(\frac { -4+d }{ 2 } \))
\(\frac { 1+c }{ 2 } \) = 0
1 + c = 0
c = 0 – 1
= – 1
The point C is (-1,2)
\(\frac { -4+d }{ 2 } \) = -1
– 4 + d = -2
d = – 2 + 4
= 2
Thus the coordinates of the vertices of ∆ ABC are A (1, – 4) B (3, 2) and C (- 1, 2)
Area of ∆ ABC = \(\frac { 1 }{ 2 } \) [x₁y₂ + x₂y₃ + x₃y₁ – (x₂y₁ + x₃y₂ + x₁y₃)]

Area of ∆ ABC = 12 sq. units

Question 6.
Find the equation of the straight lines passing through (- 3, 10) whose sum of the intercepts is 8.
Answer:
Let the “x” intercept be “a” and y intercept be “b”
Sum of the intercepts = 8
a + b = 8 ⇒ b = 8 – a
Equation of a line is \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1 ⇒ \(\frac { x }{ a } \) + \(\frac { y }{ 8-a } \) = 1
It passes through (-3,10)
\(\frac { -3 }{ a } \) + \(\frac { 10 }{ 8-a } \) = 1
\(\frac { -3(8-a)+10a }{ a(8-a) } \) = 1
-24 + 3a + 10a = 8a – a²
-24 + 13a = 8a – a²
a² + 5a – 24 = 0 ⇒ (a + 8) (a – 3) = 0
a + 8 = 0 (or) a – 3 = 0 ⇒ a = -8 (or) a = 3
The equation of a line is a
a = -8
\(\frac { x }{ -8 } \) + \(\frac { y }{ 8+8 } \) = 1
\(\frac { x }{ -8 } \) + \(\frac { y }{ 16 } \) = 1
-2x + y = 16
2x – y + 16 = 0
a = 3
\(\frac { x }{ 3 } \) + \(\frac { y }{ 5 } \) = 1
5x + 3y = 15
5x + 3y – 15 = 0
The equation of the lines are 2x – y + 16 = 0 (or) 5x + 3y – 15 = 0.

Question 7.
If (5, – 3), (- 5, 3), (6, 6) are the mid points of the sides of a triangle, find the equation of the sides.
Answer:
Since D, E, F are the mid points of ∆ ABC
EF || AB, FD || BC and DE || AC
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of EF = \(\frac { 6-3 }{ 6+5 } \) = \(\frac { 3 }{ 11 } \)
Since EF || AB; Slope of AB = \(\frac { 3 }{ 11 } \)
Equation of AB is
y – y₁ = m (x – x₁)
y + 3 = \(\frac { 3 }{ 11 } \) (x – 5)
3x – 15 = 11y + 33
3x – 11y – 15 – 33 = 0
3x – 11y – 48 = 0
Slope of DE = Slope of AC
Slope of DE = \(\frac { 3+3 }{ -5-5 } \) = \(\frac { 6 }{ -10 } \) = –\(\frac { 6 }{ 10 } \) = –\(\frac { 3 }{ 5 } \)
Slope of AC = – \(\frac { 3 }{ 5 } \)
Equation of AC is
y – y₁ = m (x – x₁)

y – 6 = – \(\frac { 3 }{ 5 } \) (x – 6) ⇒ 5y – 30 = -3x + 18
3x + 5y – 30 – 18 = 0 ⇒ 3x + 5y – 48 = 0
Slope of DF = Slope of BC
Slope of DF = \(\frac { 6+3 }{ 6-5 } \) = \(\frac { 9 }{ 1 } \) = 9
Slope of BC = 9
Equation of the line BC is
y – y₁ = m(x – x₁)
y – 3 = 9 (x + 5) ⇒ 9x + 45 = y – 3
9x – y + 45 + 3 = 0 ⇒ 9x – y + 48 = 0
Equation of the sides are
3x – 11y – 48 = 0 ; 9x – y + 48 = 0 and 3x + 5y – 48 = 0

Question 8.
Find the equation of the straight line passing through the point of intersection of the lines 5x – 8y + 23 = 0 and 7x + 6y – 71 = 0 and is perpendicular to the line joining the points (5,1) and (-2, 2)
Answer:

Substitute the value of x in (1)
5 (5) – 8y = – 23 ⇒ 25 – 8y = – 23
-8y = – 23 – 25 ⇒ -8y = – 48
y = \(\frac { 48 }{ 8 } \) = 6
The point of intersection is (5,6)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of the line joining the points (5,1) and (-2,2) = \(\frac { 2-1 }{ -2-5 } \)
= \(\frac { 1 }{ -7 } \) = – \(\frac { 1 }{ 7 } \)
Slope of the perpendicular line is = 7
Equation of a line is
y – y₁ = m(x – x₁) ⇒ y – 6 = 7 (x – 5)
y – 6 = 7x – 35 ⇒ -7x + y – 6 + 35 = 0
7x – y – 29 = 0

Question 9.
Find the equation of the line passing through the point of intersection of 4x – y – 3 = 0 and x + y – 2 = 0 and perpendicular to 2x – 5y + 3 = 0.
Answer:

x = \(\frac { 5 }{ 5 } \) = 1
Substitute the value of x = 1 in (2)
1 + y = 2
y = 2 – 1 = 1
The point of intersection is (1, 1)
Any line perpendicular to 2x – 5y + 3 = 0 is
5x + 2y + k = 0
It passes through (1,1)
5(1) + 2(1) + k = 0 ⇒ 5 + 2 + k = 0
7 + k = 0 ⇒ k = -7
The line is 5x + 2y – 7 = 0

Question 10.
Find the equation of the line through the point of intersection of the lines 2x + y – 5 = 0 and x + y – 3 = 0 and bisecting the line segment joining the points (3, – 2) and (- 5, 6).
Answer:

x = 2
Substitute the value of x = 2 in (2)
2 + y = 3
y = 3 – 2 = 1
The point of intersection is (2, 1)
Mid point of the line joining the points (3,-2) and (-5,6)

Mid point of the line
Equation of the line joining the points (2, 1) and (-1,2) is

x – 2 = -3 (y – 1)
x – 2 = -3y + 3
x + 3y – 5 = 0
The equation of the line is x + 3y – 5 = 0

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

Question 1.
Find the area of the triangle formed by the points
(i) (1,-1), (-4, 6) and (-3, -5)
(ii) (-10, -4), (-8, -1) and (-3, -5)
Solution:
(i) Let the vertices A (1, -1), B (-4, 6) and C (-3, -5)


= \(\frac { 1 }{ 2 } \) [(6 + 20 + 3) – (4 – 18 – 5)] = \(\frac { 1 }{ 2 } \) [29 – (-19)] = \(\frac { 1 }{ 2 } \) [29 + 19]
= \(\frac { 1 }{ 2 } \) × 48 = 24 sq. units.
Area of ∆ABC = 24 sq. units

(ii) Let the vertices be A(-10, -4), B(-8 -1) and C(-3, -5)

Area of ∆ABC = \(\frac { 1 }{ 2 } \)[(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]
= \(\frac { 1 }{ 2 } \) [(50 + 3 + 32) – (12 + 40 + 10)]

= \(\frac { 1 }{ 2 } \) [85 – (62)] = \(\frac { 1 }{ 2 } \) [23] = 11.5
Area of ∆ACB = 11.5 sq.units

Question 2.
Determine whether the sets of points are collinear?
(i) (-\(\frac { 1 }{ 2 } \),3)
(ii) (a,b + c), (b,c + a) and (c,a + b)
Solution:
(i) Let the points be A (-\(\frac { 1 }{ 2 } \),3), B (-5, 6) and C(-8, 8)
Area of ∆ABC = \(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₃ + x₁y₃)]
= \(\frac { 1 }{ 2 } \) [(- 3 – 40 – 24) – (-15 – 48 – 4)]

= \(\frac { 1 }{ 2 } \) [-67 + 67] = \(\frac { 1 }{ 2 } \) × 0 = 0
Area of a ∆ is 0.
∴ The three points are collinear.

(ii) Let the points be A (a, b + c), B (b, c + a) and C (c, a + b)
Area of the triangle = \(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]

Since the area of a triangle is 0.
∴ The given points are collinear.

Question 3.
Vertices of given triangles are taken in order and their areas are provided aside. In each case, find the value of ‘p’

Solution:
(i) Let the vertices be A (0,0) B (p, 8), c (6, 2)
Area of a triangle = 20 sq. units
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 20

\(\frac { 1 }{ 2 } \) [(0 + 2p + 0) – (0 + 48 + 0)] = 20
\(\frac { 1 }{ 2 } \) [2p – 48] = 20
2p – 48 = 40 ⇒ 2p = 40 + 48
p = \(\frac { 88 }{ 2 } \) = 44
The value of p = 44

(ii) Let the vertices be A (p, p), B (5, 6) and C (5, -2)
Area of a triangle = 32 sq. units
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 32

\(\frac { 1 }{ 2 } \) [6p – 10 + 5p) – (5p + 30 – 2p)] = 32
\(\frac { 1 }{ 2 } \) [11 p – 10 – 3p – 30] = 32
11p – 10 – 3p – 30 = 64
8p – 40 = 64
8p = 64 + 40 ⇒ 8p = 104
p = \(\frac { 104 }{ 8 } \) = 13
The value of p = 13

Question 4.
In each of the following, find the value of ‘a’ for which the given points are collinear.
(i) (2,3), (4, a) and (6, -3)
(ii) (a, 2 – 2a), (-a + 1, 2a) and (-4 -a, 6 – 2a).
Solution:
(i) Let the points be A (2, 3), B(4, a) and C(6, -3).
Since the given points are collinear.
Area of a triangle = 0
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 0)

\(\frac { 1 }{ 2 } \) [(2a – 12 + 18) – (12 + 6a – 6)] = 0
2a + 6 – (6 + 6a) = 0
2a + 6 – 6 – 6a = 0
-4a = 0 ⇒ a = \(\frac { 0 }{ 4 } \) = 0
The value of a = 0

(ii) Let the points be A (a, 2 – 2a), B (-a + 1, 2a) C (-4 -a, 6 – 2a).
Since the given points are collinear.
Area of a ∆ = 0

6a² – 2a – 2 – (-2a² – 6a + 2) = 0
6a² – 2a – 2 + 2a² + 6a – 2 = 0
8a² + 4a – 4 = 0 (Divided by 4)
2a² + a – 1 = 0
2a² + 2a – a – 1 = 0
2a (a + 1) – 1 (a + 1) = 0

(a + 1) (2a – 1) = 0
a + 1 = 0 (or) 2a – 1 = 0
a = -1 (or) 2a = 1 ⇒ a = \(\frac { 1 }{ 2 } \)
The value of a = -1 (or) \(\frac { 1 }{ 2 } \)

Question 5.
Find the area of the quadrilateral whose vertices are at
(i) (-9, -2), (-8, -4), (2, 2) and (1, -3)
(ii) (-9, 0), (-8,6), (-1, -2) and (-6, -3)
Solution:
(i) Let the vertices A (-9, -2), B(-8, -4), C(2, 2) and D(1, -3).
Plot the vertices in a graph.

[Note: Consider the points in counter clock wise order]

Area of the Quadrilateral ABDC = \(\frac { 1 }{ 2 } \) [36 + 24 + 2 – 4 – (16 – 4 – 6 – 18)]
= \(\frac { 1 }{ 2 } \) [58 – (-12)] – \(\frac { 1 }{ 2 } \)[58 + 12]
= \(\frac { 1 }{ 2 } \) × 70 = 35 sq. units 2
Area of the Quadrilateral = 35 sq. units

(ii) Let the vertices A(-9, 0), B(-8, 6), C(-1, -2) and D(-6, -3)
Plot the vertices in a graph and take them in counter – clock wise order.
Area of the Quadrilateral DCB

= \(\frac { 1 }{ 2 } \) [33 + 35] = \(\frac { 1 }{ 2 } \) × 68 = 34 sq. units
Area of the Quadrilateral = 34 sq. units

Question 6.
Find the value of k, if the area of a quadrilateral is 28 sq. units, whose vertices are (-4, -2), (-3, k), (3, -2) and (2, 3)
Solution:
Let the vertices A (-A, -2), B (-3, k), C (3, -2) and D (2, 3)
Area of the Quadrilateral = 28 sq. units
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)] = 28

-7k + 21 = 56
-7k = 56 – 21
-7k = 35 ⇒ 7k = – 35
k = – \(\frac { 35 }{ 7 } \) = -5
The value of k = -5

Question 7.
If the points A(-3, 9), B(a, b) and C(4, -5) are collinear and if a + b = 1, then find a and b.
Solution:
Since the three points are collinear
Area of a ∆ = 0

-3b – 5a + 36 – 9a – 4b – 15 = 0
-7b – 14a + 21 = 0
(÷ by 7) – b – 2a + 3 = 0
2a + b – 3 = 0

Substitute the value of a = 2 in (2) ⇒ 2 + b = 1
b = 1 – 2 = -1
The value of a = 2 and b = -1

Question 8.
Let P(11, 7), Q(13.5, 4) and R(9.5, 4) be the midpoints of the sides AB, BC and AC respectively of ∆ABC . Find the coordinates of the vertices A, B and C. Hence find the area of ∆ABC and compare this with area of ∆PQR.
Solution:
Let the vertices of the ∆ABC be A(x₁,y₁), B(x₂,y₂), C(x₃,y₃)

Question 9.
In the figure, the quadrilateral swimming pool shown is surrounded by concrete patio. Find the area of the patio.

Solution:

= \(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)]
= \(\frac { 1 }{ 2 } \) [(16 + 80 + 36 + 80) – (-64 – 24 – 100 – 24)]
= \(\frac { 1 }{ 2 } \) [212 – (-212)]

= \(\frac { 1 }{ 2 } \) [212 + 212] = \(\frac { 1 }{ 2 } \) [424] = 212 sq. units

= \(\frac { 1 }{ 2 } \) [90 – (-90)]

= \(\frac { 1 }{ 2 } \) [90 + 90]
= \(\frac { 1 }{ 2 } \) × 180 = 90 sq. units
Area of the patio = Area of the Quadrilateral ABCD – Area of the Quadrilateral EFGH
= (212 – 90) sq. units
Area of the patio = 122 sq. units

Question 10.
A triangular shaped glass with vertices at A(-5, -4), B(l, 6) and C(7, -4) has to be painted. If one bucket of paint covers 6 square feet, how many buckets of paint will be required to paint the whole glass, if only one coat of paint is applied.
Solution:
Given the vertices of the triangular glass is A (-5, -4), B (1, 6), and C (7, -4)


= \(\frac { 1 }{ 2 } \) [(20 + 42 – 4) – (-28 – 4 – 30)]
= \(\frac { 1 }{ 2 } \) [58 – (-62)]
= \(\frac { 1 }{ 2 } \) [58 + 62]
= \(\frac { 1 }{ 2 } \) × 120 = 60 sq. feet
Number of cans to paint 6 square feet = 1
∴ Number of cans = \(\frac { 60 }{ 6 } \) = 10 ⇒ Number of cans = 10

Question 11.
In the figure, find the area of
(i) triangle AGF
(ii) triangle FED
(iii) quadrilateral BCEG.

Solution:
Area of a triangle = \(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]
(i) Area of ∆AGF = \(\frac { 1 }{ 2 } \) [(-2.5 – 13.5 – 6) – (-13.5 – 1 – 15)]
= \(\frac { 1 }{ 2 } \) [-22 – (-29.5)]

= \(\frac { 1 }{ 2 } \) [-22 + 29.5]
= \(\frac { 1 }{ 2 } \) × 7.5 = 3.75 sq.units

(ii) Area of ∆FED = \(\frac { 1 }{ 2 } \) [(-2 + 4.5 + 3) – (4.5 + 1 – 6)]
= \(\frac { 1 }{ 2 } \) [5.5 – (-0.5)]

= \(\frac { 1 }{ 2 } \) [5.5 + 0.5] = \(\frac { 1 }{ 2 } \) × 6 = 3 sq.units

(iii)

= \(\frac { 1 }{ 2 } \) [(4 + 2 + 0.75 + 9) – (-4 -1.5 – 4.5 -2)]
= \(\frac { 1 }{ 2 } \) [15.75 + 12]
= \(\frac { 1 }{ 2 } \) [27.75] = 13.875
= 13.88 sq. units

Students can download Maths Chapter 1 Relations and Functions Ex 1.4 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 1 Relations and Functions Ex 1.4

Question 1.
Determine whether the graph given below represent functions. Give reason for your answers concerning each graph.




Answer:

The vertical line cuts the graph at A and B. The given graph does not represent a function.

The vertical line cuts the graph at most one point P. The given graph represent a function.

The vertical line cuts the graph at three points S,T and U. The given graph does not represent a function.

The vertical line cuts the graph at most one point D. The given graph represents a function.

Question 2.
Let f: A → B be a function defined by
f(x) = \(\frac { x }{ 2 } \) – 1, where A = {2, 4,6,10,12},
B = {0,1,2,4,5,9}. Represent f by
(i) set of ordered pairs
(ii) a table
(iii) an arrow diagram
(iv) a graph
Answer:
A = {2,4,6, 10, 12}
B = {0,1, 2, 4, 5, 9}
f(x) = \(\frac { x }{ 2 } \) – 1
f(2) = \(\frac { 2 }{ 2 } \) – 1 = 1 – 1 = 0
f(4) = \(\frac { 4 }{ 2 } \) – 1 = 2 – 1 = 1
f(6) = \(\frac { 6 }{ 2 } \) – 1 = 3 – 1 = 2
f(10) = \(\frac { 10 }{ 2 } \) – 1 = 5 – 1 = 4
f(12) = \(\frac { 12 }{ 2 } \) – 1 = 6 – 1 = 5

(i) Set of ordered pairs
f = {(2, 0) (4, 1) (6, 2) (10, 4) (12, 5}

(ii) Table

X	2	4	6	10	12
f(x)	0	1	2	4	5

(iii) Arrow diagram

(iv) Graph

Question 3.
Represent the function f = {(1,2), (2,2), (3,2), (4,3),(5,4)} through (i) an arrow diagram (it) a table form (iii) a graph.
Answer:
f = {(1, 2) (2, 2) (3, 2) (4, 3) (5,4)}
Let A = {1,2, 3, 4, 5}
B = {2, 3, 4}

(i) Arrow diagram

(ii) Table form

X	1	2	3	4	5
f(x)	2	2	2	3	4

(iii) Graph

Question 4.
Show that the function f : N → N defined by f(x) = 2x – 1 is one-one but not onto.
Answer:
f: N → N
N = {1,2,3,4,5,… }
f(x) = 2x – 1
f(1) = 2(1) – 1 = 2 – 1 = 1
f(2) = 2(2) – 1 = 4 – 1 = 3
f(3) = 2(3) – 1 = 6 – 1 = 5
f(4) = 2(4) – 1 = 8 – 1 = 7
f(5) = 2(5) – 1 = 10 – 1 = 9
f = {(1,1) (2, 3) (3, 5) (4, 7) (5,9) …..}

(i) Different elements has different images. This function is one to one function.
(ii) Here Range is not equal to co-domain. This function not an onto function.
∴ The given function is one-one but not an onto.

Question 5.
Show that the function f: N ⇒ N defined by f(m) = m² + m + 3 is one-one function.
Answer:
N = {1,2,3, 4,5, ….. }
f(m) = m² + m + 3
f(1) = 1² + 1 + 3 = 5
f(2) = 2² + 2 + 3 = 9
f(3) = 3² + 3 + 3 = 15
f(4) = 4² + 4 + 3 = 23
f = {(1,5) (2, 9) (3, 15) (4, 23)}

From the diagram we can understand different elements in (N) in the domain, there are different images in (N) co-domain.
∴ The function is a one-one function.

Question 6.
Let A = {1, 2, 3, 4) and B = N. Letf: A → B be
defined by f(x) = x³ then,
(i) find the range off
(ii) identify the tpe of function
Solution:
A = {1, 2, 3, 4}
B = N
f: A → B,f(x) = x³
(i) f(1) = 1³ = 1
f(2) = 2³ = 8
f(3) = 3³ = 27
f(4) = 4³ = 64
(ii) Therange of f = {1, 8, 27, 64 )
(iii) It is one-one and into function.

Question 7.
In each of the following cases state whether the function is bijective or not. Justify your answer.
(i) f: R → R defined by f (x) = 2x + 1
(ii) f: R → R defined by f(x) = 3 – 4x²
Answer:
(i) f(x) = 2x + 1
f(0) = 2(0) + 1 = 0 + 1 = 1
f(1) = 2(1) + 1 = 2 + 1 = 3
f(2) = 2(2) + 1 = 4 + 1 = 5
f(3) = 2(3) + 1 = 6 + 1 = 7
Different elements has different images
∴ It is an one-one function.
It is also an onto function. The function is one-one and onto function.
∴ It is a bijective function.

(ii) f(x) = 3 – 4x²
f(1) = 3 – 4(1)²
= 3 – 4 = -1
f(2) = 3 – 4(2)² = 3 – 16 = – 13
f(3) = 3 – 4(3)² = 3 – 36 = – 33
f(4) = 3 – 4(4²) = 3 – 64 = – 61
It is not a bijective function. The positive numbers “R” do not have negative pre – image in X in R.

Question 8.
Let A= {-1,1}and B = {0,2}.
If the function f: A → B defined by
f(x) = ax + b is an onto function? Find a and b.
Answer:
A = {-1, 1}; B = {0,2}
f(x) = ax + b
f(-1) = a(-1) + b
0 = -a + b
a – b = 0 ….(1)
f(1) = a(1) + b
2 = a + b
a + b = 2 ….(2)
Solving (1) and (2) we get

Substitute a = 1 in (1)
The value of a = 1 and b = 1

Question 9.
If the function f is defined by

find the value of
(i) f(3)
(ii) f(0)
(iii) f(1. 5)
(iv) f(2) + f(-2)
Answer:
f(x) = x + 2 when x = {2,3,4,……}
f(x) = 2
f(x) = x – 1 when x = {-2}
(i) f(x) = x + 2
f(3) = 3 + 2 = 5

(ii) f(x) = 2
f(0) = 2

(iii) f(x) = x – 1
f(-1.5) = -1.5 – 1 = -2.5

(iv) f(x) = x + 2
f(2) = 2 + 2 = 4
f(x) = x – 1
f(-2) = – 2 – 1 = – 3
f(2) + f(-2) = 4 – 3
= 1

Question 10.
A function f: [-5, 9] → R is defined as follows:

Answer:
f(x) = 6x + 1 ; x = {-5,-4,-3,-2,-1,0,1}
f(x) = 5x² – 1 ; x = {2, 3, 4, 5}
f(x) = 3x – 4 ; x = {6, 7, 8, 9}

(i) f(-3) + f(2)
f(x) = 6x + 1
f(-3) = 6(-3) + 1 = -18 + 1 = -17
f(x) = 5x² – 1
f(2) = 5(2)² – 1 = 20 – 1 = + 19
f(-3) + f(2) = – 17 + 19
= 2

(ii) f(7) – f(1)
f(x) = 3x – 4
f(7) = 3(7) – 4 = 21 – 4 = 17
f(x) = 6x + 1
f(1) = 6(1) + 1 = 6 + 1 = 7
f(7) – f(1) = 17 – 7
= 10

(iii) 2f(4) + f(8)
f(x) = 5x² – 1
f(4) = 5(4)2 – 1 = 5(16) – 1
= 80 – 1 = 79
f(x) = 3x – 4
f(8) = 3(8) – 4 = 24 – 4 = 20
2f(4) + f(8) = 2(79) + 20
= 158 + 20
= 178

f(x) = 6x + 1
f(-2) = 6(-2) + 1 = -12 + 1 = -11
f(x) = 3x – 4
f(6) = 3(6) – 4 = 18 – 4 = 14
f(x) = 5x² – 1
f(4) = 5(4)² – 1 = 5(16) – 1
= 80 – 1 = 79
f(x) = 6x + 1
f(-2) = 6(-2) + 1 = -12 + 1 = -11

Question 11.
The distance S an object travels under the influence of gravity in time t seconds is given by S(t) = \(\frac{1}{2}\) gt² + at + b where, (g is the acceleration due to gravity), a, b are constants. Check if the function S (t)is one-one.
Solution:
S(t) = \(\frac{1}{2}\) gt² + at + b
Let t be 1, 2, 3, ………, seconds
S(1) = \(\frac{1}{2}\) g(1²) + a(1) + b = \(\frac{1}{2}\) g + a + b
S(2) = \(\frac{1}{2}\) g(2²) + a(2) + b = 2g + 2a + b
Yes, for every different values of t, there will be different values as images. And there will be different preimages for the different values of the range. Therefore it is one-one function.

Question 12.
The function ‘t’ which maps temperature in Celsius (C) into temperature in Fahrenheit (F) is defined by
t(C) = F where F = \(\frac { 9 }{ 5 } \) C + 32. Find,
(i) t(0)
(ii) t(28)
(iii) t(-10)
(iv) the value of C when t(C) = 212
(v) the temperature when the Celsius value is equal to the Fahrenheit value.
Answer:
Given t(C) = \(\frac { 9C }{ 5 } \) + 32

(i) t(0) = \(\frac { 9(0) }{ 5 } \) + 32
= 32° F

(ii) t(28) = \(\frac { 9(28) }{ 5 } \) + 32
= \(\frac { 252 }{ 5 } \) + 32
= 50.4 + 32
= 82.4° F

(iii) t(-10) = \(\frac { 9(-10) }{ 5 } \) + 32
= -18 + 32
= 14° F

(iv) t(C) = 212
\(\frac { 9C }{ 5 } \) + 32 = 212
\(\frac { 9C }{ 5 } \) = 212 – 32
= 180
9C = 180 × 5
C = \(\frac{180 \times 5}{9}\)
= 100° C

(v) consider the value of C be “x”
t(C) = \(\frac { 9C }{ 5 } \) + 32
x = \(\frac { 9x }{ 5 } \) + 32
5x = 9x + 160
-160 = 9x – 5x
-160 = 4x
x = \(\frac { -160 }{ 4 } \) = -40
The temperature when the Celsius value is equal to the fahrenheit value is -40°

Composition of two Functions

Let f: A → B and g: B → C be two functions. Then the composition of f and g denoted by gof is defined as the function gof (x) = g[f(x)] for all x ∈ A.

Composition of three Functions

Let A, B, C, D be four sets and let f: A → B; g : B → C and h : C → D be three functions, using composite functions fog and goh, we get two new functions like (fog) oh and fo (goh).
Note: Composition of three function is always associative.