Exercise 3.2 - Chapter 3 - Analytical Geometry - 11th Business Maths Guide Samacheer Kalvi Solutions

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Exercise 3.2

 Text Book Back Questions and Answers

Question 1.
Find the angle between the lines whose slopes are 1/2 and 3.
Solution:
Given that m₁ = 1/2 and m₂ = 3.
Let θ be the angle between the lines then

tan θ = 1
tan θ = tan 45°
∴ θ = 45°

Question 2.
Find the distance of the point (4, 1) from the line 3x – 4y + 12 = 0.
Solution:
The length of perpendicular from a point (x₁, y₁) to the line ax + by + c = 0 is d =

∴ The distance of the point (4, 1) to the line 3x – 4y + 12 = 0 is
[Here (x₁, y₁) = (4, 1), a = 3, b = -4, c = 12]

Question 3.
Show that the straight lines x + y – 4 = 0, 3x + 2 = 0 and 3x – 3y + 16 = 0 are concurrent.
Solution:
The lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0, a₃x + b₃y + c₃ = 0 are concurrent if

The given lines x + y – 4 = 0, 3x + 0y + 2 = 0, 3x – 3y + 16 = 0

= 1(0 + 6) – 1(48 – 6) – 4(-9 – 0)
= 6 – (42) + 36
= 42 – 42
= 0
The given lines are concurrent.

Question 4.
Find the value of ‘a’ for which the straight lines 3x + 4y = 13; 2x – 7y = -1 and ax – y – 14 = 0 are concurrent.
Solution:
The lines 3x + 4y = 13, 2x – 7y = -1 and ax – y – 14 = 0 are concurrent.

⇒ 3(98 + 1) – 4(-28 – a) – 13(-2 + 7a) = 0
⇒ 3(99) + 112 + 4a + 26 – 91a = 0
⇒ 297 + 112 + 26 + 4a – 91a = 0
⇒ 435 – 87a = 0
⇒ -87a = -435

Question 5.
A manufacturer produces 80 TV sets at a cost of 2,20,000 and 125 TV sets at a cost of 2,87,500. Assuming the cost curve to be linear, find the linear expression of the given information. Also, estimate the cost of 95 TV sets.
Solution:
Let x represent the TV sets, and y represent the cost.

The equation of straight line expressing the given information as a linear equation in x and y is

1(y – 2,20,000) = (x – 80)1500
y – 2,20,000 = 1500x – 80 × 1500
y = 1500x – 1,20,000 + 2,20,000
y = 1500x + 1,00,000 which is the required linear expression.
When x = 95,
y = 1,500 × 95 + 1,00,000
= 1,42,500 + 1,00,000
= 2,42,500
∴ The cost of 95 TV sets is 2,42,500.