Exercise 3.1 - Chapter 3 Analytical Geometry 11th Business Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Chapter 3 - Analytical Geometry - 11th Business Maths Guide Samacheer Kalvi Solutions - Text Book Back Questions and Answers

Text Book Back Questions and Answers

Question 1.
Find the locus of a point which is equidistant from (1, 3) and x-axis.
Solution:

Let P(x₁, y₁) be any point on the locus.
Let A be the point (1, 3)
The distance from the x-axis on the moving pint P(x₁, y₁) is y₁.
Given that AP = y₁

∴ The locus of the point (x₁, y₁) is x² – 2x – 6y + 10 = 0

Question 2.
A point moves so that it is always at a distance of 4 units from the point (3, -2).
Solution:
Let P(x₁, y₁) be any point on the locus.
Let A be the point (3, -2)
Given that PA = 4
PA² = 16

∴ The locus of the point (x₁, y₁) is x² + y² – 6x + 4y – 3 = 0

Question 3.
If the distance of a point from the points (2, 1) and (1, 2) are in the ratio 2 : 1, then find the locus of the point.
Solution:
Let P(x₁, y₁) be any point on the locus.
Let A(2, 1) and B(1, 2) be the given point.
Given that PA : PB = 2 : 1

∴ The locus of the point (x₁, y₁) is 3x² + 3y² – 4x – 14y + 15 = 0

Question 4.
Find a point on the x-axis which is equidistant from the points (7, -6) and (3, 4).
Solution:
Let P(x₁, 0) be any point on the x-axis.
Let A(7, -6) and B(3, 4) be the given points.
Given that PA = PB
PA² = PB²

Question 5.
If A(-1, 1) and B(2, 3) are two fixed points, then find the locus of a point P so that the area of triangle APB = 8 sq.units.
Solution:
Let the point P(x₁, y₁).
Fixed points are A(-1, 1) and B(2, 3).
Given area (formed by these points) of the triangle APB = 8
⇒ 1/2 [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 8
⇒ 1/2 [x₁(1 – 3) + (-1) (3 – y₁) + 2(y₁ – 1)] = 8
⇒ 1/2 [-2x₁ – 3 + y₁ + 2y₁ – 2] = 8
⇒ 1/2 [-2x₁ + 3y₁ – 5] = 8
⇒ -2x₁ + 3y₁ – 5 = 16
⇒ -2x₁ + 3y₁ – 21 = 0
⇒ 2x₁ – 3y₁ + 21 = 0
∴ The locus of the point P(x₁, y₁) is 2x – 3y + 21 = 0.