The equation of straight line passing through the point (1, 2) and parallel to the line y = 3x + 1 is

  1. y + 2 = x + 3
  2. y + 2 = 3(x + 3)
  3. y - 2 = 3(x - 3)
  4. None of these

Answer (Detailed Solution Below)

Option 4 : None of these
Free
CRPF Head Constable & ASI Steno (Final Revision): Mini Mock Test
21.2 K Users
40 Questions 40 Marks 45 Mins

Detailed Solution

Download Solution PDF

Concept:

Equation of a line parallel to a given line ax + by + c = 0 is

ax + by + λ = 0,

where λ is a constant

Calculation:

We have,y = 3x + 1

⇒ 3x - y + 1 = 0        

Equation of a line parallel to above line is 

3x - y + λ = 0       -----(1)

It is passing through point (1, 2)

⇒ (3 × 1) - 2 + λ = 0

⇒ 1 + λ = 0

⇒ λ = -1

Put this value in equation (1)

⇒ 3x - y - 1 = 0

⇒y - 2 = 3x - 1 - 2

⇒ y - 2 = 3x - 3

⇒y - 2 = 3(x - 1)

Latest Indian Coast Guard Navik GD Updates

Last updated on Jul 4, 2025

-> The Indian Coast Guard Navik GD Application Correction Window is open now. Candidates can make the changes in the Application Forms through the link provided on the official website of the Indian Navy.

-> A total of 260 vacancies have been released through the Coast Guard Enrolled Personnel Test (CGEPT) for the 01/2026 and 02/2026 batch.

-> Candidates can apply online from 11th to 25th June 2025.

-> Candidates who have completed their 10+2 with Maths and Physics are eligible for this post. 

-> Candidates must go through the Indian Coast Guard Navik GD previous year papers.

More Lines Questions

Get Free Access Now
Hot Links: teen patti fun teen patti sweet teen patti casino download teen patti gold new version 2024