Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
128 views
in Mathematics by (113k points)
closed by
Find the equation of the line perpendicular to the line x - 3y + 5 = 0 and passes through the point (2, -4).
1. 3x + y - 2 = 0
2. 3x - y - 10 = 0
3. x - 3y - 14 = 0
4. x + 3y + 10 = 0

1 Answer

0 votes
by (114k points)
selected by
 
Best answer
Correct Answer - Option 1 : 3x + y - 2 = 0

Concept:

The general equation of a line is y = mx + c 

Where m is the slope and c is any constant

  • The slope of parallel lines is equal.
  • The slope of the perpendicular line have their product = -1
 

Equation of a line with slope m and passing through (x1, y1)

(y - y1) = m (x - x1)

Calculation:

Given line x - 3y + 5 = 0

⇒ y = \(1\over3\)x + \(5\over3\)

⇒ Slope(m1) = \(1\over3\) and c1 = \(5\over3\)

Now for the slope of the perpendicular line (m2)

m1 × m2 = -1

⇒ \(1\over3\) × m2 = -1 

⇒ m2 = -3

Perpendicular line has the slope -3 and passes through (2, -4)

∴ Equation of the perpendicular line is

(y - y1) = m (x - x1)

⇒ y - (-4) = -3 (x - 2)

⇒ y + 3x - 2 = 0

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...