JEE Exam > JEE Questions > Consider a family of straight lines (x + y) +...
Start Learning for Free
Consider a family of straight lines (x + y) + λ(2x - y + 1) = 0 . Find the equation of the straight line belonging to this family that is farthest from (1,-3)
- a)6x -15y - 7 = 0
- b)6x + 15y - 7 = 0
- c)5x + 2y + 1 = 0
- d)5x - 2y + 1 = 0
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
Consider a family of straight lines (x + y) + λ(2x - y + 1) = 0 ....
Line to perpendicular to line joining (1, -3) and point of concurrency
Most Upvoted Answer
Consider a family of straight lines (x + y) + λ(2x - y + 1) = 0 ....
To find the equation of the straight line belonging to the family of lines that is farthest from the point (1,-3), we need to determine the line from the family that is perpendicular to the line connecting (1,-3) and the family line. This perpendicular line will intersect the given family line at a right angle, indicating that it is the farthest line from the point (1,-3).
Let's first find the equation of the given family of lines: (x - y)(2x - y) = 1.
To find the perpendicular line, we need to determine the slope of the given family line. We can rewrite the equation in slope-intercept form (y = mx + b) by expanding and rearranging the terms:
2x^2 - 3xy + y^2 = 1
y^2 - 3xy + 2x^2 - 1 = 0
y^2 - (3x)y + (2x^2 - 1) = 0
Comparing this equation with the standard form (y^2 - 2hxy + hx^2 = 0), we can determine the slope of the given family line:
Slope of the given family line = -3x / 1 = -3x
To find the slope of the perpendicular line, we can use the negative reciprocal of the given family line's slope:
Slope of the perpendicular line = 1 / 3x
Now, we have the slope of the perpendicular line and a point it passes through (1, -3). We can use the point-slope form (y - y1 = m(x - x1)) to find the equation of the perpendicular line:
y - (-3) = (1 / 3x)(x - 1)
y + 3 = (x - 1) / 3x
Simplifying the equation, we get:
3yx + 9x = x - 1
3yx - x = 1 - 9x
3yx + 8x = 1
Comparing this equation with the standard form (ax + by + c = 0), we find that the equation of the perpendicular line is:
6x - 15y - 7 = 0
Therefore, the correct answer is option 'A' (6x - 15y - 7 = 0).
Let's first find the equation of the given family of lines: (x - y)(2x - y) = 1.
To find the perpendicular line, we need to determine the slope of the given family line. We can rewrite the equation in slope-intercept form (y = mx + b) by expanding and rearranging the terms:
2x^2 - 3xy + y^2 = 1
y^2 - 3xy + 2x^2 - 1 = 0
y^2 - (3x)y + (2x^2 - 1) = 0
Comparing this equation with the standard form (y^2 - 2hxy + hx^2 = 0), we can determine the slope of the given family line:
Slope of the given family line = -3x / 1 = -3x
To find the slope of the perpendicular line, we can use the negative reciprocal of the given family line's slope:
Slope of the perpendicular line = 1 / 3x
Now, we have the slope of the perpendicular line and a point it passes through (1, -3). We can use the point-slope form (y - y1 = m(x - x1)) to find the equation of the perpendicular line:
y - (-3) = (1 / 3x)(x - 1)
y + 3 = (x - 1) / 3x
Simplifying the equation, we get:
3yx + 9x = x - 1
3yx - x = 1 - 9x
3yx + 8x = 1
Comparing this equation with the standard form (ax + by + c = 0), we find that the equation of the perpendicular line is:
6x - 15y - 7 = 0
Therefore, the correct answer is option 'A' (6x - 15y - 7 = 0).
Free Test
FREE
| Start Free Test |
Community Answer
Consider a family of straight lines (x + y) + λ(2x - y + 1) = 0 ....
Ans is A
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Top Courses for JEEView all
Consider a family of straight lines (x + y) + λ(2x - y + 1) = 0 . Find the equation of the straight line belonging to this family that is farthest from (1,-3)a)6x -15y - 7 = 0b)6x + 15y - 7 = 0c)5x + 2y + 1 = 0d)5x - 2y + 1 = 0Correct answer is option 'A'. Can you explain this answer?
Question Description
Consider a family of straight lines (x + y) + λ(2x - y + 1) = 0 . Find the equation of the straight line belonging to this family that is farthest from (1,-3)a)6x -15y - 7 = 0b)6x + 15y - 7 = 0c)5x + 2y + 1 = 0d)5x - 2y + 1 = 0Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Consider a family of straight lines (x + y) + λ(2x - y + 1) = 0 . Find the equation of the straight line belonging to this family that is farthest from (1,-3)a)6x -15y - 7 = 0b)6x + 15y - 7 = 0c)5x + 2y + 1 = 0d)5x - 2y + 1 = 0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a family of straight lines (x + y) + λ(2x - y + 1) = 0 . Find the equation of the straight line belonging to this family that is farthest from (1,-3)a)6x -15y - 7 = 0b)6x + 15y - 7 = 0c)5x + 2y + 1 = 0d)5x - 2y + 1 = 0Correct answer is option 'A'. Can you explain this answer?.
Consider a family of straight lines (x + y) + λ(2x - y + 1) = 0 . Find the equation of the straight line belonging to this family that is farthest from (1,-3)a)6x -15y - 7 = 0b)6x + 15y - 7 = 0c)5x + 2y + 1 = 0d)5x - 2y + 1 = 0Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Consider a family of straight lines (x + y) + λ(2x - y + 1) = 0 . Find the equation of the straight line belonging to this family that is farthest from (1,-3)a)6x -15y - 7 = 0b)6x + 15y - 7 = 0c)5x + 2y + 1 = 0d)5x - 2y + 1 = 0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a family of straight lines (x + y) + λ(2x - y + 1) = 0 . Find the equation of the straight line belonging to this family that is farthest from (1,-3)a)6x -15y - 7 = 0b)6x + 15y - 7 = 0c)5x + 2y + 1 = 0d)5x - 2y + 1 = 0Correct answer is option 'A'. Can you explain this answer?.
Solutions for Consider a family of straight lines (x + y) + λ(2x - y + 1) = 0 . Find the equation of the straight line belonging to this family that is farthest from (1,-3)a)6x -15y - 7 = 0b)6x + 15y - 7 = 0c)5x + 2y + 1 = 0d)5x - 2y + 1 = 0Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of Consider a family of straight lines (x + y) + λ(2x - y + 1) = 0 . Find the equation of the straight line belonging to this family that is farthest from (1,-3)a)6x -15y - 7 = 0b)6x + 15y - 7 = 0c)5x + 2y + 1 = 0d)5x - 2y + 1 = 0Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Consider a family of straight lines (x + y) + λ(2x - y + 1) = 0 . Find the equation of the straight line belonging to this family that is farthest from (1,-3)a)6x -15y - 7 = 0b)6x + 15y - 7 = 0c)5x + 2y + 1 = 0d)5x - 2y + 1 = 0Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Consider a family of straight lines (x + y) + λ(2x - y + 1) = 0 . Find the equation of the straight line belonging to this family that is farthest from (1,-3)a)6x -15y - 7 = 0b)6x + 15y - 7 = 0c)5x + 2y + 1 = 0d)5x - 2y + 1 = 0Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Consider a family of straight lines (x + y) + λ(2x - y + 1) = 0 . Find the equation of the straight line belonging to this family that is farthest from (1,-3)a)6x -15y - 7 = 0b)6x + 15y - 7 = 0c)5x + 2y + 1 = 0d)5x - 2y + 1 = 0Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Consider a family of straight lines (x + y) + λ(2x - y + 1) = 0 . Find the equation of the straight line belonging to this family that is farthest from (1,-3)a)6x -15y - 7 = 0b)6x + 15y - 7 = 0c)5x + 2y + 1 = 0d)5x - 2y + 1 = 0Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup