JEE Exam > JEE Questions > The point moves such that the area of the tri...
Start Learning for Free
The point moves such that the area of the triangle formed by it with the points (1, 5) and (3, -7) is 21 sq.units. The locus of the point is
- a)6x + y - 32 = 0
- b)6x - y + 32 = 0
- c)x + 6y - 32 = 0
- d)6x - y - 32 = 0
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
The point moves such that the area of the triangle formed by it with t...
Given information:
- Two fixed points (1,5) and (3,-7)
- Area of the triangle formed by the point and the two fixed points is 21 sq.units
To find:
- Locus of the point
Solution:
Let the variable point be (x,y)
Area of the triangle formed by the three points = 1/2 * base * height
where base is the distance between (1,5) and (3,-7), which is sqrt((3-1)^2 + (-7-5)^2) = sqrt(170)
and height is the perpendicular distance from (x,y) to the line passing through (1,5) and (3,-7)
Equation of the line passing through (1,5) and (3,-7)
=> (y - 5)/(x - 1) = (-7 - 5)/(3 - 1)
=> y - 5 = -6(x - 1)
=> y = -6x + 11
Perpendicular distance from (x,y) to the line y = -6x + 11
= |(-6x + 11)x + (-1)y + (6x - 11)*5| / sqrt((-6)^2 + (-1)^2)
= |(-6x^2 + 37x - 56)| / sqrt(37)
Area of the triangle = 1/2 * sqrt(170) * |(-6x^2 + 37x - 56)| / sqrt(37) = 21
=> |(-6x^2 + 37x - 56)| = 21 * sqrt(37/170)
=> -6x^2 + 37x - 56 = 21 * sqrt(37/170) or -(-6x^2 + 37x - 56) = 21 * sqrt(37/170)
=> -6x^2 + 37x - 56 = 6.16 or 6x^2 - 37x + 56 = 6.16
=> -6x^2 + 37x - 62.16 = 0 or 6x^2 - 37x + 49.84 = 0
Using the quadratic formula, we can solve for x in both equations
For the first equation,
a = -6, b = 37, c = -62.16
=> x = (37 + sqrt(37^2 - 4 * -6 * -62.16)) / (2 * -6) or (37 - sqrt(37^2 - 4 * -6 * -62.16)) / (2 * -6)
=> x = 2.67 or 5.56
For the second equation,
a = 6, b = -37, c = 49.84
=> x = (-(-37) + sqrt((-37)^2 - 4 * 6 * 49.84)) / (2 * 6) or (-(-37) - sqrt((-37)^2 - 4 * 6 * 49.84)) / (2 * 6)
=> x = 2.67 or 5.56
Therefore, the possible values of x are 2.67 and 5.56
Substituting these values in the equation of
- Two fixed points (1,5) and (3,-7)
- Area of the triangle formed by the point and the two fixed points is 21 sq.units
To find:
- Locus of the point
Solution:
Let the variable point be (x,y)
Area of the triangle formed by the three points = 1/2 * base * height
where base is the distance between (1,5) and (3,-7), which is sqrt((3-1)^2 + (-7-5)^2) = sqrt(170)
and height is the perpendicular distance from (x,y) to the line passing through (1,5) and (3,-7)
Equation of the line passing through (1,5) and (3,-7)
=> (y - 5)/(x - 1) = (-7 - 5)/(3 - 1)
=> y - 5 = -6(x - 1)
=> y = -6x + 11
Perpendicular distance from (x,y) to the line y = -6x + 11
= |(-6x + 11)x + (-1)y + (6x - 11)*5| / sqrt((-6)^2 + (-1)^2)
= |(-6x^2 + 37x - 56)| / sqrt(37)
Area of the triangle = 1/2 * sqrt(170) * |(-6x^2 + 37x - 56)| / sqrt(37) = 21
=> |(-6x^2 + 37x - 56)| = 21 * sqrt(37/170)
=> -6x^2 + 37x - 56 = 21 * sqrt(37/170) or -(-6x^2 + 37x - 56) = 21 * sqrt(37/170)
=> -6x^2 + 37x - 56 = 6.16 or 6x^2 - 37x + 56 = 6.16
=> -6x^2 + 37x - 62.16 = 0 or 6x^2 - 37x + 49.84 = 0
Using the quadratic formula, we can solve for x in both equations
For the first equation,
a = -6, b = 37, c = -62.16
=> x = (37 + sqrt(37^2 - 4 * -6 * -62.16)) / (2 * -6) or (37 - sqrt(37^2 - 4 * -6 * -62.16)) / (2 * -6)
=> x = 2.67 or 5.56
For the second equation,
a = 6, b = -37, c = 49.84
=> x = (-(-37) + sqrt((-37)^2 - 4 * 6 * 49.84)) / (2 * 6) or (-(-37) - sqrt((-37)^2 - 4 * 6 * 49.84)) / (2 * 6)
=> x = 2.67 or 5.56
Therefore, the possible values of x are 2.67 and 5.56
Substituting these values in the equation of
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
|
|
Similar JEE Doubts
Top Courses for JEEView all
The point moves such that the area of the triangle formed by it with the points (1, 5) and (3, -7) is 21 sq.units. The locus of the point isa)6x + y - 32 = 0b)6x - y + 32 = 0c)x + 6y - 32 = 0d)6x - y - 32 = 0Correct answer is option 'A'. Can you explain this answer?
Question Description
The point moves such that the area of the triangle formed by it with the points (1, 5) and (3, -7) is 21 sq.units. The locus of the point isa)6x + y - 32 = 0b)6x - y + 32 = 0c)x + 6y - 32 = 0d)6x - y - 32 = 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 The point moves such that the area of the triangle formed by it with the points (1, 5) and (3, -7) is 21 sq.units. The locus of the point isa)6x + y - 32 = 0b)6x - y + 32 = 0c)x + 6y - 32 = 0d)6x - y - 32 = 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 The point moves such that the area of the triangle formed by it with the points (1, 5) and (3, -7) is 21 sq.units. The locus of the point isa)6x + y - 32 = 0b)6x - y + 32 = 0c)x + 6y - 32 = 0d)6x - y - 32 = 0Correct answer is option 'A'. Can you explain this answer?.
The point moves such that the area of the triangle formed by it with the points (1, 5) and (3, -7) is 21 sq.units. The locus of the point isa)6x + y - 32 = 0b)6x - y + 32 = 0c)x + 6y - 32 = 0d)6x - y - 32 = 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 The point moves such that the area of the triangle formed by it with the points (1, 5) and (3, -7) is 21 sq.units. The locus of the point isa)6x + y - 32 = 0b)6x - y + 32 = 0c)x + 6y - 32 = 0d)6x - y - 32 = 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 The point moves such that the area of the triangle formed by it with the points (1, 5) and (3, -7) is 21 sq.units. The locus of the point isa)6x + y - 32 = 0b)6x - y + 32 = 0c)x + 6y - 32 = 0d)6x - y - 32 = 0Correct answer is option 'A'. Can you explain this answer?.
Solutions for The point moves such that the area of the triangle formed by it with the points (1, 5) and (3, -7) is 21 sq.units. The locus of the point isa)6x + y - 32 = 0b)6x - y + 32 = 0c)x + 6y - 32 = 0d)6x - y - 32 = 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 The point moves such that the area of the triangle formed by it with the points (1, 5) and (3, -7) is 21 sq.units. The locus of the point isa)6x + y - 32 = 0b)6x - y + 32 = 0c)x + 6y - 32 = 0d)6x - y - 32 = 0Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The point moves such that the area of the triangle formed by it with the points (1, 5) and (3, -7) is 21 sq.units. The locus of the point isa)6x + y - 32 = 0b)6x - y + 32 = 0c)x + 6y - 32 = 0d)6x - y - 32 = 0Correct answer is option 'A'. Can you explain this answer?, a detailed solution for The point moves such that the area of the triangle formed by it with the points (1, 5) and (3, -7) is 21 sq.units. The locus of the point isa)6x + y - 32 = 0b)6x - y + 32 = 0c)x + 6y - 32 = 0d)6x - y - 32 = 0Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of The point moves such that the area of the triangle formed by it with the points (1, 5) and (3, -7) is 21 sq.units. The locus of the point isa)6x + y - 32 = 0b)6x - y + 32 = 0c)x + 6y - 32 = 0d)6x - y - 32 = 0Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The point moves such that the area of the triangle formed by it with the points (1, 5) and (3, -7) is 21 sq.units. The locus of the point isa)6x + y - 32 = 0b)6x - y + 32 = 0c)x + 6y - 32 = 0d)6x - y - 32 = 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