JEE Exam > JEE Questions > The base of triangle is passing through a fix...
Start Learning for Free
The base of triangle is passing through a fixed point (a,b) and its sides are bisected at right angles by lines, y2 - 4xy - 5x2 = 0, then the locus of the vertex of triangle is
- a)2x2 + 2y2 + (3a + 2b)x + (2a - 3b)y = 0
- b)2x2 + 2y2 + (3a + 2b)x + (2a + 3b)y = 0
- c)2x2 - 2h2 + (3a + 2b)x + (2a + 3b)y = 0
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The base of triangle is passing through a fixed point (a,b) and its si...
y2 - 4xy - 5x2 = 0
⇒ (y - 5x) (y + x) = 0
y - 5x, cuts AB at M at right angle
and y + x = 0 cuts AC at N at right angle
By collinearity of (x1, y1), (x2, y2), (a, b) we get h = x, k = y
2x2 + 2y2 + (3a + 2b)x + (2a - 3b)y = 0
Most Upvoted Answer
The base of triangle is passing through a fixed point (a,b) and its si...
Problem:
The base of a triangle is passing through a fixed point (a,b) and its sides are bisected at right angles by lines, y^2 - 4xy - 5x^2 = 0. Find the locus of the vertex of the triangle.
Solution:
Let A(a,b) be the fixed point on the base of the triangle. Let B and C be the points on the bisectors of the sides AB and AC respectively such that AB = BC.
Let the equation of the bisector of AB be y = mx, where m is the slope of the bisector. Since AB is bisected at right angles by the bisector, the slope of AB is -1/m. Therefore, the equation of AB is y - b = (-1/m)(x - a).
Similarly, let the equation of the bisector of AC be y = nx, where n is the slope of the bisector. The equation of AC is y - b = nx - a.
Since AB = BC, the coordinates of C are given by the intersection of the bisector of AC and the perpendicular bisector of AB. Therefore, the coordinates of C are ((a + mb + nb/m - na)/(m + n), (b + ma - na/m + nb)/(m + n)).
Let P be any point on the locus of the vertex of the triangle. Then, the slopes of AP, BP and CP must be mutually perpendicular.
- Slope of AP = (y - b)/(x - a)
- Slope of BP = (y - b)/(x - a) + m
- Slope of CP = (y - ((a + mb + nb/m - na)/(m + n)))/(x - ((b + ma - na/m + nb)/(m + n))) + n
Since the slopes are mutually perpendicular, we have:
- (y - b)/(x - a) * [(y - b)/(x - a) + m] = -1
- (y - b)/(x - a) * [(y - ((a + mb + nb/m - na)/(m + n)))/(x - ((b + ma - na/m + nb)/(m + n))) + n] = -1
- [(y - ((a + mb + nb/m - na)/(m + n)))/(x - ((b + ma - na/m + nb)/(m + n))) + n] * [(y - b)/(x - a) + m] = -1
Simplifying the above equations, we get:
- y^2 + (m + 2a - 2mb)x - (a^2 + b^2) = 0
- y^2 + 2(na - mb + n^2a - m^2b - nab)/(m + n)x - ((a + mb + nb/m - na)/(m + n))^2 - (b + ma - na/m + nb)^2/(m + n)^2 = 0
- y^2 + 2(na - mb + n^2a - m^2b - nab)/(m + n)x - ((a + mb + nb/m - na)/(m + n))^2 - (b + ma - na/m + nb)^2/(m + n)^2 - (mb - na)/(m + n) = 0
Comparing
The base of a triangle is passing through a fixed point (a,b) and its sides are bisected at right angles by lines, y^2 - 4xy - 5x^2 = 0. Find the locus of the vertex of the triangle.
Solution:
Let A(a,b) be the fixed point on the base of the triangle. Let B and C be the points on the bisectors of the sides AB and AC respectively such that AB = BC.
Let the equation of the bisector of AB be y = mx, where m is the slope of the bisector. Since AB is bisected at right angles by the bisector, the slope of AB is -1/m. Therefore, the equation of AB is y - b = (-1/m)(x - a).
Similarly, let the equation of the bisector of AC be y = nx, where n is the slope of the bisector. The equation of AC is y - b = nx - a.
Since AB = BC, the coordinates of C are given by the intersection of the bisector of AC and the perpendicular bisector of AB. Therefore, the coordinates of C are ((a + mb + nb/m - na)/(m + n), (b + ma - na/m + nb)/(m + n)).
Let P be any point on the locus of the vertex of the triangle. Then, the slopes of AP, BP and CP must be mutually perpendicular.
- Slope of AP = (y - b)/(x - a)
- Slope of BP = (y - b)/(x - a) + m
- Slope of CP = (y - ((a + mb + nb/m - na)/(m + n)))/(x - ((b + ma - na/m + nb)/(m + n))) + n
Since the slopes are mutually perpendicular, we have:
- (y - b)/(x - a) * [(y - b)/(x - a) + m] = -1
- (y - b)/(x - a) * [(y - ((a + mb + nb/m - na)/(m + n)))/(x - ((b + ma - na/m + nb)/(m + n))) + n] = -1
- [(y - ((a + mb + nb/m - na)/(m + n)))/(x - ((b + ma - na/m + nb)/(m + n))) + n] * [(y - b)/(x - a) + m] = -1
Simplifying the above equations, we get:
- y^2 + (m + 2a - 2mb)x - (a^2 + b^2) = 0
- y^2 + 2(na - mb + n^2a - m^2b - nab)/(m + n)x - ((a + mb + nb/m - na)/(m + n))^2 - (b + ma - na/m + nb)^2/(m + n)^2 = 0
- y^2 + 2(na - mb + n^2a - m^2b - nab)/(m + n)x - ((a + mb + nb/m - na)/(m + n))^2 - (b + ma - na/m + nb)^2/(m + n)^2 - (mb - na)/(m + n) = 0
Comparing
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
The base of triangle is passing through a fixed point (a,b) and its sides are bisected at right angles by lines, y2 - 4xy - 5x2 = 0, then the locus of the vertex of triangle isa)2x2 + 2y2 + (3a + 2b)x + (2a - 3b)y = 0b)2x2 + 2y2 + (3a + 2b)x + (2a + 3b)y = 0c)2x2 - 2h2 + (3a + 2b)x + (2a + 3b)y = 0d)None of theseCorrect answer is option 'A'. Can you explain this answer?
Question Description
The base of triangle is passing through a fixed point (a,b) and its sides are bisected at right angles by lines, y2 - 4xy - 5x2 = 0, then the locus of the vertex of triangle isa)2x2 + 2y2 + (3a + 2b)x + (2a - 3b)y = 0b)2x2 + 2y2 + (3a + 2b)x + (2a + 3b)y = 0c)2x2 - 2h2 + (3a + 2b)x + (2a + 3b)y = 0d)None of theseCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The base of triangle is passing through a fixed point (a,b) and its sides are bisected at right angles by lines, y2 - 4xy - 5x2 = 0, then the locus of the vertex of triangle isa)2x2 + 2y2 + (3a + 2b)x + (2a - 3b)y = 0b)2x2 + 2y2 + (3a + 2b)x + (2a + 3b)y = 0c)2x2 - 2h2 + (3a + 2b)x + (2a + 3b)y = 0d)None of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The base of triangle is passing through a fixed point (a,b) and its sides are bisected at right angles by lines, y2 - 4xy - 5x2 = 0, then the locus of the vertex of triangle isa)2x2 + 2y2 + (3a + 2b)x + (2a - 3b)y = 0b)2x2 + 2y2 + (3a + 2b)x + (2a + 3b)y = 0c)2x2 - 2h2 + (3a + 2b)x + (2a + 3b)y = 0d)None of theseCorrect answer is option 'A'. Can you explain this answer?.
The base of triangle is passing through a fixed point (a,b) and its sides are bisected at right angles by lines, y2 - 4xy - 5x2 = 0, then the locus of the vertex of triangle isa)2x2 + 2y2 + (3a + 2b)x + (2a - 3b)y = 0b)2x2 + 2y2 + (3a + 2b)x + (2a + 3b)y = 0c)2x2 - 2h2 + (3a + 2b)x + (2a + 3b)y = 0d)None of theseCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The base of triangle is passing through a fixed point (a,b) and its sides are bisected at right angles by lines, y2 - 4xy - 5x2 = 0, then the locus of the vertex of triangle isa)2x2 + 2y2 + (3a + 2b)x + (2a - 3b)y = 0b)2x2 + 2y2 + (3a + 2b)x + (2a + 3b)y = 0c)2x2 - 2h2 + (3a + 2b)x + (2a + 3b)y = 0d)None of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The base of triangle is passing through a fixed point (a,b) and its sides are bisected at right angles by lines, y2 - 4xy - 5x2 = 0, then the locus of the vertex of triangle isa)2x2 + 2y2 + (3a + 2b)x + (2a - 3b)y = 0b)2x2 + 2y2 + (3a + 2b)x + (2a + 3b)y = 0c)2x2 - 2h2 + (3a + 2b)x + (2a + 3b)y = 0d)None of theseCorrect answer is option 'A'. Can you explain this answer?.
Solutions for The base of triangle is passing through a fixed point (a,b) and its sides are bisected at right angles by lines, y2 - 4xy - 5x2 = 0, then the locus of the vertex of triangle isa)2x2 + 2y2 + (3a + 2b)x + (2a - 3b)y = 0b)2x2 + 2y2 + (3a + 2b)x + (2a + 3b)y = 0c)2x2 - 2h2 + (3a + 2b)x + (2a + 3b)y = 0d)None of theseCorrect 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 base of triangle is passing through a fixed point (a,b) and its sides are bisected at right angles by lines, y2 - 4xy - 5x2 = 0, then the locus of the vertex of triangle isa)2x2 + 2y2 + (3a + 2b)x + (2a - 3b)y = 0b)2x2 + 2y2 + (3a + 2b)x + (2a + 3b)y = 0c)2x2 - 2h2 + (3a + 2b)x + (2a + 3b)y = 0d)None of theseCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The base of triangle is passing through a fixed point (a,b) and its sides are bisected at right angles by lines, y2 - 4xy - 5x2 = 0, then the locus of the vertex of triangle isa)2x2 + 2y2 + (3a + 2b)x + (2a - 3b)y = 0b)2x2 + 2y2 + (3a + 2b)x + (2a + 3b)y = 0c)2x2 - 2h2 + (3a + 2b)x + (2a + 3b)y = 0d)None of theseCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for The base of triangle is passing through a fixed point (a,b) and its sides are bisected at right angles by lines, y2 - 4xy - 5x2 = 0, then the locus of the vertex of triangle isa)2x2 + 2y2 + (3a + 2b)x + (2a - 3b)y = 0b)2x2 + 2y2 + (3a + 2b)x + (2a + 3b)y = 0c)2x2 - 2h2 + (3a + 2b)x + (2a + 3b)y = 0d)None of theseCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of The base of triangle is passing through a fixed point (a,b) and its sides are bisected at right angles by lines, y2 - 4xy - 5x2 = 0, then the locus of the vertex of triangle isa)2x2 + 2y2 + (3a + 2b)x + (2a - 3b)y = 0b)2x2 + 2y2 + (3a + 2b)x + (2a + 3b)y = 0c)2x2 - 2h2 + (3a + 2b)x + (2a + 3b)y = 0d)None of theseCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The base of triangle is passing through a fixed point (a,b) and its sides are bisected at right angles by lines, y2 - 4xy - 5x2 = 0, then the locus of the vertex of triangle isa)2x2 + 2y2 + (3a + 2b)x + (2a - 3b)y = 0b)2x2 + 2y2 + (3a + 2b)x + (2a + 3b)y = 0c)2x2 - 2h2 + (3a + 2b)x + (2a + 3b)y = 0d)None of theseCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup