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The circumcentre of a triangle lies at the origin and its centroid is the mid point of the line segment joining the points (a2 + 1, a2 + 1) and (2a, –2a), a ≠ 0. Then for any a, the orthocentre of this triangle lies on the line :
- a)y - (a2 + 1) x = 0
- b)y + x = 0
- c)(a - 1)2x - (a + 1)2 y = 0
- d)y - 2ax = 0
Correct answer is option 'C'. Can you explain this answer?
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The circumcentre of a triangle lies at the origin and its centroid is ...
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To find the coordinates of the centroid, we need to find the average of the x-coordinates and the average of the y-coordinates of the three vertices of the triangle.
Let the coordinates of the first vertex be (a^2, 1).
Let the coordinates of the second vertex be (2a, 3a).
Let the coordinates of the third vertex be (0, 0) since the circumcentre is at the origin.
The x-coordinate of the centroid is the average of the x-coordinates:
(x1 + x2 + x3)/3 = (a^2 + 2a + 0)/3 = (a^2 + 2a)/3
The y-coordinate of the centroid is the average of the y-coordinates:
(y1 + y2 + y3)/3 = (1 + 3a + 0)/3 = (1 + 3a)/3
Since the centroid is the midpoint of the line segment joining the points (a^2, 1) and (2a, 3a), we can set up the following equation:
[(a^2 + 2a)/3] = (a^2 + 2a)/2
[(1 + 3a)/3] = (1 + 3a)/2
Simplifying these equations, we get:
(a^2 + 2a)/3 = (a^2 + 2a)/2
(1 + 3a)/3 = (1 + 3a)/2
Cross multiplying, we get:
2(a^2 + 2a) = 3(a^2 + 2a)
3(1 + 3a) = 2(1 + 3a)
Expanding and simplifying, we get:
2a^2 + 4a = 3a^2 + 6a
3 + 9a = 2 + 6a
Moving all terms to one side, we get:
a^2 - 2a = 0
3a - 2a = -1
Factoring out a from the first equation, we get:
a(a - 2) = 0
a = 0 or a = 2
So, the possible values for a are a = 0 or a = 2.
If a = 0, then the coordinates of the centroid are:
(x, y) = ((0^2 + 2(0))/3, (1 + 3(0))/3) = (0, 1/3)
If a = 2, then the coordinates of the centroid are:
(x, y) = ((2^2 + 2(2))/3, (1 + 3(2))/3) = (8/3, 7/3)
Therefore, the possible coordinates of the centroid are (0, 1/3) and (8/3, 7/3).
Let the coordinates of the first vertex be (a^2, 1).
Let the coordinates of the second vertex be (2a, 3a).
Let the coordinates of the third vertex be (0, 0) since the circumcentre is at the origin.
The x-coordinate of the centroid is the average of the x-coordinates:
(x1 + x2 + x3)/3 = (a^2 + 2a + 0)/3 = (a^2 + 2a)/3
The y-coordinate of the centroid is the average of the y-coordinates:
(y1 + y2 + y3)/3 = (1 + 3a + 0)/3 = (1 + 3a)/3
Since the centroid is the midpoint of the line segment joining the points (a^2, 1) and (2a, 3a), we can set up the following equation:
[(a^2 + 2a)/3] = (a^2 + 2a)/2
[(1 + 3a)/3] = (1 + 3a)/2
Simplifying these equations, we get:
(a^2 + 2a)/3 = (a^2 + 2a)/2
(1 + 3a)/3 = (1 + 3a)/2
Cross multiplying, we get:
2(a^2 + 2a) = 3(a^2 + 2a)
3(1 + 3a) = 2(1 + 3a)
Expanding and simplifying, we get:
2a^2 + 4a = 3a^2 + 6a
3 + 9a = 2 + 6a
Moving all terms to one side, we get:
a^2 - 2a = 0
3a - 2a = -1
Factoring out a from the first equation, we get:
a(a - 2) = 0
a = 0 or a = 2
So, the possible values for a are a = 0 or a = 2.
If a = 0, then the coordinates of the centroid are:
(x, y) = ((0^2 + 2(0))/3, (1 + 3(0))/3) = (0, 1/3)
If a = 2, then the coordinates of the centroid are:
(x, y) = ((2^2 + 2(2))/3, (1 + 3(2))/3) = (8/3, 7/3)
Therefore, the possible coordinates of the centroid are (0, 1/3) and (8/3, 7/3).
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The circumcentre of a triangle lies at the origin and its centroid is the mid point of the line segment joining the points (a2 + 1, a2 + 1) and (2a, –2a), a ≠ 0. Then for any a, the orthocentre of this triangle lies on the line :a)y - (a2 + 1) x = 0b)y + x = 0c)(a - 1)2x - (a + 1)2 y = 0d)y - 2ax = 0Correct answer is option 'C'. Can you explain this answer?
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The circumcentre of a triangle lies at the origin and its centroid is the mid point of the line segment joining the points (a2 + 1, a2 + 1) and (2a, –2a), a ≠ 0. Then for any a, the orthocentre of this triangle lies on the line :a)y - (a2 + 1) x = 0b)y + x = 0c)(a - 1)2x - (a + 1)2 y = 0d)y - 2ax = 0Correct answer is option 'C'. 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 circumcentre of a triangle lies at the origin and its centroid is the mid point of the line segment joining the points (a2 + 1, a2 + 1) and (2a, –2a), a ≠ 0. Then for any a, the orthocentre of this triangle lies on the line :a)y - (a2 + 1) x = 0b)y + x = 0c)(a - 1)2x - (a + 1)2 y = 0d)y - 2ax = 0Correct answer is option 'C'. 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 circumcentre of a triangle lies at the origin and its centroid is the mid point of the line segment joining the points (a2 + 1, a2 + 1) and (2a, –2a), a ≠ 0. Then for any a, the orthocentre of this triangle lies on the line :a)y - (a2 + 1) x = 0b)y + x = 0c)(a - 1)2x - (a + 1)2 y = 0d)y - 2ax = 0Correct answer is option 'C'. Can you explain this answer?.
The circumcentre of a triangle lies at the origin and its centroid is the mid point of the line segment joining the points (a2 + 1, a2 + 1) and (2a, –2a), a ≠ 0. Then for any a, the orthocentre of this triangle lies on the line :a)y - (a2 + 1) x = 0b)y + x = 0c)(a - 1)2x - (a + 1)2 y = 0d)y - 2ax = 0Correct answer is option 'C'. 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 circumcentre of a triangle lies at the origin and its centroid is the mid point of the line segment joining the points (a2 + 1, a2 + 1) and (2a, –2a), a ≠ 0. Then for any a, the orthocentre of this triangle lies on the line :a)y - (a2 + 1) x = 0b)y + x = 0c)(a - 1)2x - (a + 1)2 y = 0d)y - 2ax = 0Correct answer is option 'C'. 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 circumcentre of a triangle lies at the origin and its centroid is the mid point of the line segment joining the points (a2 + 1, a2 + 1) and (2a, –2a), a ≠ 0. Then for any a, the orthocentre of this triangle lies on the line :a)y - (a2 + 1) x = 0b)y + x = 0c)(a - 1)2x - (a + 1)2 y = 0d)y - 2ax = 0Correct answer is option 'C'. Can you explain this answer?.
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