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The incentre of an equilateral triangle is (1, 1) and the equation of one side is 3x + 4y + 3 = 0. Then the equation of the circumcircle of the triangle is
- a)x2 + y2 – 2x – 2y – 2 = 0
- b)x2 + y2 – 2x – 2y – 14 = 0
- c)x2 + y2– 2x – 2y + 2 = 0
- d)x2 + y2 – 2x – 2y + 14 = 0
Correct answer is option 'B'. Can you explain this answer?
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The incentre of an equilateral triangle is (1, 1) and the equation of ...
For equalateral triangle R = 2r = 2 × 2 = 4 equation of circumcircle (x – 1)2 + (y – 1)2 = 42 ; x2 + y2 – 2x – 2y – 14 = 0
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The incentre of an equilateral triangle is (1, 1) and the equation of ...
The incentre of an equilateral triangle is the intersection point of the angle bisectors of the triangle. Since the incentre of the triangle is given as (1, 1), we can find the equations of the angle bisectors passing through this point.
Let the equation of one side of the equilateral triangle be 3x - 4y + 3 = 0. To find the equation of the angle bisector passing through the point (1, 1), we can use the point-slope form of a line.
The slope of the given side is -3/4. The slope of the angle bisector is the negative reciprocal of the slope of the side. Therefore, the slope of the angle bisector is 4/3.
Using the point-slope form, the equation of the angle bisector passing through (1, 1) is:
y - 1 = (4/3)(x - 1)
3y - 3 = 4x - 4
4x - 3y + 1 = 0
Similarly, we can find the equations of the other two angle bisectors passing through the incentre. Since the triangle is equilateral, the angle bisectors will be perpendicular to each other. Therefore, the other two angle bisectors will have slopes -3/4 and -4/3.
The equations of the other two angle bisectors passing through the incentre are:
3x + 4y - 7 = 0
-4x + 3y + 1 = 0
To find the circumcircle of the triangle, we need to find the intersection points of these three angle bisectors. Solving the system of equations, we find the coordinates of the three intersection points are:
(-1, 2)
(2, -1)
(4, 7)
The circumcircle of the triangle passes through these three points. Using the general equation of a circle, the equation of the circumcircle is:
(x - h)^2 + (y - k)^2 = r^2
where (h, k) is the center of the circle and r is the radius.
Using one of the intersection points, for example, (-1, 2), we can substitute these values into the equation and solve for r:
(-1 - h)^2 + (2 - k)^2 = r^2
Substituting the values of (-1, 2) and (h, k) = (1, 1):
(-1 - 1)^2 + (2 - 1)^2 = r^2
4 + 1 = r^2
r^2 = 5
Therefore, the equation of the circumcircle is:
(x - 1)^2 + (y - 1)^2 = 5
So, the correct answer is: x^2 + y^2 - 2x - 2y + 1 = 5.
Let the equation of one side of the equilateral triangle be 3x - 4y + 3 = 0. To find the equation of the angle bisector passing through the point (1, 1), we can use the point-slope form of a line.
The slope of the given side is -3/4. The slope of the angle bisector is the negative reciprocal of the slope of the side. Therefore, the slope of the angle bisector is 4/3.
Using the point-slope form, the equation of the angle bisector passing through (1, 1) is:
y - 1 = (4/3)(x - 1)
3y - 3 = 4x - 4
4x - 3y + 1 = 0
Similarly, we can find the equations of the other two angle bisectors passing through the incentre. Since the triangle is equilateral, the angle bisectors will be perpendicular to each other. Therefore, the other two angle bisectors will have slopes -3/4 and -4/3.
The equations of the other two angle bisectors passing through the incentre are:
3x + 4y - 7 = 0
-4x + 3y + 1 = 0
To find the circumcircle of the triangle, we need to find the intersection points of these three angle bisectors. Solving the system of equations, we find the coordinates of the three intersection points are:
(-1, 2)
(2, -1)
(4, 7)
The circumcircle of the triangle passes through these three points. Using the general equation of a circle, the equation of the circumcircle is:
(x - h)^2 + (y - k)^2 = r^2
where (h, k) is the center of the circle and r is the radius.
Using one of the intersection points, for example, (-1, 2), we can substitute these values into the equation and solve for r:
(-1 - h)^2 + (2 - k)^2 = r^2
Substituting the values of (-1, 2) and (h, k) = (1, 1):
(-1 - 1)^2 + (2 - 1)^2 = r^2
4 + 1 = r^2
r^2 = 5
Therefore, the equation of the circumcircle is:
(x - 1)^2 + (y - 1)^2 = 5
So, the correct answer is: x^2 + y^2 - 2x - 2y + 1 = 5.
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The incentre of an equilateral triangle is (1, 1) and the equation of one side is 3x + 4y + 3 = 0. Then the equation of the circumcircle of the triangle isa)x2 + y2 – 2x – 2y – 2 = 0b)x2 + y2 – 2x – 2y – 14 = 0c)x2 + y2– 2x – 2y + 2 = 0d)x2 + y2– 2x – 2y + 14 = 0Correct answer is option 'B'. Can you explain this answer?
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The incentre of an equilateral triangle is (1, 1) and the equation of one side is 3x + 4y + 3 = 0. Then the equation of the circumcircle of the triangle isa)x2 + y2 – 2x – 2y – 2 = 0b)x2 + y2 – 2x – 2y – 14 = 0c)x2 + y2– 2x – 2y + 2 = 0d)x2 + y2– 2x – 2y + 14 = 0Correct answer is option 'B'. 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 incentre of an equilateral triangle is (1, 1) and the equation of one side is 3x + 4y + 3 = 0. Then the equation of the circumcircle of the triangle isa)x2 + y2 – 2x – 2y – 2 = 0b)x2 + y2 – 2x – 2y – 14 = 0c)x2 + y2– 2x – 2y + 2 = 0d)x2 + y2– 2x – 2y + 14 = 0Correct answer is option 'B'. 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 incentre of an equilateral triangle is (1, 1) and the equation of one side is 3x + 4y + 3 = 0. Then the equation of the circumcircle of the triangle isa)x2 + y2 – 2x – 2y – 2 = 0b)x2 + y2 – 2x – 2y – 14 = 0c)x2 + y2– 2x – 2y + 2 = 0d)x2 + y2– 2x – 2y + 14 = 0Correct answer is option 'B'. Can you explain this answer?.
The incentre of an equilateral triangle is (1, 1) and the equation of one side is 3x + 4y + 3 = 0. Then the equation of the circumcircle of the triangle isa)x2 + y2 – 2x – 2y – 2 = 0b)x2 + y2 – 2x – 2y – 14 = 0c)x2 + y2– 2x – 2y + 2 = 0d)x2 + y2– 2x – 2y + 14 = 0Correct answer is option 'B'. 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 incentre of an equilateral triangle is (1, 1) and the equation of one side is 3x + 4y + 3 = 0. Then the equation of the circumcircle of the triangle isa)x2 + y2 – 2x – 2y – 2 = 0b)x2 + y2 – 2x – 2y – 14 = 0c)x2 + y2– 2x – 2y + 2 = 0d)x2 + y2– 2x – 2y + 14 = 0Correct answer is option 'B'. 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 incentre of an equilateral triangle is (1, 1) and the equation of one side is 3x + 4y + 3 = 0. Then the equation of the circumcircle of the triangle isa)x2 + y2 – 2x – 2y – 2 = 0b)x2 + y2 – 2x – 2y – 14 = 0c)x2 + y2– 2x – 2y + 2 = 0d)x2 + y2– 2x – 2y + 14 = 0Correct answer is option 'B'. Can you explain this answer?.
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The incentre of an equilateral triangle is (1, 1) and the equation of one side is 3x + 4y + 3 = 0. Then the equation of the circumcircle of the triangle isa)x2 + y2 – 2x – 2y – 2 = 0b)x2 + y2 – 2x – 2y – 14 = 0c)x2 + y2– 2x – 2y + 2 = 0d)x2 + y2– 2x – 2y + 14 = 0Correct answer is option 'B'. Can you explain this answer?, a detailed solution for The incentre of an equilateral triangle is (1, 1) and the equation of one side is 3x + 4y + 3 = 0. Then the equation of the circumcircle of the triangle isa)x2 + y2 – 2x – 2y – 2 = 0b)x2 + y2 – 2x – 2y – 14 = 0c)x2 + y2– 2x – 2y + 2 = 0d)x2 + y2– 2x – 2y + 14 = 0Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of The incentre of an equilateral triangle is (1, 1) and the equation of one side is 3x + 4y + 3 = 0. Then the equation of the circumcircle of the triangle isa)x2 + y2 – 2x – 2y – 2 = 0b)x2 + y2 – 2x – 2y – 14 = 0c)x2 + y2– 2x – 2y + 2 = 0d)x2 + y2– 2x – 2y + 14 = 0Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The incentre of an equilateral triangle is (1, 1) and the equation of one side is 3x + 4y + 3 = 0. Then the equation of the circumcircle of the triangle isa)x2 + y2 – 2x – 2y – 2 = 0b)x2 + y2 – 2x – 2y – 14 = 0c)x2 + y2– 2x – 2y + 2 = 0d)x2 + y2– 2x – 2y + 14 = 0Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
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