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Find the circumcentre of a triangle whose vertices are A = (-2, -3) , B = (-1, 0) and C = (7, -6) .
- a)(3, -3)
- b)(-3, -3)
- c)(-3, 3)
- d)(3, 0)
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Find the circumcentre of a triangle whose vertices are A = (-2, -3) , ...
In a circumcentre the distance from point of intersection of all the perpendicular bisector to all the vertices is equal,
Let a point be P and vertices of P be (x, y) and vertices of Δ be A, B and C,
⇒ PA = PB = PC
By distance formula, PA2 = PB2 = PC2 = R2
For PA2 = PB2,
⇒ (x + 2)2 + (y + 3)2 = (x + 1)2 + y2
⇒ x + 3y = -6 ----(1)
For PA2 = PC2,
⇒ (x + 2)2 + (y + 3)2 = (x – 7)2 + (y + 6)2
⇒ 3x – y = 12 ----(2)
On solving equation (1) and (2), we get x = 3 and y = -3
∴ Circumcentre (x, y) = (3, -3)
Let a point be P and vertices of P be (x, y) and vertices of Δ be A, B and C,
⇒ PA = PB = PC
By distance formula, PA2 = PB2 = PC2 = R2
For PA2 = PB2,
⇒ (x + 2)2 + (y + 3)2 = (x + 1)2 + y2
⇒ x + 3y = -6 ----(1)
For PA2 = PC2,
⇒ (x + 2)2 + (y + 3)2 = (x – 7)2 + (y + 6)2
⇒ 3x – y = 12 ----(2)
On solving equation (1) and (2), we get x = 3 and y = -3
∴ Circumcentre (x, y) = (3, -3)
Most Upvoted Answer
Find the circumcentre of a triangle whose vertices are A = (-2, -3) , ...
The circumcentre of a triangle is the centre of the circle that passes through all three vertices of the triangle. To find the circumcentre, we can use the concept of perpendicular bisectors.
1. Find the slopes of the sides of the triangle:
- The slope of side AB = (0 - (-3)) / (-1 - (-2)) = 3/1 = 3
- The slope of side BC = (-6 - 0) / (7 - (-1)) = -6/8 = -3/4
- The slope of side AC = (-6 - (-3)) / (7 - (-2)) = -3/9 = -1/3
2. Find the midpoints of the sides of the triangle:
- The midpoint of side AB = ( (-2 + (-1))/2 , (-3 + 0)/2 ) = (-3/2, -3/2)
- The midpoint of side BC = ( (-1 + 7)/2 , (0 + (-6))/2 ) = (3, -3)
- The midpoint of side AC = ( (-2 + 7)/2 , (-3 + (-6))/2 ) = (5/2, -9/2)
3. Find the perpendicular bisectors of the sides of the triangle:
- The perpendicular bisector of side AB has a slope of -1/3 and passes through the midpoint (-3/2, -3/2).
- The equation of this perpendicular bisector is y - (-3/2) = (-1/3)(x - (-3/2)).
Simplifying, we get y + 3/2 = (-1/3)(x + 3/2).
Rearranging, we get 3y + 9/2 = -x - 3/2.
Bringing all terms to one side, we get x + 3y = -6.
- The perpendicular bisector of side BC has a slope of 3 and passes through the midpoint (3, -3).
- The equation of this perpendicular bisector is y - (-3) = 3(x - 3).
Simplifying, we get y + 3 = 3x - 9.
Rearranging, we get 3x - y = 12.
- The perpendicular bisector of side AC has a slope of -3/4 and passes through the midpoint (5/2, -9/2).
- The equation of this perpendicular bisector is y - (-9/2) = (-3/4)(x - 5/2).
Simplifying, we get y + 9/2 = (-3/4)(x - 5/2).
Rearranging, we get 3y + 9/2 = -4x + 10/2.
Bringing all terms to one side, we get 4x + 3y = 10.
4. Solve the system of equations formed by the perpendicular bisectors:
Solving the system of equations x + 3y = -6, 3x - y = 12, and 4x + 3y = 10, we find the point of intersection to be (3, -3).
5. Therefore, the circumcentre of the triangle ABC is (3
1. Find the slopes of the sides of the triangle:
- The slope of side AB = (0 - (-3)) / (-1 - (-2)) = 3/1 = 3
- The slope of side BC = (-6 - 0) / (7 - (-1)) = -6/8 = -3/4
- The slope of side AC = (-6 - (-3)) / (7 - (-2)) = -3/9 = -1/3
2. Find the midpoints of the sides of the triangle:
- The midpoint of side AB = ( (-2 + (-1))/2 , (-3 + 0)/2 ) = (-3/2, -3/2)
- The midpoint of side BC = ( (-1 + 7)/2 , (0 + (-6))/2 ) = (3, -3)
- The midpoint of side AC = ( (-2 + 7)/2 , (-3 + (-6))/2 ) = (5/2, -9/2)
3. Find the perpendicular bisectors of the sides of the triangle:
- The perpendicular bisector of side AB has a slope of -1/3 and passes through the midpoint (-3/2, -3/2).
- The equation of this perpendicular bisector is y - (-3/2) = (-1/3)(x - (-3/2)).
Simplifying, we get y + 3/2 = (-1/3)(x + 3/2).
Rearranging, we get 3y + 9/2 = -x - 3/2.
Bringing all terms to one side, we get x + 3y = -6.
- The perpendicular bisector of side BC has a slope of 3 and passes through the midpoint (3, -3).
- The equation of this perpendicular bisector is y - (-3) = 3(x - 3).
Simplifying, we get y + 3 = 3x - 9.
Rearranging, we get 3x - y = 12.
- The perpendicular bisector of side AC has a slope of -3/4 and passes through the midpoint (5/2, -9/2).
- The equation of this perpendicular bisector is y - (-9/2) = (-3/4)(x - 5/2).
Simplifying, we get y + 9/2 = (-3/4)(x - 5/2).
Rearranging, we get 3y + 9/2 = -4x + 10/2.
Bringing all terms to one side, we get 4x + 3y = 10.
4. Solve the system of equations formed by the perpendicular bisectors:
Solving the system of equations x + 3y = -6, 3x - y = 12, and 4x + 3y = 10, we find the point of intersection to be (3, -3).
5. Therefore, the circumcentre of the triangle ABC is (3
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Find the circumcentre of a triangle whose vertices are A = (-2, -3) , B = (-1, 0) and C = (7, -6) .a)(3, -3)b)(-3, -3)c)(-3, 3)d)(3, 0)Correct answer is option 'A'. Can you explain this answer?
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