find the circumcentre of A(2,-5),B(6,8) C(2,8)
Answers
Given,
Coordinates of the three vertices of ∆ABC are = A(2,-5),B(6,8) C(2,8)
To find,
The coordinate of the circumcentre of ∆ABC.
Solution,
Let, the coordinates of the circumcentre ∆ABC = Point O = x,y
Now, OA = OB = OC = Radius of the circle
Length of OA = ✓(x-2)²+(y+5)² units
Length of OB = ✓(x-6)²+(y-8)² units
Length of OC = ✓(x-2)²+(y-8)² units
Now,
OB= OC
OB² = OC²
(x-6)²+(y-8)² = (x-2)²+(y-8)²
(x-6)² = (x-2)²
x²-12x+36 = x²-4x+4
-12x+4x = 4-36
-8x = -32
x = 4
OA² = OB²
(x-2)²+(y+5)² = (x-6)²+(y-8)²
(4-2)²+(y+5)² = (4-6)²+(y-8)²
4+y²+10y+25 = 4+y²-16y+64
10y+25 = -16y+64
26y = 39
y = 1.5
Hence, the coordinate of the circumcentre is (4,1.5)
SOLUTION
TO DETERMINE
The circumcentre of A(2,-5),B(6,8), C(2,8)
CONCEPT TO BE IMPLEMENTED
Let ABC is a right angle triangle with ∠C as right angle and AB as hypotenuse. If we draw a circumcircle of the triangle then AB is the diameter of the circle and midpoint of AB is circumcentre of the circle
EVALUATION
Here the given points are A(2,-5) , B(6,8), C(2,8)
Now
So From above
∴ ABC is a right angle triangle with ∠C as right angle and AB as hypotenuse
Hence midpoint of AB is the circumcentre
Hence the circumcentre
Figure : For figure refer to the attachment
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