Math, asked by SAB0108, 5 months ago

find the circumcentre of A(2,-5),B(6,8) C(2,8)

Answers

Answered by Anonymous
6

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)

Answered by pulakmath007
6

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

 \sf{AB =   \sqrt{ {(6 - 2)}^{2}  +{ (8 + 5)}^{2} } }

 \implies \sf{AB =   \sqrt{ {4}^{2}  +{13}^{2} } }

 \therefore \:  \:  \sf{AB =   \sqrt{185}  }

 \sf{BC =   \sqrt{ {(6 - 2)}^{2}  +{ (8  - 8)}^{2} } }

 \implies \sf{BC =   \sqrt{ {4}^{2}  +{ 0}^{2} } }

 \therefore \:  \:  \sf{BC =  4 }

 \sf{AC =   \sqrt{ {(2- 2)}^{2}  +{ (8  +  5)}^{2} } }

 \implies \sf{AC =   \sqrt{ {0}^{2}  +{13}^{2} } }

 \therefore \:  \:  \sf{AC =   13}

So From above

 \sf{ {AB }^{2}  ={BC}^{2}  + {AC }^{2} }

∴ 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

 \displaystyle =  \sf{ \bigg( \:  \frac{2 + 6}{2}  \:,   \: \frac{ - 5+ 8}{2}  \:   \bigg)}

 \displaystyle =  \sf{ \bigg( \:  \frac{8}{2}  \:,   \: \frac{ 3}{2}  \:   \bigg)}

 \displaystyle =  \sf{ \bigg( \:  4  \:,   \: \frac{ 3}{2}  \:   \bigg)}

 \displaystyle =  \sf{ \bigg( \: 4 \:,   \: 1.5  \:   \bigg)}

Figure : For figure refer to the attachment

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