Question

The vertices of a triangle are (5, 1),
(11, 1) and (11, 9). Find the co-ordinates of
the circumcentre of the triangle.​

Answer 1

$\sf \huge \underline{solution}$

Vertices of the triangle are A(5,1), B(11,1) and C (11,9). Let P(x, y) be the circumcenter of the triangle.

 So, PA = PB = PC  (Circumcenter is equidistant from the vertices of the triangle)

=>(PA)2 = (PB)2

=>{(x – 5)2 + (y – 1)2} = {(x –11)2 + (y - 1)2}

= {(x –11)2 + (y - 1)2}=>x2 + 25 – 10x + y2 + 1 – 2y = x2 + 121 – 22x + y2 + 1 - 2y

=> 12x = 96

=> x = 8_______(1)

(1)  As,(PA)2 = (PC)2=>{ (x –5)2 + (y – 1)2 } = { (x – 11)2 + (y - 9)2 }=> x2 + 25 – 10x + y2 +1 –2y = x2 + 121 – 22x + y2 + 81 - 18y=> 25+1 -10x-2y = 121+81-22x -18y

=> 26-10x-2y= 202 -22x-18y

=> 22x-10x +18y-2y = 202 -26

=> 12x +16y = 276 .....(using 1)

=> 16y = 276-96=> 16y = 180

y = 180/16

=>y = 45/4

Answer 2

Step-by-step explanation:

Vertices of tge triangle are A (5 , 1 ) , B ( 11 , 1 ) , C ( 11 , 9 )

Let P (x,y) be the circumference of the triangle

So, PA = PB = PC ( circumference is the equidisant from the vertices of the triangle )

---------> (PA)2 =( PB )2

----------> €(X - 5 )2 + (Y - 1 ) 2 )

= X2 + 25 - 10x + y2 + 1 - 2y = X2 + 121 -22x +y2+1 - 2y

««12x = 96

therefore , x = 8 ..............(1)

As ,( PA )2 = (PC)2

««€(x - 5) 2 + (y - 1 ) 2 ))= €(x - 11)2 + (y - 9)2)

### X2 + 25 - 10x + y2 +1 - 2y = X2 + 121 - 22x + y2 + 81 - 18y

= 25 +1 - 10x - 2y = 121 + 81 - 22x - 18y

= 26 - 10x - 2y = 202 - 22x - 18y

= 22x - 10x + 18y - 2y = 202 - 26

= 12x + 16y = 276

= 12(8)+ 16y = 276 (using 1)

= 16 + 16y = 276

= 16y = 276 - 16

= 16y = 180

Therefore :

y = 180/16

and , y= 45/ 4

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