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Question: If equation of three sides of a triangle are \(x = 2,\) \(y + 1 = 0\) and \(x + 2y = 4\) then co-ord...

If equation of three sides of a triangle are x=2,x = 2, y+1=0y + 1 = 0 and x+2y=4x + 2y = 4 then co-ordinates of

circumcentre of this triangle are.

A

(4, 0)

B

(2, –1)

C

(0, 4)

D

(–1, 2)

Answer

(4, 0)

Explanation

Solution

Equations of three sides of a triangle are

x=2,y+1=0x = 2 , y + 1 = 0and x+2y=4x + 2 y = 4

Co-ordinates of point of intersection of the x=2x = 2 and y+1=0y + 1 = 0 is (2, –1)

Co-ordinates of point of intersection of x=2x = 2 and x+2y=4x + 2 y = 4 is (2, 1)

Co-ordinates of point of intersection of y+1=0y + 1 = 0 and x+2y=4x + 2 y = 4 is (6, –1)

Let Co-ordinates of circumcentre is (x, y)

(x2)2+(y+1)2=(x2)2+(y1)2( x - 2 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = ( x - 2 ) ^ { 2 } + ( y - 1 ) ^ { 2 }

(y+1)2=(y1)2( y + 1 ) ^ { 2 } = ( y - 1 ) ^ { 2 }; y2+2y+1=y22y+1y ^ { 2 } + 2 y + 1 = y ^ { 2 } - 2 y + 1

4y=04 y = 0, y=0y = 0 and (x2)2+(y1)2=(x6)2+(y+1)2( x - 2 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = ( x - 6 ) ^ { 2 } + ( y + 1 ) ^ { 2 }

In this equation put y=0y = 0

(x2)2+(01)2=(x6)2+(0+1)2( x - 2 ) ^ { 2 } + ( 0 - 1 ) ^ { 2 } = ( x - 6 ) ^ { 2 } + ( 0 + 1 ) ^ { 2 }

(x2)2+1=(x6)2+1( x - 2 ) ^ { 2 } + 1 = ( x - 6 ) ^ { 2 } + 1; (x2)2(x6)2=0( x - 2 ) ^ { 2 } - ( x - 6 ) ^ { 2 } = 0

(x2+x6)(x2x+6)=0( x - 2 + x - 6 ) ( x - 2 - x + 6 ) = 0

4(2x8)=04 ( 2 x - 8 ) = 08(x4)=08 ( x - 4 ) = 0; x4=0x - 4 = 0x=4x = 4

∴ Co-ordinates of circumcentre is (4, 0).