A) (4, 0)
B) (2, -1)
C) (0, 4)
D) (-1, 2)
Correct Answer: A
Solution :
Equations of three sides of a triangle are \[x=2,\,\,y+1=0\]and \[x+2y=4\] Co-ordinates of point of intersection of the \[x=2\] and \[y+1=0\] is (2, -1) Co-ordinates of point of intersection of \[x=2\] and \[x+2y=4\] is (2, 1) Co-ordinates of point of intersection of \[y+1=0\] and \[x+2y=4\] is (6, -1) Let Co-ordinates of circumcentre is (x, y) \[{{(x-2)}^{2}}+{{(y+1)}^{2}}={{(x-2)}^{2}}+{{(y-1)}^{2}}\] \[{{(y+1)}^{2}}={{(y-1)}^{2}}\]; \[{{y}^{2}}+2y+1={{y}^{2}}-2y+1\] \[4y=0\], \[y=0\] and \[{{(x-2)}^{2}}+{{(y-1)}^{2}}={{(x-6)}^{2}}+{{(y+1)}^{2}}\] In this equation put \[y=0\] \[{{(x-2)}^{2}}+{{(0-1)}^{2}}={{(x-6)}^{2}}+{{(0+1)}^{2}}\] \[{{(x-2)}^{2}}+1={{(x-6)}^{2}}+1\]; \[{{(x-2)}^{2}}-{{(x-6)}^{2}}=0\] \[(x-2+x-6)(x-2-x+6)=0\] \[4(2x-8)=0\]Þ \[8(x-4)=0\]; \[x-4=0\] Þ \[x=4\] \ Co-ordinates of circumcentre is (4, 0).
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