Coordinate Geometry Question 150
Question: If equation of three sides of a triangle are $ x=2, $
$ y+1=0 $ and $ x+2y=4 $ then co-ordinates of circumcentre of this triangle are
[AMU 2005]
Options:
A) (4, 0)
B) (2, -1)
C) (0, 4)
D) (-1, 2)
Show Answer
Answer:
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 $
therefore $ 8(x-4)=0 $ ; $ x-4=0 $
therefore $ x=4 $ Co-ordinates of circumcentre is (4, 0).
BETA
