Find the circumcentre of the triangle formed by the points0a, 3), (0,

1.	Find the circumcentre of the triangle formed by the points0a, 3), (0, -2), 3,1) [March 2018AP, May 2016Ts, 2006]
Answer» LetA(1,3),B(0,-2),C(-3,1).Slope ofAB=(-2–3)/(0–1) =-5/-1=5. Mid point of AB ( (1+0)/2,(3–2)/2 ) = (1/2,1/2). Equation of perpendicular bisector of AB is … y - 1/2 = (-1/5)(x - 1/2) => x + 5y -3=0 ……(1) Slope of BC=( 1+2)/(-3–0) =3/-3= -1 Mid point of BC ( (0–3)/2, (-2+1)/2 ) = (-3/2,-1/2) Equation of perpendicular bisector of B is……. y - (-1/2) =1{x-(-3/2)} =>y + 1/2 = 1(x + 3/2) ie x-y+1 = 0…..(2) Solving (1) and (2) we get x=-1/3 and y = 2/3. So the circumcenre is (-1/3, 2/3) do u another method

Find the circumcentre of the triangle formed by the points0a, 3), (0, -2), 3,1) [March 2018AP, May 2016Ts, 2006]

Answer»

LetA(1,3),B(0,-2),C(-3,1).Slope ofAB=(-2–3)/(0–1) =-5/-1=5.

Mid point of AB ( (1+0)/2,(3–2)/2 ) = (1/2,1/2).

Equation of perpendicular bisector of AB is … y - 1/2 = (-1/5)(x - 1/2) => x + 5y -3=0 ……(1)

Slope of BC=( 1+2)/(-3–0) =3/-3= -1

Mid point of BC ( (0–3)/2, (-2+1)/2 ) = (-3/2,-1/2)

Equation of perpendicular bisector of B is……. y - (-1/2) =1{x-(-3/2)} =>y + 1/2 = 1(x + 3/2)

ie x-y+1 = 0…..(2)

Solving (1) and (2) we get x=-1/3 and y = 2/3.

So the circumcenre is (-1/3, 2/3)

do u another method