Question

Find the co-ordinates of the circumcentre of the triangle whose vertices are A(– 2, 3), B(6, – 1), C(4, 3).

Answer 1

Here, A(– 2, 3), B(6, – 1), C(4, 3) are the verticals of ΔABC.
Let F be the circumcentre of ΔABC.
Let FD and FE be the perpendicular bisectors of the sides BC and AC respectively.
∴ D and E are the midpoints of side BC and AC respectively.

∴ D = `((6 + 4)/2, (-1 + 3)/2)`

∴ D = (5, 1)

and E = `((-2 + 4)/2, (3 + 3)/2)`

∴ E = (1, 3)

Now, slope of BC = `(-1 - 3)/(6 - 4)` = – 2

∴ slope of FD = `1/2`        ...[∵ FD ⊥ BC]

Since, FD passes through (5, 1) and has slope `1/2`

∴ Equation of FD is

y – 1 = `1/2(x - 5)`

∴ 2(y – 1) = x – 5
∴ x – 2y – 3 = 0         ....(i)
Since, both the points A and C have same y co-ordinates i.e. 3
∴ the points A and C lie on the line y = 3.
Since, FE passes through E(1, 3).
∴ the equation of FE is x = 1.      …(ii)
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value of x in (i), we get
1 – 2y – 3 = 0
∴ y = – 1
∴ Co-ordinates of circumcentre F ≡ (1, – 1).