Question

Find the co-ordinates of the circumcenter 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 vertices 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 = `(3 - (-1))/(4 - 6)`

= `4/(-2)`

= – 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

∴ 2y – 2 = x – 5

∴ x – 2y – 3 = 0    ...(i)

Since both the points A and C have the same y co-ordinates i.e. 3,

the given points lie on the line y = 3.

Since the equation 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).