D(– 1, 8), E(4, – 2), F(– 5, – 3) are midpoints of sides BC, CA and AB of ΔABC Find: co-ordinates of the circumcentre of ΔABC.

#### Solution

Here, A(0, – 13) B(– 10, 7), C(8, 9) are the vertices of ΔABC.

Let F be the circumcentre of ΔABC.

Let FD and FE be perpendicular bisectors of the sides BC and AC respectively.

∴ D and E are the midpoints of side BC and AC.

∴ D = `((-10 + 8)/2, (7 + 9)/2)`

∴ D = (– 1 , 8) and E = `((0 + 8)/2, (-13 + 9)/2)`

∴ E = (4, – 2)

Now, slope of BC = `(7 - 9)/(-10 - 8) = 1/9`

∴ slope of FD = – 9 ...[∵ FD ⊥ BC]

Since, FD passes through (– 1, 8) and has slope – 9

∴ Equation of FD is

y – 8 = – 9 (x + 1)

∴ y – 8 = – 9 x – 9

∴ y = – 9x – 1 ...(i)

Also, slope of AC = `(-13 - 9)/(0 - 8) = 11/4`

∴ Slope of FE = `(-4)/11` ...[∵ FE ⊥ AC]

Since, FE passes through (4, – 2) and has slope `(-4)/11`

∴ Equation of FE is

y + 2 = `(-4)/11 (x - 4)`

∴ 11(y + 2) = – 4 (x – 4)

∴ 11y + 22 = – 4x + 16

∴ 4x + 11y = – 6 ...(ii)

To find co-ordinates of circumcentre,

we have to solve equations (i) and (ii).

Substituting the value of y in (ii), we get

4x + 11(– 9x – 1) = – 6

∴ 4x – 99x – 11 = – 6

∴ –95x = 5

∴ x = `(-1)/19`

Substituting the value of x in (i), we get

y = `-9(-1/19) - 1 = (-10)/19`

∴ Co-ordinates of circumcentre F ≡ `((-1)/19, (-10)/19)`.