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)`.
