# 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. - Mathematics and Statistics

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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).

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#### APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 11th Standard Maharashtra State Board
Chapter 5 Locus and Straight Line
Exercise 5.4 | Q 10. (b) | Page 78