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Question
D(−1, 8), E(4, −2), F(−5, −3) are midpoints of sides BC, CA and AB of ∆ABC Find co-ordinates of the circumcenter 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 = −9x − 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 coordinates 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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