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Q.

Find the circumcentre of the triangle whose sides are 3x – y – 5 = 0, x + 2y – 4 = 0 and 5x + 3y + 1= 0.

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Detailed Solution

Given line are
3x – y – 5 = 0 ... (1)
x + 2y – 4 = 0 ... (2)
5x + 3y + 1 = 0 ... (3)
The point of intersection of (1) & (2) is A(2, 1)
The point of intersection of (2) & (3) is B(–2, 3)
The point of intersection of (3) &(1) is C(1, –2)
 Let S(α,β) be circumcentre of the triangle ABC
We know that SA = SB = SC
SA2=SB2=SC2
 Now SA2=SB2
(α2)2+(β1)2=(α+2)2+(β3)2α2+A24α+β2+12β =α2+α̸+4α+β2+96β8α4β+8=0 2αβ+2=0 ..... (4)
 And SB2=SC2
(α+2)2+(β3)2=(α1)2+(β+2)2α2+4+4α+β2+96β=α2+12α+β2+4+4β6α10β+8=03α5β+4=0 .(5)
Solve (4) & (5)
      α       β    11    2         2     15     4         3     5
α6=β2=17;α=67,β=27
 Circumcentre S(α,β)=67,27

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