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
$\Rightarrow {SA}^{2} = {SB}^{2} = {SC}^{2}$
$Now {SA}^{2} = {SB}^{2}$
$\begin{array}{l} \Rightarrow (α - 2)^{2} + (β - 1)^{2} = (α + 2)^{2} + (β - 3)^{2} \\ \Rightarrow α^{2} + A^{2} - 4 α + β^{2} + 1 - 2 β \\ = α^{2} + α̸ + 4 α + β^{2} + 9 - 6 β \\ \Rightarrow 8 α - 4 β + 8 = 0 \\ 2 α - β + 2 = 0 . . . . . (4) \end{array}$
$And {SB}^{2} = {SC}^{2}$
$\begin{array}{l} \Rightarrow (α + 2)^{2} + (β - 3)^{2} = (α - 1)^{2} + (β + 2)^{2} \\ \Rightarrow α^{2} + 4 + 4 α + β^{2} + 9 - 6 β \\ = α^{2} + 1 - 2 α + β^{2} + 4 + 4 β \\ \Rightarrow 6 α - 10 β + 8 = 0 \\ \Rightarrow 3 α - 5 β + 4 = 0 \dots . (5) \end{array}$
Solve (4) & (5)
$\begin{array}{l} α β 1 \\ - 1 2 2 - 1 \\ - 5 4 3 - 5 \end{array}$
$\frac{α}{6} = \frac{β}{- 2} = \frac{1}{- 7}; α = \frac{- 6}{7}, β = \frac{2}{7}$
$∴ Circumcentre S (α, β) = (\frac{- 6}{7}, \frac{2}{7})$