Question

Find the circumcenter of the triangle whose sides are given by

x + y + 2 = 0, 5 x - y - 2 = 0

and

x - 2 y + 5 = 0

.

Accepted Answer

Given lines are $x + y + 2 = 0$ ...(i), $5 x - y - 2 = 0$ ...(ii), $x - 2 y + 5 = 0$ ...(iii)

Let, point of intersection of (i) and (ii) is $A = (0, - 2)$

Let, point of intersection of (ii) and (iii) is $B = (1, 3)$

Let, point of intersection of (i) and (iii) is $C = (- 3, 1)$

Let, $S = (α, β)$ the orthocentre of $Δ A B C$ , then $S A = S B = S C$

$\Rightarrow S A^{2} = S B^{2} = S C^{2}$

$\Rightarrow (α - 0)^{2} + (β + 2)^{2} = (α - 1)^{2} + (β - 3)^{2} = (α + 3)^{2} + (β - 1)^{2}$

$\Rightarrow α^{2} + β^{2} + 4 β + 4 = α^{2} + β^{2} - 2 α - 6 β + 10 = α^{2} + β^{2} + 6 α - 2 β + 10$

Let, $S A = S B$

$\Rightarrow S A^{2} = S B^{2}$

$\Rightarrow α^{2} + β^{2} + 4 β + 4 = α^{2} + β^{2} - 2 α - 6 β + 10$

$\Rightarrow 2 α + 10 β - 6 = 0$

$\Rightarrow α + 5 β - 3 = 0$ ...(iv)

Let, $S A = S C$

$S A^{2} = S C^{2}$

$\Rightarrow α^{2} + β^{2} + 4 β + 4 = α^{2} + β^{2} + 6 α - 2 β + 10$

$\Rightarrow 6 α - 6 β + 6 = 0$

$\Rightarrow α - β + 1 = 0$ ...(v)

Solving (iv) and (v), we get $6 β - 4 = 0$

$\Rightarrow 6 β = 4$

$\Rightarrow β = \frac{2}{3}$

From (v), $α - \frac{2}{3} + 1 = 0$

$\Rightarrow α = - \frac{1}{3}$

$∴$ Circumcentre $S = (- \frac{1}{3}, \frac{2}{3})$

Telangana Board Solutions for Chapter: The Straight Line, Exercise 5: Exercise 3(e)

Telangana Board Mathematics Solutions for Exercise - Telangana Board Solutions for Chapter: The Straight Line, Exercise 5: Exercise 3(e)

Questions from Telangana Board Solutions for Chapter: The Straight Line, Exercise 5: Exercise 3(e) with Hints & Solutions

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