Cartesian plane

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The circum centre of the triangle whose sides are $3 x - y - 5 = 0, x + 2 y - 4 = 0$ , $5 x + 3 y + 1 = 0$ is $(α, β)$ then $α^{2} + β^{2} =$

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Solution

Given lines are
$3 x - y - 5 = 0$ …… (1)
$x + 2 y - 4 = 0$ …… (2)
$5 x + 3 y + 1 = 0$ …… (3)
The point of intersections of above lines are vertices of triangle
Hence the vertices are $A (2, 1), B (- 2, 3), C (1, - 2)$
Suppose that the point $S (α, β)$ is the circum center of the triangle, so that $SA = SB = SC$
Solving the equation $SA = SB = SC$ . We get $(α, β) = (\frac{- 6}{7}, \frac{2}{7})$
Therefore, $α^{2} + β^{2} = \frac{36}{49} + \frac{4}{49} = \frac{40}{49} = 0 .816$

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Cartesian plane

Remember concepts with our Masterclasses.

The circum centre of the triangle whose sides are 3x−y−5=0,x+2y−4=0,5x+3y+1=0 is α,β then α2+β2=

Ready to Test Your Skills?

free study material

The circum centre of the triangle whose sides are $3 x - y - 5 = 0, x + 2 y - 4 = 0$ , $5 x + 3 y + 1 = 0$ is $(α, β)$ then $α^{2} + β^{2} =$