Question

find the circumcentre of triangle whose sides are 3x-y-5=0,x+2y-4=0 and 5x+3y+1=0​

Answer 1

Answer:

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The circumcenter of the triangle is (x, y) = (2, 1)

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Given line are

3x-y-5=0...(1)

x-2y-4=0...(2)

5x+3y+1=0...(3)

The point of intersection of (1) and (2) by solving, is A(2,1)

The point of intersection of (2) and (3) by solving, is B(-2,3)

The point of intersection of (3) and (1) by solving, is C(1,-2)

Let s(a,b) be circumcentre of triangle ABC

we know that SA=SB=SC

=>SA^2=SB^2=SC^2

NOW SA^2=SB^2

(a-1)^2+(b-1)^2 = (a+2)^2+(b-3)^2

by solving we get,

2a-b+2=0...(4)

AND SB^2=SC^2

(a+2)^2+(b-3)^2=(a-1)^2+(b+2)^2

by solving we get,

3a-5b+4=0....(5)

now solve 4 and 5 eqn, we get a = -6/7, b=2/7

Circumcentre S(a,b) = (-6/7,2/7)