CBSE BOARD XII, asked by pujitharao000, 6 months ago

find the circumcentre of the triangle whose vertices are given by x+y+2=0, 5x-y-2=0 and x-2y +5=0​

Answers

Answered by roshanmishra3209
1

Answer:

  1. x+y+2=0,
  2. 5x-y-2=0
  3. x-2y +5=0​

3x−y−5=0....(i)

x+2y−4=0.....(ii)

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

First find the vertices of the △

Solving (i) and (ii) simultaneously

x  

1

​  

=2,y  

1

​  

=1  ∴A(2,1)

Similarly solving (ii) and (iii) simultaneously

x  

2

​  

=−2,y  

2

​  

=3

∴ B(−2,3)

Finally from (i) and (iii) x  

3

​  

=1,y  

3

​  

=−2 C(1,−2)

Let the equation of circle through A,B,C be x  

2

+y  

2

+2gx+2fy+c=0

put the coordinates of A(2,1)

4+1+4g+2g+c=0

4g+2f+c=−5....(iv)

Similarly by putting coordinates of B(−2,3) in the circle equation we get

4g−6g−c=13....(v)

and putting C(1,−2)  

2g−4f+c=−5....(vi)

Solving (iv), (v), (vi) simultaneously

f=−  

7

2

​  

,g=  

7

6

​  

,c=−  

7

5.5

​  

 

The circumcentre of the circle ≡(−g,−f)

≡(−  

7

6

​  

,  

7

2

​  

)

Explanation:

Answered by saradabolisetti
0

Answer:

-1/3,2/3

Explanation:

we can do sa=sb=sc

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