Let C(α,β) be the circumcenter of the triangle formed by the lines
4x+3y=69
4y−3x=17
x+7y=61
Then (α−β)2+α+β is equal to
Explanation
We have,
4x+3y=69
4y−3x=17
x+7y=61
On solving (i) and (iii), we get
x=12, and y=7
So, A≡(12,7)
On solving (ii) and (iii), we get
x=5 and y=8
So, B=(5,8)
Hence, circumcentre=(212+5,27+8)
≡(217,215)
∴ α=217,β=215
∴ (α−β)2+(α+β)=(217−215)2+(217+215)2
=(1)2+(16)=17
Explanation
We have,
4x+3y=69
4y−3x=17
x+7y=61
On solving (i) and (iii), we get
x=12, and y=7
So, A≡(12,7)
On solving (ii) and (iii), we get
x=5 and y=8
So, B=(5,8)
Hence, circumcentre=(212+5,27+8)
≡(217,215)
∴ α=217,β=215
∴ (α−β)2+(α+β)=(217−215)2+(217+215)2
=(1)2+(16)=17
