BITSAT 2005 Solved Question Paper

Given x1,x2 are the roots of the equation
x2+2x−3=0
⇒x2+3x−x−3=0
⇒x(x+3)−1(x+3)=0
⇒(x−1)(x+3)=0
⇒x1=−3,x2=1
and y1,y2 are the roots of the equation
y2+4y−12=0
⇒y2+6y−2y−12=0
⇒y(y+6)−2(y+6)=0
⇒(y−2)(y+6)=0
⇒y1=−6,y2=2
∴ Points are P(−3,−6) and Q(1,2).
Since, P and Q are the end points of a diameter.
∴ Centre = mid point of PQ
=(

−3+1

2

,

−6+2

2

)
=(−1,−2)