Question

11. (1) Given equations are
[
begin{array}{l}
a x+b y=3
b x-a y=4
end{array}
]
and ( x^{2}+y^{2}=1 )
(iii)
On squaring Eqs. (i) and (ii) and then adding, we get
[
a^{2} x^{2}+b^{2} y^{2}+2 a x b y+a^{2} y^{2}+b^{2} x^{2}
]
( -2 a x b y=9+16 )
( Rightarrow a^{2}left(x^{2}+y^{2}right)+b^{2}left(x^{2}+y^{2}right)+2 a x b y )
( -2 a x b y=25 )
( Rightarrow a^{2} times 1+b^{2} times 1=25 )
( left[text { put } x^{2}+y^{2}=1right] )
( therefore quad a^{2}+b^{2}=25 )

# 11. If ax + by=3, bx -ay=4 and x + y2 =1, then the value of a +b is (1) 25 (2) 26 (3) 27 (4) 28

Solution