If `x^2-2p x y-y^2=0` and `x^2-2q x y-y^2=0` bisect angles between each other, then find the condition.

Options

A. `p+q=1`
B. `pq=1`
C. `pq+1`
D. `p^(2)+pq+q^(2)=0`

Answer

C

Explanation

The equation of the bisectors of the angles between the lines `x^(2)-2pxy-y^(2)=0`, is
`(x^(2)-y^(2))/(1-(-1))=(xy)/(-p)rArr(x^(2)-y^(2))/(2)=(-xy)/(p)rArrpx^(2)+2xy-py^(2)=0`
This must be same as `x^(2)-2qxy-y^(2)=0`
`therefore" "(p)/(1)=(2)/(_2q)=(-p)/(-1)rArrpq+1=0`