If the equation of the pair of straight lines passing through the point `(1,1)` , one making an angle `theta` with the positive direction of the x-axis and the other making the same angle with the positive direction of the y-axis, is `x^2-(a+2)x y+y^2+a(x+y-1)=0,a!=2,` then the value of `sin2theta` is `a-2` (b) `a+2` `2(a+2)` (d) `2/a`

Options

A. `a-2`
B. `a+2`
C. `(2)/(a+2)`
D. `(2)/(a)`

Answer

C

Explanation

The joint equation of the given lines is
`x^(2)-(a+2)xy+y^(2)+a(x+y-1)=0`
`rArr" "(x-y)^(2)-a(xy-x-y+1)=0`
`rArr" "{(x-1)-(y-1)}^(2)-a(x-1)(y-1)=0`
Shifting origin at (1, 1) it reduces to `(X-Y)^(2)-aXY=0`, or `X^(2)-XY(a+2)+Y^(2)=0`. Lines represented by this equation make angle `theta` and `(pi)/(2)-theta` with x-axis.
`therefore" "tan theta+tan((pi)/(2)-theta)=a+2`
`rArr" "tan theta=cot theta=a+2`
`rArr" "(1)/(sin theta cos theta)=a+2rArr sin 2 theta=(2)/(a+2)`