A variable line L is drawn through O (0,0) to meet the lines L 1 ​ :y−x−10=0 and L 2 ​ :y−x−20=0 at the points A and B respectively. A point P is taken on L such that OP 2 ​ = OA 1 ​ + OB 1 ​ and P,A,B lies on same side of origin O. The locus of P is

Options

A. 3x+3y=40
B. 3x+3y+40 =0
C. 3x-3y=40
D. 3y-3x=40

Answer

D

Explanation

Let the parametric equation of the line drawn be
`(x)/("cos" theta) = (y)/("sin" theta) =r`
`"or " x = r "cos" theta, y = r"sin" theta`
Putting it in `L_(1)`, we get
`r "sin" theta, = r"cos" theta +10`
`"or "(1)/(OA) = ("sin" theta-"cos" theta)/(10)`
Similarly, putting the general point of drawn line in the equation of `L_(2)` , we get
`(1)/(OB) = ("sin" theta- "cos" theta)/(20)`
`"Let "P-=(h,k) " and " OP =r." Then, "r "cos"theta = h, r " sin"theta = k. " We have "`
`(2)/(r) = ("sin" theta-"cos" theta)/(10) + ("sin" theta-"cos" theta)/(20)`
`"or " 40 = 3r "sin" theta-3r "cos" theta`
or 3y-3x=40