a variable line `L` is drawn trough `O(0,0)` to meet the lines `L_1:y-x-10=0` and `L_2:y-x-20=0` at point `A&B` respectively .A point `P` is taken on line `L` the `(1) ` if `2/(OP)=1/(OA)+1/(OB)` then locus of `P` is `(2)` if `(OP)^2=(OA)*(OB)` then locus of `P` is `(3)` if `1/(OP)^2=1/(OA)^2+1/(OB)^2` then locus of point `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