A point `P(x,y)` moves that the sum of its distance from the lines `2x-y-3=0` and `x+3y+4=0 ` is `7`. The area bounded by locus `P` is (in sq. unit)

Options

A. 70
B. `70sqrt(2)`
C. `35sqrt(2)`
D. 140

Answer

B

Explanation

Sum of distances of P from lines 2x-y-3=0 and x+3y+4=0 is 7.
`therefore (|2x-y-3|)/(sqrt(5)) + (|x+3y+4|)/(sqrt(10)) = 7`
`"or " +-sqrt(2)(2x-y-3)+-(x+3y-4)=7sqrt(10)`
The are of parallelogram formed by the lines `a_(1)x + b_(1)y+c_(1)=0,a_(2)x + b_(2)y + d_(1)=0, a_(1)x + b_(1)y+c_(2) = 0 " and " a_(2)x + b_(2)y +d_(2) = 0` is
`|((c_(1)-c_(2))(d_(1)-d_(2)))/(a_(1)b_(2) - a_(2)b_(1))|` sq. units.
So, area `= (14^(2) xx 10)/(14sqrt(2)) = 70sqrt(2)`