The pair equation of the lines passing through the origin and having slopes 3 and `-(1)/(3)`, is

Options

A. `3y^(2) + 8xy -3x^(2) = 0`
B. `3x^(2) + 8xy + 3y^(2) = 0`
C. `3y^(2) - 8xy - 3x^(2) = 0`
D. `3x^(2) + 8xy - 3y^(2) = 0`

Answer

D

Explanation

Here , `m_(1) = 3 , m_(2) = (-1)/(3)`
Hence , the lines are y = 3x , `y = (-1)/(3) x`
On multiplying both the lines , we get
`(y - 3x) (3y + x) = 0`
`implies 3x ^(2) + 8xy - 3y^(2) = 0`