If one of the lines given by the equation `2x^2+p x y+3y^2=0` coincide with one of those given by `2x^2+q x y-3y^2=0` and the other lines represented by them are perpendicular, then `p=5` (b) `p=-5` `q=-1` (d) `q=1`

Options

A. `a=-5, b=1`
B. `a=5, b=-1`
C. `a=5, b=1`
D. none of these

Answer

C

Explanation

Let `y=mx` be the common line and let `y=m_(1)x` and `y=m_(2)x` be the other lines given by `2x^(2)+axy+3y^(2)=0` and `2x^(2)+bxy-3y^(2)=0` respectively. Then,
`m+m_(1)=-(a)/(3),mm_(1)=(2)/(3),m+m_(2)=(b)/(3)andmm_(2)=-(2)`
`therefore" "(mm_(1))(mm_(2))=(2)/(3)xx-(2)/(3)`
`rArr" "m^(2)(m_(1)m_(2))=-(4)/(9)" "[because m_(1)m_(2)=-1"(Given)"]`
`rArr" "m^(2)=(4)/(9)rArrm = pm (2)/(3)`
When `m=(2)/(3):`
`mm_(1)=(2)/(3) and mm_(2)=(2)/(3) rArrm_(1)=1 and, m_(2)=-1`
`therefore" "m+m_(1)=-(a)/(3) and m+m_(2)=(b)/(3)`
`rArr" "a=-5and b=-`
When `m=-(2)/(3):`
`mm_(1)=(2)/(3) and mm_(2)=-(2)/(3)rArr m_(1)=-1 and , m_(2)=1`
`therefore" "m+m_(1)=-(a)/(3) and m+m_(2)=(b)/(3)`
`rArr" "a=5 and b=1`