The straight lines represented by `x^2+m x y-2y^2+3y-1=0` meet at

Options

A. `(1//3,-2//3)`
B. `(-1//3,-2//3)`
C. `(1//3,2//3)`
D. None of these

Answer

C

Explanation

We have , `f(x,y)=x^2+mxy-2y^2+3y-1=0` .....(i)
this equation will represent pair of straight lines , if `1xx(-2)xx(-1)+2xx(3)/(2)xx0xx(m)/(2)-(-2)xx0-(-1)xx(m^2)/(4)=0`
`rArr m=1,-1`
The point of intersection of the pairs of lines represented by Eq.(i) is `((bg-hf)/(h^2-ab),(af-gh)/(h^2-ab))`
`=((-2xx0-(m)/(2)xx(3)/(2))/(((m)/(2))^2-1xx-2),(1xx(3)/(2)-0)/(((m)/(2))^2-1xx-2))`
`=((-(3m)/(4))/((m^2+8)/(4)),((3)/(2))/((m^2+8)/(4)))=((-3m)/(m^2+8),(6)/(m^2+8))`
Where `m=1`, then point of intersection is `((-3)/(1+8),(6)/(1+8))i.e., (-(1)/(3),(2)/(3))`
Where `m=-1` the point of intersection is `((3)/(1+8),(6)/(1+8))i.e.,((1)/(3),(2)/(3))`.
In this case , the required point is obtained by solving equations `2x-y=0 and -x-4y+3=0`. Clearly , such a point is `(1//3,2//3)`.