The straight lines represented by `x^2+m x y-2y^2+3y-1=0` meet at (a) `(-1/3,2/3)` (b) `(-1/3,-2/3)`(c) `(1/3,2/3)` (d) none of these

Options

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

Answer

A, C

Explanation

We have,
`f(x,y)=x^(2)+mxy-2y^(2)+3y-1=0" ….(i)"`
This equation will represent a pair of straight lines, if
`1xx --2xx-1+2xx(3)/(2)xx0xx(m)/(2)-(9)/(4)-(-2)xx0(-1)xx(m^(3))/(4)=0`
`rArr" "m=1,-1`
The points of intersection of the pairs of lines represented by
(i) are obtained by solving the equations
`(delf)/(delx)=0 and (delf)/(dely)=0` or , `2x+my=0` and `mx-4y+3=0`
`ul("CASE I")" When m = 1"`
In this case, the requried point is obtained by solving `2x+y=0` and `x-4y+3=0`.
Clearly, such a point is `(-1//3,2//3)`
`ul("CASE II")" When m = "-1`
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)`