The sides of a triangle have the combined equation `x^2-3y^2-2x y+8y-4=0` . The third side, which is variable, always passes through the point `(-5,-1)` . Find the range of values of the slope of the third line such that the origin is an interior point of the triangle.

Options

A. `2 lt m lt 5`
B. `-1 lt m lt 1/5`
C. `-5 lt m lt -1`
D. `-7 lt m lt -5`

Answer

B

Explanation

`x^(2)-2xy-3y^(2)+8y-4=0`
`rArr(x+y-2)(x-3y+2)=0`
& equation of sides are `L_(1) , x + y - 2 = 0` & `L_(2): x - 3 y + 2 =0`

Any line through `(-5,-1)` having slope m is
`L_(3) : y + 1 = m(x+5)`
If `L_(3)` is parallel to `L_(1)` (i.e, slope `m = -1`)then none of `Delta` is possibe & when slope of `L_(3)` is `m = (1)/(5)` then origin will lie on third side.
`:. -1 lt m lt 1/5`