How to solve this?

piyush_nitt wrote:Pls help[/list]

IMO A...

statement 1..negative slope means it goes through quads 2 and 4 in all cases...

statment 2...all this tells you is that y int is -6..x is 0, we do not know slope..insuff

It is A
a line with negative slope will be slanting from left (up) to right (down)...
where ever you move the line parallely it will still pass through II quadrant.

if you don get, i can draw and upload

1) is sufficient.

A line with a slope of -1/6 will be going from upper left to lower right. It must cross through quadrants II and iV (if it crosses through 0,0 then it will only cross through quadrants II and iV).

The equation for this line is y = -1/6x = C (where C is a constant).

Quadrant II is y is pos when x is neg.
-1/6 * a negative number will be positive.

And since lines extend to infinity, but C is a constant, there will have to be value of x when it is negative where -1/6 x will be a positive number greater than C and therefore the line will break into quadrant II.

2) alone is insufficient as the y intercept is just one point and that doesn't tell us enough about a line.

So, answer is A

What is OA?

-Carrie

OA Is A

Thanks guys for all help !