Hey ccassel,

The answer is A.

Quadrant II is the quadrant in which x values are negative and y values are positive.

Statement 1 tells you the slope is -1/6. Any line with a negative slope MUST pass through Quadrant II (and Quadrant IV, by the way). Also (just FYI), any line with a positive slope MUST pass through Quadrants I and III. Try drawing several examples of each on some paper, and you'll see these rules always hold.

Statement 2 gives you a negative y-intercept. That's not enough, and to see why, just use the following two examples:

a. Line K has a positive x-intercept. By connecting the two intercepts, you'll see that the line DOES NOT intersect Quadrant II.
b. Line K has a negative x-intercept. By connecting the two intercepts, you'll see that the line DOES intersect Quadrant II.

Hopefully this makes sense! Let me know if you have any other questions.

Thank you for the explanation.

I mixed up quadrant 1 with quadrant 2; hence, giving me the wrong answer.

Cheers.

No problem!

There's a cheesy little trick I always use to remember where each quadrant is: You want to start where everything is positive. So the upper-right quadrant is Quadrant I. From there, you just go counterclockwise.

For statement 2:

Correct me if I'm wrong, but if line K is x = 0, then it would be a vertical line that won't cross any of the quadrants, and its y-intercept could be any positive and negative number.

Right..?

2. The y-intercept of k is -6

In statement 2, y = mx + -6. Therefore, the line crosses the y-axis below the x-axis and we dont know the slope so we cannot determine sufficiently if the line k intersects quadrant 2.

To answer your question, if the equation of line k was x=0 then it would be a verticle line running directly up the y-axis where x equals 0. It would hit points (0,-2), (0,-1), (0,1), (0,2), (0,3), etc. along the line going in both directions infinately. y=0 would be the exact opposite running along the x-axis infinately. Neither of these equations would touch any quadrants.