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by ccassel » Wed Mar 30, 2011 11:11 am
Hi,

Can anyone explain the answer to this question?

In the rectangular coordinate system does the line K (not shown) intersect quadrant II?

1. The slope of K is -1/6
2. The y-intercept of k is -6


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by Rich@VeritasPrep » Wed Mar 30, 2011 11:28 am
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.
Rich Zwelling
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by ccassel » Wed Mar 30, 2011 11:46 am
Thank you for the explanation.

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

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by Rich@VeritasPrep » Wed Mar 30, 2011 12:31 pm
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.
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by MAAJ » Wed Mar 30, 2011 1:39 pm
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..?
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by ccassel » Wed Mar 30, 2011 2:17 pm
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.
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