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## Comment on

Passing Through Quadrant II## how do I know if the slope is

thanks

## Move along a line from left

Move along a line from left to right. If you rise as you move from left to rise, then the slope is positive. If you get lower as you move from left to right, then the slope is negative.

More here: https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...

## For statement 1, line could

## That's also true!

That's also true!

Cheers,

Brent

## Shouldn't the answer be E as

## If the slope is negative,

If the slope is negative, then the line will (eventually) pass through 3 quadrants (one of which will be quadrant II)

Cheers,

Brent

## Brent, Can you please explain

## Good question!

Good question!

I believe you're thinking about a line segment (which has a finite length).

If we were dealing with a line segment with a negative slope, then it would be possible for that line segment to pass through 1 or 2 or 3 quadrants.

However, the question involves a line, and lines go on forever without end.

So, if we take the line that appears at 1:52 in the video, and extend it forever in both directions, the line will eventually cross into quadrant II.

Does that help?

Cheers,

Brent

## Ah..Got it. That makes sense

## Hi,

I want to clarify the STATEMENT 1.

The second red line you draw that crossing Quand II. But it's wrong because X IS NOT = 0 ?

As I know from the theory:

"To find the y-intercept , plug x=0 into the equation"

## Hi Sam,

Hi Sam,

For statement 1, the second line I drew has a negative y-intercept. We know this because the point shown on the y-axis has a negative y-coordinate. Also noticed that the same point has 0 for its x-coordinate.

So for example, the point shown on the y-axis COULD have the coordinates (0, -3). If this were the case, then the y-intercept would be -3. More importantly, the x-coordinate of that point is 0.

To further my point, that's recognize that the equation of that second line COULD be y = -2x - 3 (this equation tells us that the slope of the line is -2, and it's y-intercept is -3.

Also, if x = 0, then y = -3, thus the point (0, -3) on the line

Does that help?