OG- diagnostic DS

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bblast: Legendary Member; Posts: 1077; Joined: Mon Dec 13, 2010 1:44 am; Thanked: 118 times; Followed by:33 members; GMAT Score:710

OG- diagnostic DS

by bblast » Tue May 17, 2011 10:52 am

In the xy-plane, if line k has negative slope and passes through the point (-5, r), is the x-intercept of line k positive?

1) slope of line k is -5
2) r > 0

oa-E

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Source: Beat The GMAT — Data Sufficiency | Tue May 17, 2011 10:52 am

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Tue May 17, 2011 11:42 am

bblast wrote:In the xy-plane, if line k has negative slope and passes through the point (-5, r), is the x-intercept of line k positive?

1) slope of line k is -5
2) r > 0

oa-E

Statement 1: Slope = -5
Statement 2: When x= -5, r > 0

The x-intercept of the line above is negative.

Statement 1: Slope = -5
Statement 2: When x= -5, r > 0

The x-intercept of the line above is positive.

The correct answer is E.

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bblast: Legendary Member; Posts: 1077; Joined: Mon Dec 13, 2010 1:44 am; Thanked: 118 times; Followed by:33 members; GMAT Score:710

by bblast » Wed May 18, 2011 8:06 am

could u please elaborate statement 2 a little bit ?

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Wed May 18, 2011 8:44 am

bblast wrote:could u please elaborate statement 2 a little bit ?

The equation of a line is y=mx + b.

Statement 1: m = -5
Thus, y = -5x + b

Statement 2: In point (-5,r), r>0.

If x = -5 and y = r in the equation y = -5x + b, we get:
r = 25 + b.

Thus, in order for r to be positive:
25+b > 0
b > - 25.

In y = -5x + b, the x-intercept = -b/(-5).
Thus:
In order for the x intercept to be positive, b>0.
In order for the x-intercept to be negative, b<0.

Combining the information above:
If b>0, then r>0 and the x-intercept is positive.
If -25<b<0, then r>0 and the x-intercept is negative.

To illustrate:
Let b=10.
The equation of the line is y = -5x + 10.
The point (-5, 35) is on the line, making r>0.
The x-intercept is (-10)/(-5) = 2.

Let b = -10.
The equation of the line is y = -5x - 10.
The point (-5, 15) is on the line, making r>0.
The x-intercept = 10/(-5) = -2.

Notice that one drawing in my original post has a negative y-intercept, the other has a positive y-intercept.

I think that drawing is much easier and faster than using algebra.

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bblast: Legendary Member; Posts: 1077; Joined: Mon Dec 13, 2010 1:44 am; Thanked: 118 times; Followed by:33 members; GMAT Score:710

by bblast » Fri May 20, 2011 10:45 am

I agree. The algebra is way too convoluted. Hope to stay away from such tricky co-ordinate problems on test day.

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pemdas: Legendary Member; Posts: 1084; Joined: Fri Apr 15, 2011 2:33 pm; Thanked: 158 times; Followed by:21 members

by pemdas » Mon Dec 12, 2011 12:52 am

on very rare occasions I could manage to solve GMAT q. on coordinate Geo timely using equation lines, may be for only simple q. I could use equations. If you use draws/images in mind, then it's easier to spot that parallel lines have the same slopes and the x-intercepts may be shifted as long as the lines are dragged along their parallel coordinate points and keep the same slope/angle property.