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Re: In the xy-plane, points A, B and C are not on the same line. [#permalink]

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19 Jan 2013, 03:46

fozzzy wrote:

In the xy-plane, points A, B and C are not on the same line. Is the slope of line BC negative?

1. The slope of line AB is -1. 2. The measure of angle ABC is 37 degrees.

Tricky one. A, B and C are not on the same line. 1) Slope of line AB is -1. No information on point C. Not sufficient. 2) Measure of Angle ABC Is 37 deg. Again no information on which way angle ABC orientation is.

Combine We know angle ABC equals 37 degrees, which is the same as saying: lines AB and BC intersect at a 37 degree angle. This means, the orientation of BC must be 37 degrees different from the orientation of AB. There are two possibilities a) BC's slope is less negative than AB. b) BC's slope is more negative than AB In either case, BC is still oriented negative slope. _________________

Re: In the xy-plane, points A, B and C are not on the same line. [#permalink]

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19 Jan 2013, 12:36

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Method 1 --See next post for Method 2

(1) The slope of line AB is -1 INSUFFICIENT: This tells line AB has -1 negative slope and hence makes \(45^{\circ}\) angle with X axis. However, it does not give any information about point C.

(2) The measure of angle ABC is 37 degrees. INSUFFICIENT: this statement doesn't give information about orientation/locations of points ABC

Combining (1) & (2) -- Here you can solve it even without drawing diagram. SUFFICIENT: With negative slope -1, line AB makes \(45^{\circ}\) angle with X axis. So any line (e.g. BC) making angle less than \(45^{\circ}\) with line AB will have negative slope, similarly line making more than \(45^{\circ}\) angle with AB will turn into positive slope. In this case BC makes angle \(37^{\circ}\) with AB, hence it has negative slope (regardless of which quadrants these points lies in)

Re: In the xy-plane, points A, B and C are not on the same line. [#permalink]

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19 Jan 2013, 12:53

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Attachment:

Line AB with negative slope.jpg [ 348.17 KiB | Viewed 8984 times ]

Method 2

(1) The slope of line AB is -1 INSUFFICIENT: This tells line AB has equation \(y = -x + constant\) (with slope m=-1) However, it does not give any information about point C.

(2) The measure of angle ABC is 37 degrees. INSUFFICIENT: this statement doesn't give information about orientation/locations of points ABC

Combining (1) & (2) -- Here you solve it by plotting diagram SUFFICIENT: As line AB has -1 i.e. negative slope, it can be represented with equation \(y = -x\) (assume constant zero). Plot line AB such that it makes \(45^{\circ}\) with X axis with negative slope. Plot point C such that BC makes \(37^{\circ}\) with AB.

You can observe that the line BC shows negative slope. No matter where you plot point C (either left or right) w.r.t. line AB, the slope of the line is always negative. Same is the case, if you plot ABC in any quadrant.

The line BC can make positive slope ONLY IF it makes MORE than \(45^{\circ}\) angle with AB.

The slope or gradient of a line describes its steepness, incline, or grade. A higher slope value indicates a steeper incline.

The slope is defined as the ratio of the "rise" divided by the "run" between two points on a line, or in other words, the ratio of the altitude change to the horizontal distance between any two points on the line.

Given two points \((x_1,y_1)\) and \((x_2,y_2)\) on a line, the slope \(m\) of the line is:

\(m=\frac{y_2-y_1}{x_2-x_1}\)

If the equation of the line is given in the Point-intercept form: \(y=mx+b\), then \(m\) is the slope. This form of a line's equation is called the slope-intercept form, because \(b\) can be interpreted as the y-intercept of the line, the y-coordinate where the line intersects the y-axis.

If the equation of the line is given in the General form:\(ax+by+c=0\), then the slope is \(-\frac{a}{b}\) and the y intercept is \(-\frac{c}{b}\).

SLOPE DIRECTION The slope of a line can be positive, negative, zero or undefined.

Positive slope Here, y increases as x increases, so the line slopes upwards to the right. The slope will be a positive number. The line below has a slope of about +0.3, it goes up about 0.3 for every step of 1 along the x-axis.

Negative slope Here, y decreases as x increases, so the line slopes downwards to the right. The slope will be a negative number. The line below has a slope of about -0.3, it goes down about 0.3 for every step of 1 along the x-axis.

Zero slope Here, y does not change as x increases, so the line in exactly horizontal. The slope of any horizontal line is always zero. The line below goes neither up nor down as x increases, so its slope is zero. Undefined slope When the line is exactly vertical, it does not have a defined slope. The two x coordinates are the same, so the difference is zero. The slope calculation is then something like \(slope=\frac{15}{0}\) When you divide anything by zero the result has no meaning. The line above is exactly vertical, so it has no defined slope.

SLOPE AND QUADRANTS:

1. If the slope of a line is negative, the line WILL intersect quadrants II and IV. X and Y intersects of the line with negative slope have the same sign. Therefore if X and Y intersects are positive, the line intersects quadrant I; if negative, quadrant III.

2. If the slope of line is positive, line WILL intersect quadrants I and III. Y and X intersects of the line with positive slope have opposite signs. Therefore if X intersect is negative, line intersects the quadrant II too, if positive quadrant IV.

3. Every line (but the one crosses origin OR parallel to X or Y axis OR X and Y axis themselves) crosses three quadrants. Only the line which crosses origin \((0,0)\) OR is parallel to either of axis crosses only two quadrants.

4. If a line is horizontal it has a slope of \(0\), is parallel to X-axis and crosses quadrant I and II if the Y intersect is positive OR quadrants III and IV, if the Y intersect is negative. Equation of such line is y=b, where b is y intersect.

5. If a line is vertical, the slope is not defined, line is parallel to Y-axis and crosses quadrant I and IV, if the X intersect is positive and quadrant II and III, if the X intersect is negative. Equation of such line is \(x=a\), where a is x-intercept.

6. For a line that crosses two points \((x_1,y_1)\) and \((x_2,y_2)\), slope \(m=\frac{y_2-y_1}{x_2-x_1}\)

7. If the slope is 1 the angle formed by the line is \(45\) degrees.

8. Given a point and slope, equation of a line can be found. The equation of a straight line that passes through a point \((x_1, y_1)\) with a slope \(m\) is: \(y - y_1 = m(x - x_1)\) _________________

Re: In the xy-plane, points A, B and C are not on the same line. [#permalink]

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06 Mar 2013, 21:55

PraPon wrote:

Attachment:

Line AB with negative slope.jpg

Method 2

(1) The slope of line AB is -1 INSUFFICIENT: This tells line AB has equation \(y = -x + constant\) (with slope m=-1) However, it does not give any information about point C.

(2) The measure of angle ABC is 37 degrees. INSUFFICIENT: this statement doesn't give information about orientation/locations of points ABC

Combining (1) & (2) -- Here you solve it by plotting diagram SUFFICIENT: As line AB has -1 i.e. negative slope, it can be represented with equation \(y = -x\) (assume constant zero). Plot line AB such that it makes \(45^{\circ}\) with X axis with negative slope. Plot point C such that BC makes \(37^{\circ}\) with AB.

You can observe that the line BC shows negative slope. No matter where you plot point C (either left or right) w.r.t. line AB, the slope of the line is always negative. Same is the case, if you plot ABC in any quadrant.

The line BC can make positive slope ONLY IF it makes MORE than \(45^{\circ}\) angle with AB.

Hence choice(C) is the answer.

Just trying to cement my understanding: So if you have slope=1, line passes thru origin and makes 45 degree with X axis and line will be going upwards as you move from -ve to +ve along teh x axis.. and if you have slope= - 1, line passes thru origin and makes 45 degree with X axis and line will be going downwards as you move from -ve to +ve along teh x axis

right? _________________

hope is a good thing, maybe the best of things. And no good thing ever dies.

Re: In the xy-plane, points A, B and C are not on the same line. [#permalink]

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06 Mar 2013, 22:30

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Sachin9 wrote:

PraPon wrote:

Attachment:

Line AB with negative slope.jpg

Method 2

(1) The slope of line AB is -1 INSUFFICIENT: This tells line AB has equation \(y = -x + constant\) (with slope m=-1) However, it does not give any information about point C.

(2) The measure of angle ABC is 37 degrees. INSUFFICIENT: this statement doesn't give information about orientation/locations of points ABC

Combining (1) & (2) -- Here you solve it by plotting diagram SUFFICIENT: As line AB has -1 i.e. negative slope, it can be represented with equation \(y = -x\) (assume constant zero). Plot line AB such that it makes \(45^{\circ}\) with X axis with negative slope. Plot point C such that BC makes \(37^{\circ}\) with AB.

You can observe that the line BC shows negative slope. No matter where you plot point C (either left or right) w.r.t. line AB, the slope of the line is always negative. Same is the case, if you plot ABC in any quadrant.

The line BC can make positive slope ONLY IF it makes MORE than \(45^{\circ}\) angle with AB.

Hence choice(C) is the answer.

Just trying to cement my understanding: So if you have slope=1, line passes thru origin and makes 45 degree with X axis and line will be going upwards as you move from -ve to +ve along teh x axis.. and if you have slope= - 1, line passes thru origin and makes 45 degree with X axis and line will be going downwards as you move from -ve to +ve along teh x axis

right?

That is correct. Consider line with positive slope as positive slash '/' and negative slope as backward slash '\'. _________________

Re: In the xy-plane, points A, B and C are not on the same line. [#permalink]

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06 Mar 2013, 23:27

fozzzy wrote:

In the xy-plane, points A, B and C are not on the same line. Is the slope of line BC negative?

(1) The slope of line AB is -1. (2) The measure of angle ABC is 37 degrees.

The angle (x) between two lines L1 and L2 with respective slopes as m1 and m2 : tanx = mod[(m1-m2)/(1+m1*m2)]

From F.S 1, we have m1=-1. Clearly Insufficient.

From F.S 2, we have x= 37 degrees. We know that value of tan(30) = 1/\(\sqrt{3}\). So tan 37 will be a little more in value than this. But we don't know anything about m1 and m2. Insufficient.

Combining both, we have tan(37) = mod[m1+1/(1-m1)]. As tan(37) has a positive value and less than 1, lets represent it by a constant value P and 0<P<1.

Now, removing the mod sign, we have either (m1+1)/(1-m1) = P or -P. In either case, we have the value of m1 as negative. Sufficient.

Re: In the xy-plane, points A, B and C are not on the same line. [#permalink]

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30 May 2014, 17:46

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Re: In the xy-plane, points A, B and C are not on the same line. [#permalink]

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09 Jul 2015, 18:16

Why can't the line AB be at a different degree? The slope is given as -1. Is it not possible to have the line/slope(-1) AB at any other degree other than 45? Thanks.

In the xy-plane, points A, B and C are not on the same line. [#permalink]

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09 Jul 2015, 18:33

Expert's post

Amit0507 wrote:

Why can't the line AB be at a different degree? The slope is given as -1. Is it not possible to have the line/slope(-1) AB at any other degree other than 45? Thanks.

No, as slope = tan (angle) where 'angle' = angle that a line makes with the positive direction of the x-axis, i.e. if \(m_{AB}\) = Slope of line AB, then, based on the attached figure, tan (45 deg) = 1 and tan (-45) or tan (135) = -1. Thus, if you are given that slope of a line is 1 or -1, the line MUST be inclined at 45 degrees or 135 degrees respectively to the positive direction of the X-axis.

I have shown in the 2nd figure, why the angle will be 45 degrees, if the slope of a line is 1.

In the xy-plane, points A, B and C are not on the same line. [#permalink]

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11 Jul 2016, 11:16

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Expert's post

This is a fantastic example of a 'just draw it and look at it' DS geometry problem! You can get away with almost no math here. Try drawing a couple of lines with a slope of -1, and labeling the points A, B, and C accordingly. You'll find that it's impossible to draw a set of points where BC has a positive slope.

Critically, on test day, you don't have to understand mathematically why that's the case, as long as you can convince yourself that it's true. Complex diagrams and explanations like the ones above are really great when you're reviewing, but they also sometimes make people think that you'll have to go through all of that math within 2 minutes on the test, which makes this problem look impossible. You don't! It's okay to just draw it, try a couple of cases, and look at it. That's just as valid a solution as the 'mathematical' one, and it gives you the same correct answer.

Here's what my scratch work looked like - this one took me about 45 seconds:

Attachments

slope_abc.png [ 9.64 KiB | Viewed 222 times ]

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