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All points (x,y) that lie below the line l, shown above, satisfy which of the following inequalities? A. y < 2x + 3 B. y < -2x + 3 C. y < -x + 3 D. y < 1/2*x + 3 E. y < -1/2*x + 3

First of all we should write the equation of the line \(l\):

We have two points: A(0,3) and B(6,0).

Equation of a line which passes through two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(\frac{y-y_1}{x-x_1}=\frac{y_1-y_2}{x_1-x_2}\)

So equation of a line which passes the points A(0,3) and B(6,0) would be: \(\frac{y-3}{x-0}=\frac{3-0}{0-6}\) --> \(2y+x-6=0\) --> \(y=-\frac{1}{2}x+3\)

Points below this line satisfy the inequality: \(y<-\frac{1}{2}x+3\)

OR The equation of line which passes through the points \(A(0,3)\) and \(B(6,0)\) can be written in the following way:

Equation of a line in point intercept form is \(y=mx+b\), where: \(m\) is the slope of the line and \(b\) is the y-intercept of the line (the value of \(y\) for \(x=0\)).

The slope of a line, \(m\), is the ratio of the "rise" divided by the "run" between two points on a line, thus \(m=\frac{y_1-y_2}{x_1-x_2}\) -->\(\frac{3-0}{0-6}=-\frac{1}{2}\) and \(b\) is the value of \(y\) when \(x=0\) --> A(0,3) --> \(b=3\).

So the equation is \(y=-\frac{1}{2}x+3\)

Points below this line satisfy the inequality: \(y<-\frac{1}{2}x+3\).

Actually one could guess that the answer is E at the stage of calculating the slope \(m=-\frac{1}{2}\), as only answer choice E has the same slope line in it.

Answer: E.

For more please check Coordinate Geometry chapter of the Math Book (link in my signature).

The line will be in the form of y=mx+b In this case we are looking for a negative m (eliminate option A and D). Finally we are looking for a line with X intercept of 6. So if you make y=0 for all remaining formulas you get

compute the slope. The slope is the change in rise over run or (y2-y1)/(x2-x1). So, slope is (3-0)/(0-6) = -1/2. (Or, (0-3)/(6-0) = -1/2. And the y-intercept of the line (where the line crosses or touches the y-axis) is +3. Thus, using the line equation y = mx + b (in which "y" and "x" is an ordinate pair for any point on the line, "m" is slope, and "b" is y-intercept), the equation of the line is y = -1/2x + 3. We're looking for points that lie below this line, so for any given value of x, the y value should be the biggest possible value without going above the line. Thus, the correct answer is choice E.

You can pick numbers to confirm. Let x = 6. We know that when x = 6, according to the line, y = 0. Let's plug x = 6 into the line equation: y = (-1/2)*6 + 3 = 0. Yep, that's right. To fall below the line, then, when x = 6, y<0.

If you understand the line equation, then as soon as you computed the slope of "-1/2", you know that the answer is E, and you're done (because none of the other choices have "-1/2").

We could have also observed that the slope is negative (because the line is "falling" reading from left to right). That observation allows us to cancel choices A and D. The slope is definitely not -1...eliminate C. Choice B is a trap for someone who reversed the given and x and y coordinates or else a trap for someone who computed slope as run/rise rather than rise/run.

Re: All points (x,y) that lie below the line l, satisfy which of [#permalink]

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28 Jun 2013, 01:30

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Expert's post

study wrote:

All points (x,y) that lie below the line l, satisfy which of the following inequalities?

A. y<2x+3 B. y<-2x+3 C. y<-x+3 D. y<1/2x+3 E. y<-1/2x+3

Attachment:

new_PS_Lines_E.JPG

For any given line L, if the intercepts are given, we can write the equation of the line as \(\frac {x} {x-intercept} + \frac {y} {y-intercept} -1 = 0\)

For the given line, it stands as L = \(\frac {x} {6} + \frac {y} {3} -1 = 0\) Now, notice that when the value of origin is plugged in (0,0), we get L as 0+0-1 --> L<0. Thus, the origin lies on the negative side of the given line. And, as origin lies below the given line, all the points in that region will make L<0 -->

Re: All points (x,y) that lie below the line l, shown above [#permalink]

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28 Jun 2013, 01:45

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study wrote:

Attachment:

Line.png

All points (x,y) that lie below the line l, shown above, satisfy which of the following inequalities?

A. y < 2x + 3 B. y < -2x + 3 C. y < -x + 3 D. y < 1/2*x + 3 E. y < -1/2*x + 3

Frankly, I did this question without any calculation. I hope my approach helps you save time.

First step: We have equation: \(y = ax + b\) in which a is the slope of the line. I see the line "l" passes through quadrant II and IV ==> The slope of line "l" should be negative ==> A, D are out immediately.

Second step. We see two points, say A (6, 0) and B (0, 3) on line "l". Let plug in one point, say A (6,0) to B, C, E ==> C is out Let plug in the second point, say B (0,3) to D & E ==> D is out

Only E remains and is correct.

Hope it helps.

PS: You can save a lot of time by using "plug in" method _________________

Please +1 KUDO if my post helps. Thank you.

"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."

Re: All points (x,y) that lie below the line l, shown above [#permalink]

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30 Jun 2013, 02:59

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study wrote:

Attachment:

Line.png

All points (x,y) that lie below the line l, shown above, satisfy which of the following inequalities?

A. y < 2x + 3 B. y < -2x + 3 C. y < -x + 3 D. y < 1/2*x + 3 E. y < -1/2*x + 3

As an alternative solution to this question i suggest to plug-in 0 for both x and y to find the x and y intercepts. From the graph it is clearly seen that the values of y should be less than 3 and the values of x should be less than 6 so ideally when we find the x and y intercepts should get the y<3 and x<6.

a) x=0 then y<3 this part works; y=0, x>-1,5 not our target; b) x=0 then y<3 this part works; y=0, x<1,5 not our target; c) x=0 then y<3 this part works; y=0, x<3 not our target; d) x=0 then y<3 this part works; y=0, x>-6 not our target; e) x=0 then y<3 this part works; y=0, x<6 BINGO!

E is the line in the graph satisfies y< -1/2*x + 3. This method seems timeconsuming but for those who forget the functions of slope and othe formulas this is very basic visual solution. It took just under 2 min, plus just from one glance it is seen that y<3 in all options so no need to spend time for y, just concentrate to find option which satisfies for x.

Hope that helps! _________________

If you found my post useful and/or interesting - you are welcome to give kudos!

Re: All points (x,y) that lie below the line l, shown above [#permalink]

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13 Oct 2013, 02:19

All points (x,y) that lie below the line l, shown above, satisfy which of the following inequalities?

A. y < 2x + 3 B. y < -2x + 3 C. y < -x + 3 D. y < 1/2*x + 3 E. y < -1/2*x + 3

The easiest approach. Firstly, you have to know that a line which slopes downwards from left to right ALWAYS has -ve slope =====>options A and D are eliminated WHY? ALL equations are of form y=mx+c, where m is slope.

Now, check y-intercept i.e., c in the options B,C and E ALL are 3 and in the diagram also y-intercept is 3 So, at this point you cannot eliminate any options on basis of y-intercept\

BUT note that in the options slopes for all 3 options is DIFFERENT-this is what you should attack. HOW? Observe diagram: two co-ordinates on the line are (0,3) and (6,0) slope=0-3/6-0=1/2 As line is -ve sloped(discussed earlier) m=-1/2 Only one options exists with this slope and its E.

Re: All points (x,y) that lie below the line l, shown above [#permalink]

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20 Oct 2013, 03:49

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Took me 5 seconds to figure out the answer: When you look at this graph, you can write it right away in this form: y=mx+b b=point on y coordinate m=negative > decreasing m=positive > increasing m=-1/2 = the line goes from b point to (2,2), as the next y,x integer cross, and (4,1) as the next, and (6,0) next (basically 1 in slope means that it goes 1 down, and 2 means it goes 2 down (if slope is negative, in positive one 1 means one up, 2 means 2 right).

So just by looking at (6,0) point and b=3 you can firmly say that the line is gonna be y=-1/2x+3.

So for instance if you draw line with b=2, and slope 5/3, you will get this green line (y=5/3x+2), and if you draw line with b=2 and slope -5/3, you will get orange line (y=-5/3x+2)

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14 Nov 2014, 07:55

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Re: All points (x,y) that lie below the line l, shown above [#permalink]

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14 Dec 2014, 11:35

Bunuel wrote:

All points (x,y) that lie below the line l, shown above, satisfy which of the following inequalities? A. y < 2x + 3 B. y < -2x + 3 C. y < -x + 3 D. y < 1/2*x + 3 E. y < -1/2*x + 3

First of all we should write the equation of the line \(l\):

We have two points: A(0,3) and B(6,0).

Equation of a line which passes through two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(\frac{y-y_1}{x-x_1}=\frac{y_1-y_2}{x_1-x_2}\)

So equation of a line which passes the points A(0,3) and B(6,0) would be: \(\frac{y-3}{x-0}=\frac{3-0}{0-6}\) --> \(2y+x-6=0\) --> \(y=-\frac{1}{2}x+3\)

Points below this line satisfy the inequality: \(y<-\frac{1}{2}x+3\)

OR The equation of line which passes through the points \(A(0,3)\) and \(B(6,0)\) can be written in the following way:

Equation of a line in point intercept form is \(y=mx+b\), where: \(m\) is the slope of the line and \(b\) is the y-intercept of the line (the value of \(y\) for \(x=0\)).

The slope of a line, \(m\), is the ratio of the "rise" divided by the "run" between two points on a line, thus \(m=\frac{y_1-y_2}{x_1-x_2}\) -->\(\frac{3-0}{0-6}=-\frac{1}{2}\) and \(b\) is the value of \(y\) when \(x=0\) --> A(0,3) --> \(b=3\).

So the equation is \(y=-\frac{1}{2}x+3\)

Points below this line satisfy the inequality: \(y<-\frac{1}{2}x+3\).

Actually one could guess that the answer is E at the stage of calculating the slope \(m=-\frac{1}{2}\), as only answer choice E has the same slope line in it.

Answer: E.

For more please check Coordinate Geometry chapter of the Math Book (link in my signature).

Hope it's clear.

HI Bunuel,

Could you please clarify following statement.

Points below this line satisfy the inequality: y<-\frac{1}{2}x+3

Re: All points (x,y) that lie below the line l, shown above [#permalink]

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29 Dec 2015, 10:40

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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