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In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, 19 May 2019, 10:05
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In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, y –x ≥ -6 and the line, that is perpendicular to x = 0 and passes through the origin?

A. 9/4
B. 27/4
C. 9
D. 27/2
E. Cannot be determined
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B
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In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, 19 May 2019, 20:59
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In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, y –x ≥ -6 and the line, that is perpendicular to x = 0 and passes through the origin?

A. 9/4
B. 27/4
C. 9
D. 27/2
E. Cannot be determined
Show more

The line x = 0 is y axis. Perpendicular to it passing through origin is the x axis.

y + 2x = 3 is the equation of a line passing through (0, 3) and (1.5, 0).
y + 2x >= 3 is the area away from (0, 0) because (0, 0) doesn't lie in this area.

y - x = -6 is the equation of a line passing through (0, -6) and (6, 0).
y - x >= -6 is the area towards (0, 0) because (0, 0) lies in this area.

Attachment:
Screenshot 2019-05-20 at 09.28.26.png
Screenshot 2019-05-20 at 09.28.26.png [ 23.2 KiB | Viewed 10184 times ]
So we are looking for the area of triangle under x axis. The length of the base is 6 - 1.5 = 4.5.
For the altitude, we need the point of intersection of the two lines which is (3, -3). So length of altitude is 3.

Area = (1/2)*4.5 * 3 = 27/4

Answer (B)
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In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, 19 May 2019, 20:37
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mangamma can you post the official answer please?
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In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, 03 Jul 2020, 08:57
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Attachment:
Screenshot 2019-05-20 at 09.28.26.png


using the same screenshot.
The points 3/2,0 for line y+2x>=3 is the x-intercept of the line. hence we get this point by equating y=0 in the equation.
The point (3, -3) is the intersection point of the lines y + 2x = 3 and y - x = -6. So, if you solve these two equations for the values of x and y, you will get the point as (3, -3).
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In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, 14 Jul 2020, 05:06
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VeritasKarishma
mangamma
In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, y –x ≥ -6 and the line, that is perpendicular to x = 0 and passes through the origin?

A. 9/4
B. 27/4
C. 9
D. 27/2
E. Cannot be determined

The line x = 0 is y axis. Perpendicular to it passing through origin is the x axis.

y + 2x = 3 is the equation of a line passing through (0, 3) and (1.5, 0).
y + 2x >= 3 is the area away from (0, 0) because (0, 0) doesn't lie in this area.

y - x = -6 is the equation of a line passing through (0, -6) and (6, 0).
y - x >= -6 is the area towards (0, 0) because (0, 0) lies in this area.

Attachment:
Screenshot 2019-05-20 at 09.28.26.png
So we are looking for the area of triangle under x axis. The length of the base is 6 - 1.5 = 4.5.
For the altitude, we need the point of intersection of the two lines which is (3, -3). So length of altitude is 3.

Area = (1/2)*4.5 * 3 = 27/4

Answer (B)
Show more

Hi Karishma,
what is the method to decide 'the area' the sign >= is referring? how did you come up with "y + 2x >= 3 is the area away from (0, 0) because (0, 0) doesn't lie in this area." and "y - x >= -6 is the area towards (0, 0) because (0, 0) lies in this area. "
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In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, 11 Aug 2020, 20:18
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Hi karishma

Could you explain the bolded statement below, please? Thank you.

y + 2x = 3 is the equation of a line passing through (0, 3) and (1.5, 0).
y + 2x >= 3 is the area away from (0, 0) because (0, 0) doesn't lie in this area.

y - x = -6 is the equation of a line passing through (0, -6) and (6, 0).
y - x >= -6 is the area towards (0, 0) because (0, 0) lies in this area.
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In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, 15 Sep 2020, 02:15
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y + 2x = 3 is the equation of a line passing through (0, 3) and (1.5, 0).
y + 2x >= 3 is the area away from (0, 0) because (0, 0) doesn't lie in this area.

y - x = -6 is the equation of a line passing through (0, -6) and (6, 0).
y - x >= -6 is the area towards (0, 0) because (0, 0) lies in this area.[/quote]


For the equation y + 2x >= 3 when you substitute x=0,y=0 you get 0 >= 3, which is not true. Therefore (0,0) doesn't lie in that area.
For the euqation y - x >= -6, after substitution you get 0 >= -6 which is true. Therefore (0,0) lies in that area.
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In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, 25 Nov 2020, 22:17
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VeritasKarishma


The line x = 0 is y axis. Perpendicular to it passing through origin is the x axis.

y + 2x = 3 is the equation of a line passing through (0, 3) and (1.5, 0).
y + 2x >= 3 is the area away from (0, 0) because (0, 0) doesn't lie in this area.

y - x = -6 is the equation of a line passing through (0, -6) and (6, 0).
y - x >= -6 is the area towards (0, 0) because (0, 0) lies in this area.

So we are looking for the area of triangle under x axis. The length of the base is 6 - 1.5 = 4.5.
For the altitude, we need the point of intersection of the two lines which is (3, -3). So length of altitude is 3.

Area = (1/2)*4.5 * 3 = 27/4

Answer (B)
Show more

Hi VeritasKarishma thank you for always helping out, after reading your explanation, I still don't get why is you calculated wit 4.5 as the base.
TIA

Why is 6-1.5?

Also, can't I calculate the big triangle then subtracts the smaller one?
(9*3/2) - (1.5*3/2) = 45/4
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In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, 26 Nov 2020, 01:43
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hkkat
VeritasKarishma


The line x = 0 is y axis. Perpendicular to it passing through origin is the x axis.

y + 2x = 3 is the equation of a line passing through (0, 3) and (1.5, 0).
y + 2x >= 3 is the area away from (0, 0) because (0, 0) doesn't lie in this area.

y - x = -6 is the equation of a line passing through (0, -6) and (6, 0).
y - x >= -6 is the area towards (0, 0) because (0, 0) lies in this area.

So we are looking for the area of triangle under x axis. The length of the base is 6 - 1.5 = 4.5.
For the altitude, we need the point of intersection of the two lines which is (3, -3). So length of altitude is 3.

Area = (1/2)*4.5 * 3 = 27/4

Answer (B)

Hi VeritasKarishma thank you for always helping out, after reading your explanation, I still don't get why is you calculated wit 4.5 as the base.
TIA

Why is 6-1.5?

Also, can't I calculate the big triangle then subtracts the smaller one?
(9*3/2) - (1.5*3/2) = 45/4
Show more


hkkat

I think your confusion lies in knowing which region we are talking about. We need to know the area of the orange region.
Attachment:
Screenshot 2020-11-26 at 14.09.16.png
Screenshot 2020-11-26 at 14.09.16.png [ 33.56 KiB | Viewed 7424 times ]

I suggest you to check out this post:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2010/1 ... s-part-ii/

You will understand how to arrive at the orange region.
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In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, 26 Nov 2020, 03:04
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mangamma
In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, y –x ≥ -6 and the line, that is perpendicular to x = 0 and passes through the origin?

A. 9/4
B. 27/4
C. 9
D. 27/2
E. Cannot be determined
Show more
VeritasKarishma Bunuel
I have a query.
Would I be correct to infer that the question is asking for area of triangular region bound by the three lines mentioned in the question?
As mentioned there are only three equations - the three equations of lines are y +2x ≥ 3, y –x ≥ -6 and x-axis, i would not have to worry about the inequality. This way i would only focus on finding area of triangle with base of length between points (6,0) & (3/2,0) and perpendicular from (3,-3) to x-axis.

Or
In other words, I need not to bother about the shape of region.
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In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, 26 Nov 2020, 22:23
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unraveled
mangamma
In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, y –x ≥ -6 and the line, that is perpendicular to x = 0 and passes through the origin?

A. 9/4
B. 27/4
C. 9
D. 27/2
E. Cannot be determined
VeritasKarishma Bunuel
I have a query.
Would I be correct to infer that the question is asking for area of triangular region bound by the three lines mentioned in the question?
As mentioned there are only three equations - the three equations of lines are y +2x ≥ 3, y –x ≥ -6 and x-axis, i would not have to worry about the inequality. This way i would only focus on finding area of triangle with base of length between points (6,0) & (3/2,0) and perpendicular from (3,-3) to x-axis.

Or
In other words, I need not to bother about the shape of region.
Show more

Though bound by 3 distinct lines will be a triangle until and unless you have some other constraints such as "for positive values of x" or infinite area is possible etc, I would still take a look at the region the inequality shows. It is just a matter of a couple of seconds if you plug in (0, 0) and see where it lies.
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In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, 27 Nov 2020, 00:14
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Concept: When we need to know the direction of a linear inequality, we put x and y = 0 in the equation of the line. If the inequality is true, then the shaded region will enclose the origin (0,0) and if it is not true, it will not enclose (0,0)

To understand the shaded region better, let us break up the diagram.


Diagram 1: In the equation y + 2x ≥ 3, the x intercept (when y = 0) is (1.5, 0), and the y intercept (when x = 0) is (0,3)

To see the direction of the shaded region, put x and y = 0 in y + 2x ≥ 3.

We get 0 ≥ 3, which is not true. So the shaded region will not enclose (0,0).



Diagram 2: In the equation y - x ≥ -6, the x intercept is (6, 0), and the y intercept is (0,-6)

Putting x and y = 0 in y - x ≥ -6. We get 0 ≥ -6, which is true. So the shaded region will enclose (0,0).


The line perpendicular to x = 0 is the y axis.


Combining the three lines, we get Diagram 3.


The Base BC = 6 - 1.5 = 4.5 units = 9/2 units.

To find the point A, which is the intersection of lines y + 2x = 3 and y - x = -6

Subtracting we get 3x = 9, or x = 3

Putting x = 3 in y - x = -6, we get y = -3

The height is therefore 3 units


Area = 1/2 * b * h = 1/2 * 9/2 * 3 = 27/4




Option B

Arun Kumar
Attachments

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1 and 2.jpg [ 1.21 MiB | Viewed 7172 times ]

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3.jpg [ 857.04 KiB | Viewed 7176 times ]

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In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, 27 Nov 2020, 02:20
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VeritasKarishma
unraveled
mangamma
In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, y –x ≥ -6 and the line, that is perpendicular to x = 0 and passes through the origin?

A. 9/4
B. 27/4
C. 9
D. 27/2
E. Cannot be determined
VeritasKarishma Bunuel
I have a query.
Would I be correct to infer that the question is asking for area of triangular region bound by the three lines mentioned in the question?
As mentioned there are only three equations - the three equations of lines are y +2x ≥ 3, y –x ≥ -6 and x-axis, i would not have to worry about the inequality. This way i would only focus on finding area of triangle with base of length between points (6,0) & (3/2,0) and perpendicular from (3,-3) to x-axis.

Or
In other words, I need not to bother about the shape of region.

Though bound by 3 distinct lines will be a triangle until and unless you have some other constraints such as "for positive values of x" or infinite area is possible etc, I would still take a look at the region the inequality shows. It is just a matter of a couple of seconds if you plug in (0, 0) and see where it lies.
Show more
Thank you VeritasKarishma,
Yes, Origin has to be taken care of.
As we already know, in this question, the lines have different slopes, so the area must be enclosed by the three and not the other way round. Also, for PS atleast, GMAT would not ask about the infinite area(if option E as in this question is not there). Would i be correct to think like that?
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In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, 27 Nov 2020, 05:46
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VeritasKarishma
mangamma
In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, y –x ≥ -6 and the line, that is perpendicular to x = 0 and passes through the origin?

A. 9/4
B. 27/4
C. 9
D. 27/2
E. Cannot be determined

The line x = 0 is y axis. Perpendicular to it passing through origin is the x axis.

y + 2x = 3 is the equation of a line passing through (0, 3) and (1.5, 0).
y + 2x >= 3 is the area away from (0, 0) because (0, 0) doesn't lie in this area.

y - x = -6 is the equation of a line passing through (0, -6) and (6, 0).
y - x >= -6 is the area towards (0, 0) because (0, 0) lies in this area.

Attachment:
Screenshot 2019-05-20 at 09.28.26.png
So we are looking for the area of triangle under x axis. The length of the base is 6 - 1.5 = 4.5.
For the altitude, we need the point of intersection of the two lines which is (3, -3). So length of altitude is 3.

Area = (1/2)*4.5 * 3 = 27/4

Answer (B)
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Hi! How do we know that this is a right triangle? We have used the formula assuming it to be right. Thank you in advance.
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In the xy-plane, what is the area of the region bounded by y +2x ≥ 3, 28 Nov 2020, 11:48
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CrackVerbalGMAT
Concept: When we need to know the direction of a linear inequality, we put x and y = 0 in the equation of the line. If the inequality is true, then the shaded region will enclose the origin (0,0) and if it is not true, it will not enclose (0,0)

To understand the shaded region better, let us break up the diagram.


Diagram 1: In the equation y + 2x ≥ 3, the x intercept (when y = 0) is (1.5, 0), and the y intercept (when x = 0) is (0,3)

To see the direction of the shaded region, put x and y = 0 in y + 2x ≥ 3.

We get 0 ≥ 3, which is not true. So the shaded region will not enclose (0,0).



Diagram 2: In the equation y - x ≥ -6, the x intercept is (6, 0), and the y intercept is (0,-6)

Putting x and y = 0 in y - x ≥ -6. We get 0 ≥ -6, which is true. So the shaded region will enclose (0,0).


The line perpendicular to x = 0 is the y axis.


Combining the three lines, we get Diagram 3.


The Base BC = 6 - 1.5 = 4.5 units = 9/2 units.

To find the point A, which is the intersection of lines y + 2x = 3 and y - x = -6

Subtracting we get 3x = 9, or x = 3

Putting x = 3 in y - x = -6, we get y = -3

The height is therefore 3 units


Area = 1/2 * b * h = 1/2 * 9/2 * 3 = 27/4




Option B

Arun Kumar
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can you explain this part? (that is perpendicular to x = 0 and passes through the origin?)
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