In the xy-plane, what is the area of the region bounded by y +2x ≥ 3,

mangamma can you post the official answer please?
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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)
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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).

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. "

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.

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.

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

hkkat

I think your confusion lies in knowing which region we are talking about. We need to know the area of the orange region.


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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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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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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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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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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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.

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

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