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Bunuel
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In the xy-plane, triangular region S is bounded by the lines x=0,y=0, 30 Nov 2016, 03:05
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35% (medium)

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76% (02:34) correct 24% (02:37) wrong based on 524 sessions

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In the xy-plane, triangular region S is bounded by the lines x=0,y=0, and 3x−2y=30. Which of the following points does NOT lie inside region S?

A. (1,−1)
B. (2,−10)
C. (3,−13)
D. (5,−5)
E. (8,−2)
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C
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In the xy-plane, triangular region S is bounded by the lines x=0,y=0, 09 Dec 2019, 11:34
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Bunuel
In the xy-plane, triangular region S is bounded by the lines x=0,y=0, and 3x−2y=30. Which of the following points does NOT lie inside region S?

A. (1,−1)
B. (2,−10)
C. (3,−13)
D. (5,−5)
E. (8,−2)

For the point to lie in the specified region
1) Point should be in 4th quadrant (all options are of 4th uadrant)
2) Since 3x−2y=30 i.e. y = (3/2)x-15 so y must be greater than or equal to (3/2)x-15

Checking options
A. (1,−1) -1 > (3/2)(1)-15 Hence, Point lies in the region hence Incorrect Option
B. (2,−10) -10 > (3/2)(2)-15 Hence, Point lies in the region hence Incorrect Option
C. (3,−13) -13 < (3/2)(3)-15 Hence, Point does NOT lie in the region hence CORRECT Option
D. (5,−5) -5 > (3/2)(5)-15 Hence, Point lies in the region hence Incorrect Option
E. (8,−2) -2 > (3/2)(8)-15 Hence, Point lies in the region hence Incorrect Option

Answer: Option C
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When x = 0, y = -15
When y = 0, x = 10

Drawing the lines, we notice that our graph is partially covering the IV quadrant.

Convert the equation 3x−2y=30 into an inequality.
Test (1, 1): 3 - 2 < 30
Test (10, -15): 30 + 30 > 30
Therefore, inequality: 3x - 2y < 30 (since the points should lie "inside". If they could lie on the border, then we would include = as well.)

All points in the shaded area should satisfy that inequality.

A. (1,−1)
3 + 2 < 30 Yes

B. (2,−10)
6 + 20 = 26 < 30 Yes

C. (3,−13)
9 + 26 = 35 < 30 No => ANSWER

D. (5,−5)
15 + 10 = 25 < 30 Yes

E. (8,−2)
24 + 4 = 28 < 30 Yes

ANSWER: C
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In the xy-plane, triangular region S is bounded by the lines x=0,y=0, 30 Nov 2016, 04:01
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This is how went with the problem,

Drawn the shaded region with using three lines x=0, y=0 and 3x-2y=30.

Went back to the answer choices and putting the x values verified the y values. All the points falling in the region should follow equation 3x-2y≤30.

Only answer choice with point (3,−13) does now follow this equation and the value of the equation comes as 35 which is beyond the shaded region.

Hope I am right ?
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In the xy-plane, triangular region S is bounded by the lines x=0,y=0, 23 Feb 2017, 08:04
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I just guessed the answer using the following method,

Using the equation 3x-2y=30, we can find x-intercept and y-intercept
if x=0, then y=-15
if y=0, then x=10
after plotting it on the plane , find the midpoint of these 2 points
(x1+x2/2, y1+y2/2)=(0+10/2,-15+0/2)=>(5,-7.5)

only one answer stands out (3,-13) -> D
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In the xy-plane, triangular region S is bounded by the lines x=0,y=0, 24 Feb 2017, 06:24
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Bunuel
In the xy-plane, triangular region S is bounded by the lines x=0,y=0, and 3x−2y=30. Which of the following points does NOT lie inside region S?

A. (1,−1)
B. (2,−10)
C. (3,−13)
D. (5,−5)
E. (8,−2)
Show more

For the point to lie in the specified region
1) Point should be in 4th quadrant (all options are of 4th uadrant)
2) Since 3x−2y=30 i.e. y = (3/2)x-15 so y must be greater than or equal to (3/2)x-15

Checking options
A. (1,−1) -1 > (3/2)(1)-15 Hence, Point lies in the region hence Incorrect Option
B. (2,−10) -10 > (3/2)(2)-15 Hence, Point lies in the region hence Incorrect Option
C. (3,−13) -13 < (3/2)(3)-15 Hence, Point does NOT lie in the region hence CORRECT Option
D. (5,−5) -5 > (3/2)(5)-15 Hence, Point lies in the region hence Incorrect Option
E. (8,−2) -2 > (3/2)(8)-15 Hence, Point lies in the region hence Incorrect Option

Answer: Option C
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In the xy-plane, triangular region S is bounded by the lines x=0,y=0, 15 Jun 2017, 22:47
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Bunuel
In the xy-plane, triangular region S is bounded by the lines x=0,y=0, and 3x−2y=30. Which of the following points does NOT lie inside region S?

A. (1,−1)
B. (2,−10)
C. (3,−13)
D. (5,−5)
E. (8,−2)

For the point to lie in the specified region
1) Point should be in 4th quadrant (all options are of 4th uadrant)
2) Since 3x−2y=30 i.e. y = (3/2)x-15 so y must be greater than or equal to (3/2)x-15

Checking options
A. (1,−1) -1 > (3/2)(1)-15 Hence, Point lies in the region hence Incorrect Option
B. (2,−10) -10 > (3/2)(2)-15 Hence, Point lies in the region hence Incorrect Option
C. (3,−13) -13 < (3/2)(3)-15 Hence, Point does NOT lie in the region hence CORRECT Option
D. (5,−5) -5 > (3/2)(5)-15 Hence, Point lies in the region hence Incorrect Option
E. (8,−2) -2 > (3/2)(8)-15 Hence, Point lies in the region hence Incorrect Option

Answer: Option C
Show more

How do we know that the for the answer choices that y must be greater than the expression 3/2x - 15? How do we know that it must be y is greater than that then say y must be less than that value?
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In the xy-plane, triangular region S is bounded by the lines x=0,y=0, 15 Jun 2017, 23:01
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Bunuel
In the xy-plane, triangular region S is bounded by the lines x=0,y=0, and 3x−2y=30. Which of the following points does NOT lie inside region S?

A. (1,−1)
B. (2,−10)
C. (3,−13)
D. (5,−5)
E. (8,−2)
Show more

@Bunuel- although we can diagram this problem, calculate the x and y intercepts which would inevitably allow us to find the answer- can we not just reason that because we are given the hypotenuse of the triangle in this particular form 3x-2y=30 that no set of coordinates within the triangle will result in a number greater than 30? And when I say hypotenuse of the triangle I want to note I am not trying to make a generalization- there was another similar problem in ps made by manhattan in which the triangle in the plane had a x line and y line and an equation by the side it formed by connecting to the x and y lines turned out not to be the hypotenuse.
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In the xy-plane, triangular region S is bounded by the lines x=0,y=0, 15 Nov 2024, 01:26
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