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Manager  G
Joined: 20 Feb 2017
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From among all the triangles that could be drawn in the coordinate  [#permalink]

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Question Stats: 51% (02:48) correct 49% (02:40) wrong based on 73 sessions

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From among all the triangles that could be drawn in the coordinate plane with vertices with integer coordinates (x,y) satisfying 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3, one triangle is to be chosen at random. If one of the edges of this triangle lies on the x-axis, what is the probability the area of the triangle is an integer ?
a) 1/3
B)1/2
C)5/9
D)2/3
E)3/4

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Math Expert V
Joined: 02 Aug 2009
Posts: 8023
From among all the triangles that could be drawn in the coordinate  [#permalink]

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Raksat wrote:
From among all the triangles that could be drawn in the coordinate plane with vertices with integer coordinates (x,y) satisfying 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3, one triangle is to be chosen at random. If one of the edges of this triangle lies on the x-axis, what is the probability the area of the triangle is an integer ?
a) 1/3
B)1/2
C)5/9
D)2/3
E)3/4

since edge is on x axis, two vertices will be on x axis that is $$( x_1,0)$$ and $$(x_2,0)$$ and x can take any of the 4 values - 0,1,2,3

I. total triangles possible = choose any two of 0,1,2,3 so 4C2=6
and for this the vertex can be any of the 3*4=12, 3 for y= 1,2, and 3 and 4 for x as 0,1,2,3
total 12*6=72

II. when area = integer
for this one of the side is EVEN as area = $$\frac{1}{2} * x*y$$
when the edge on x axis is odd ...
(0,0) and (1,0); (1,0) and (2,0); (2,0) and (3,0); (0,0) and (3,0); so 4 cases
y has to be even so 2 so (0,2), (1,2),(2,2),(3,2) so 4 cases
total 4*4=16
when the edge on x axis is even ...
(0,0) and (2,0); (1,0) and (3,0); so 2 cases
y can be anything so 2 so 1,2 and 3 and x 0,1,2,3 so 3*4 = 12 cases
total 2*12=24
Overall = 16+24=40

prob = $$\frac{40}{72}=\frac{5}{9}$$
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VP  D
Joined: 09 Mar 2016
Posts: 1230
Re: From among all the triangles that could be drawn in the coordinate  [#permalink]

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chetan2u wrote:
Raksat wrote:
From among all the triangles that could be drawn in the coordinate plane with vertices with integer coordinates (x,y) satisfying 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3, one triangle is to be chosen at random. If one of the edges of this triangle lies on the x-axis, what is the probability the area of the triangle is an integer ?
a) 1/3
B)1/2
C)5/9
D)2/3
E)3/4

since edge is on x axis, two vertices will be on x axis that is $$( x_1,0)$$ and $$(x_2,0)$$ and x can take any of the 4 values - 0,1,2,3

I. total triangles possible = choose any two of 0,1,2,3 so 4C2=6
and for this the vertex can be any of the 3*4=12, 3 for y= 1,2, and 3 and 4 for x as 0,1,2,3
total 12*6=72

II. when area = integer
for this one of the side is EVEN as area = $$\frac{1}{2} * x*y$$
when the edge on x axis is odd ...
(0,0) and (1,0); (1,0) and (2,0); (2,0) and (3,0); (0,0) and (3,0); so 4 cases
y has to be even so 2 so (0,2), (1,2),(2,2),(3,2) so 4 cases
total 4*4=16
when the edge on x axis is even ...
(0,0) and (2,0); (1,0) and (3,0); so 2 cases
y can be anything so 2 so 1,2 and 3 and x 0,1,2,3 so 3*4 = 12 cases
total 2*12=24
Overall = 16+24=40

prob = $$\frac{40}{72}=\frac{5}{9}$$

hi chetan2u, nice explanation though i do have questions if you dont mind can you please help me to understand your solution ... i definately dont understand one thing how can we create a triangle with the coordinates in which

x or y is 0 for instance you wrote these coordinates, see in red (0,0) and (1,0); (1,0) and (2,0); (2,0) and (3,0); (0,0) and (3,0); so 4 cases
y has to be even so 2 so (0,2), (1,2),(2,2),(3,2) so 4 cases

see attached diagram this is how i understood the problem as you see for one of the edges of triangles to be on X axis i got folowing options (3:3), (2:3) (2;2) , (2:1), (1:1)

Also i didnt get this "and for this the vertex can be any of the 3*4=12, 3 for y= 1,2, and 3 and 4 for x as 0,1,2,3
total 12*6=72
" can you please somehow rephrase the explanation the only thing i got is that triangle has 3 vertices ....

And one more question how did you manage to solve it without visualizing the problem ?

i would appreciate your taking time to explain thank you! Attachments Coordinate geometry +probability.png [ 67.88 KiB | Viewed 962 times ]

VP  D
Joined: 09 Mar 2016
Posts: 1230
From among all the triangles that could be drawn in the coordinate  [#permalink]

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chetan2u wrote:
Raksat wrote:
From among all the triangles that could be drawn in the coordinate plane with vertices with integer coordinates (x,y) satisfying 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3, one triangle is to be chosen at random. If one of the edges of this triangle lies on the x-axis, what is the probability the area of the triangle is an integer ?
a) 1/3
B)1/2
C)5/9
D)2/3
E)3/4

since edge is on x axis, two vertices will be on x axis that is $$( x_1,0)$$ and $$(x_2,0)$$ and x can take any of the 4 values - 0,1,2,3

I. total triangles possible = choose any two of 0,1,2,3 so 4C2=6
and for this the vertex can be any of the 3*4=12, 3 for y= 1,2, and 3 and 4 for x as 0,1,2,3
total 12*6=72

II. when area = integer
for this one of the side is EVEN as area = $$\frac{1}{2} * x*y$$
when the edge on x axis is odd ...
(0,0) and (1,0); (1,0) and (2,0); (2,0) and (3,0); (0,0) and (3,0); so 4 cases
y has to be even so 2 so (0,2), (1,2),(2,2),(3,2) so 4 cases
total 4*4=16
when the edge on x axis is even ...
(0,0) and (2,0); (1,0) and (3,0); so 2 cases
y can be anything so 2 so 1,2 and 3 and x 0,1,2,3 so 3*4 = 12 cases
total 2*12=24
Overall = 16+24=40

prob = $$\frac{40}{72}=\frac{5}{9}$$

how is it possible to draw a triangle with these coordinates (0,0) and (1,0) ? can someone draw such triangle ? Math Expert V
Joined: 02 Aug 2009
Posts: 8023
Re: From among all the triangles that could be drawn in the coordinate  [#permalink]

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1
dave13 wrote:
chetan2u wrote:
Raksat wrote:
From among all the triangles that could be drawn in the coordinate plane with vertices with integer coordinates (x,y) satisfying 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3, one triangle is to be chosen at random. If one of the edges of this triangle lies on the x-axis, what is the probability the area of the triangle is an integer ?
a) 1/3
B)1/2
C)5/9
D)2/3
E)3/4

since edge is on x axis, two vertices will be on x axis that is $$( x_1,0)$$ and $$(x_2,0)$$ and x can take any of the 4 values - 0,1,2,3

I. total triangles possible = choose any two of 0,1,2,3 so 4C2=6
and for this the vertex can be any of the 3*4=12, 3 for y= 1,2, and 3 and 4 for x as 0,1,2,3
total 12*6=72

II. when area = integer
for this one of the side is EVEN as area = $$\frac{1}{2} * x*y$$
when the edge on x axis is odd ...
(0,0) and (1,0); (1,0) and (2,0); (2,0) and (3,0); (0,0) and (3,0); so 4 cases
y has to be even so 2 so (0,2), (1,2),(2,2),(3,2) so 4 cases
total 4*4=16
when the edge on x axis is even ...
(0,0) and (2,0); (1,0) and (3,0); so 2 cases
y can be anything so 2 so 1,2 and 3 and x 0,1,2,3 so 3*4 = 12 cases
total 2*12=24
Overall = 16+24=40

prob = $$\frac{40}{72}=\frac{5}{9}$$

how is it possible to draw a triangle with these coordinates (0,0) and (1,0) ? can someone draw such triangle ? Hi..
Now coordinate(0,0) & (1,0) will give two vertices say A and B now C can be any of the 12 points above. So C can be (0,1) or (1,1) or (2,1), or (3,1) or (0,2) or (1,2) and so on and all these C coordinates will give different triangles
Similarly for (0,0) and (0,2) , again C can be any of the coordinates as shown above
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Re: From among all the triangles that could be drawn in the coordinate  [#permalink]

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thank you got it ! could you please rephrase the following sentence if you don't mind "and for this the vertex can be any of the 3*4=12, 3 for y= 1,2, and 3 and 4 for x as 0,1,2,3 total 12*6=72"

just need to understand it Intern  B
Joined: 30 Jan 2018
Posts: 17
Re: From among all the triangles that could be drawn in the coordinate  [#permalink]

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chetan2u wrote:
Raksat wrote:
From among all the triangles that could be drawn in the coordinate plane with vertices with integer coordinates (x,y) satisfying 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3, one triangle is to be chosen at random. If one of the edges of this triangle lies on the x-axis, what is the probability the area of the triangle is an integer ?
a) 1/3
B)1/2
C)5/9
D)2/3
E)3/4

since edge is on x axis, two vertices will be on x axis that is $$( x_1,0)$$ and $$(x_2,0)$$ and x can take any of the 4 values - 0,1,2,3

I. total triangles possible = choose any two of 0,1,2,3 so 4C2=6
and for this the vertex can be any of the 3*4=12, 3 for y= 1,2, and 3 and 4 for x as 0,1,2,3
total 12*6=72

II. when area = integer
for this one of the side is EVEN as area = $$\frac{1}{2} * x*y$$
when the edge on x axis is odd ...
(0,0) and (1,0); (1,0) and (2,0); (2,0) and (3,0); (0,0) and (3,0); so 4 cases
y has to be even so 2 so (0,2), (1,2),(2,2),(3,2) so 4 cases
total 4*4=16
when the edge on x axis is even ...
(0,0) and (2,0); (1,0) and (3,0); so 2 cases
y can be anything so 2 so 1,2 and 3 and x 0,1,2,3 so 3*4 = 12 cases
total 2*12=24
Overall = 16+24=40

prob = $$\frac{40}{72}=\frac{5}{9}$$

Hi chetan2u

If you don't mind could you pls explain a bit more how you are calculating the total number of triangles.

Thank you in advance.
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GPA: 3.97
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Re: From among all the triangles that could be drawn in the coordinate  [#permalink]

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chetan2u wrote:
Raksat wrote:
From among all the triangles that could be drawn in the coordinate plane with vertices with integer coordinates (x,y) satisfying 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3, one triangle is to be chosen at random. If one of the edges of this triangle lies on the x-axis, what is the probability the area of the triangle is an integer ?
a) 1/3
B)1/2
C)5/9
D)2/3
E)3/4

since edge is on x axis, two vertices will be on x axis that is $$( x_1,0)$$ and $$(x_2,0)$$ and x can take any of the 4 values - 0,1,2,3

I. total triangles possible = choose any two of 0,1,2,3 so 4C2=6
and for this the vertex can be any of the 3*4=12, 3 for y= 1,2, and 3 and 4 for x as 0,1,2,3
total 12*6=72

II. when area = integer
for this one of the side is EVEN as area = $$\frac{1}{2} * x*y$$
when the edge on x axis is odd ...
(0,0) and (1,0); (1,0) and (2,0); (2,0) and (3,0); (0,0) and (3,0); so 4 cases
y has to be even so 2 so (0,2), (1,2),(2,2),(3,2) so 4 cases
total 4*4=16
when the edge on x axis is even ...
(0,0) and (2,0); (1,0) and (3,0); so 2 cases
y can be anything so 2 so 1,2 and 3 and x 0,1,2,3 so 3*4 = 12 cases
total 2*12=24
Overall = 16+24=40

prob = $$\frac{40}{72}=\frac{5}{9}$$

Hi Chetan,
Can you please elaborate on the line " for this the vertex can be any of the 3*4=12, 3 for y= 1,2, and 3 and 4 for x as 0,1,2,3
total 12*6=72".

I am nt very clear how you got 12 combinations for those 6 ways Re: From among all the triangles that could be drawn in the coordinate   [#permalink] 13 Oct 2018, 22:33
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