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# If vertices of a triangle have coordinates (-2,2), (3,2) and

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If vertices of a triangle have coordinates (-2,2), (3,2) and  [#permalink]

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10 Aug 2009, 01:25
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If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

(1) |y-2|=1
(2) Angle at the vertex (x,y) equals to 90 degrees

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If vertices of a triangle have coordinates (-2,2), (3,2) and  [#permalink]

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02 Sep 2010, 04:45
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noboru wrote:
If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

(1) |y-2|=1
(2) angle at the vertex (x,y) equals to 90 degrees

Given two points A(-2,2) and B(3,2). Question: Area ABC=?, where C(x,y).

(1) |y-2|=1 --> $$y=3$$ or $$y=1$$ --> vertex C could be anywhere on the blue line $$y=3$$ or anywhere on the red line $$y=1$$. But in ANY case the are of ABC will be the same --> $$area=\frac{1}{2}*base*height$$ so $$base=AB=5$$ and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> $$area=\frac{1}{2}*base*height=\frac{5}{2}$$. Sufficient.

(2) angle at the vertex C(x,y) equals to 90 degrees --> ABC is a right triangle with hypotenuse AB --> consider AB to be diameter of a circle and in this case C could be anywhere on the circle and it will be right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle), thus height of ABC will be different for different location of point C, resulting the different areas (see two possible locations of of C: C3 and C4, heights of ABC3 and ABC4 are different). Not sufficient.

Hope it helps.

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Re: vertices of a triangle  [#permalink]

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04 Aug 2010, 07:42
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1
I think i understand why (2) is not sufficient. If we take the points (-2,2), (3,2) as a diameter of a circle and the point (x,y) lies on the perimeter of the circle, then any triangle formed is a right-angled triangle but with different height, therefore producing different areas. What do you think?
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Re: vertices of a triangle  [#permalink]

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10 Aug 2009, 01:53
1
noboru wrote:
If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

A) |y-2|=1
B) angle at the vertex (x,y) equals to 90 degrees

PS: 1 day to go!!!

first: good luck and wish you a 750+

from 1

suff either ways if we took the distance bet (-2,2),(3,2) as the base the height is either ways 1 unit

from 2

though i am not quite sure , id say insuff

A
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Re: vertices of a triangle  [#permalink]

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10 Aug 2009, 02:40
yezz wrote:
noboru wrote:
If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

A) |y-2|=1
B) angle at the vertex (x,y) equals to 90 degrees

PS: 1 day to go!!!

first: good luck and wish you a 750+

from 1

suff either ways if we took the distance bet (-2,2),(3,2) as the base the height is either ways 1 unit

from 2

though i am not quite sure , id say insuff

A

hi, thanks!

OK, lets do this one:

I agree: from 1 is suff since you know the height.

But, from 2: I think that there´s only one triangle you can draw that, with 2 points fixed (-2,2 and 3,2), the other vertex has 90 degrees...
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Re: vertices of a triangle  [#permalink]

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Updated on: 10 Aug 2009, 07:21
noboru wrote:
yezz wrote:
noboru wrote:
If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

A) |y-2|=1
B) angle at the vertex (x,y) equals to 90 degrees

PS: 1 day to go!!!

first: good luck and wish you a 750+

from 1

suff either ways if we took the distance bet (-2,2),(3,2) as the base the height is either ways 1 unit

from 2

though i am not quite sure , id say insuff

A

hi, thanks!

OK, lets do this one:

I agree: from 1 is suff since you know the height.

But, from 2: I think that there´s only one triangle you can draw that, with 2 points fixed (-2,2 and 3,2), the other vertex has 90 degrees...

actually there r 2 ( yours and its reflection ) however, can we get the area ( ie either the lenght of the 2 perpen sides or the known base and the hight from the right angle) we need to know x,y.

i tried using the slopes of the 2 perpend sides however i couldnt deduct any values...???

also , after some thought i think there is more than 2 possible right angled triangles with (-2,2) call it point a and (3,2) call it point b, point c ( the right angeld can be closer to a or to be as well

Originally posted by yezz on 10 Aug 2009, 02:48.
Last edited by yezz on 10 Aug 2009, 07:21, edited 1 time in total.
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Re: vertices of a triangle  [#permalink]

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10 Aug 2009, 04:05
Tricky one.. what the source

I think B is also Suff

heres how

with 2 points fixed (-2,2 and 3,2) the distance between the x axis is 5. Now we take 2 points x,y and that will make the line with co-ordinates (-2,2 and 3,2) as the hypotenuse

Now we know for a right angled triangle with hypotenuse 5 the other 2 sides must be 3,4
$$3^2 + 4^2 = 5^2$$

So we can can get the area.
What do you think guys.

Good luck noburu.. u threw some good CR questions in the forum. Cheers!
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Re: vertices of a triangle  [#permalink]

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10 Aug 2009, 05:01
Tricky one.. what the source

I think B is also Suff

heres how

with 2 points fixed (-2,2 and 3,2) the distance between the x axis is 5. Now we take 2 points x,y and that will make the line with co-ordinates (-2,2 and 3,2) as the hypotenuse

Now we know for a right angled triangle with hypotenuse 5 the other 2 sides must be 3,4
$$3^2 + 4^2 = 5^2$$

So we can can get the area.
What do you think guys.

Good luck noburu.. u threw some good CR questions in the forum. Cheers!

i am definetly with you, but OA seems to be A

As i have already said, if 2 points are fixed, and you wanna draw a right angle with the other vertex, there´s only one option (ok, yezz, u win, actually 2 taking into account its reflection).
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Re: vertices of a triangle  [#permalink]

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10 Aug 2009, 07:44
C

Statement 1:
Thrid vertex is (x,3) or (x,-3)

The area of the triangle varies if x is 0 or x is 2. So not sufficient

Statement 2: If the angle is right angle.
We can formulate one equation
The 3 points can be points on the circle with (-2,2) and (3,2) as the ends of the diameter.

We cannot solve for x,y using this 1 equation.
Bother St 1 and St 2:
We can calculate the area with both the equations, we can substitued y as 3 or -3. The area will be same for both values of y.
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Re: vertices of a triangle  [#permalink]

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04 Aug 2010, 10:32
I go with B

From the question stem we get a fix line that could be either the base, hypothenuse or a side of the mentioned triangle. To find out which part of the triangle this line is we need one value for x and one for Y.

From statement 1, Y can get two values for Y, and it mentions nothing with regards to X . Not suff

From statement 2, we get that the angle in the triangle has a 90 degree angle, and since we have already the sides of each of the triangle´s base and height, it is sufficient to determine the area.

Hope was clear
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Re: vertices of a triangle  [#permalink]

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09 Aug 2010, 08:43
ok before i move the answer i see lots of A's over here
i am confused on
|y-2|=1
this means y can take 2 values in two cases which means 2 answers
so is it really sufficient?
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Re: vertices of a triangle  [#permalink]

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11 Aug 2010, 22:56
My Opinion:
1. Since Y can take two values, this is not suff. as many others posted.
2. It is sufficient because there will be only two triangles possible with the same base having third angle as 90 degrees. Say one above the base and one below and both will have same area.

@Pipp: How can you draw two triangles with different height in one circle?
I think the circumcircle for triangles with different heights will be different. Am I wrong?
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Re: vertices of a triangle  [#permalink]

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01 Sep 2010, 11:18
noboru wrote:
Tricky one.. what the source

I think B is also Suff

heres how

with 2 points fixed (-2,2 and 3,2) the distance between the x axis is 5. Now we take 2 points x,y and that will make the line with co-ordinates (-2,2 and 3,2) as the hypotenuse

Now we know for a right angled triangle with hypotenuse 5 the other 2 sides must be 3,4
$$3^2 + 4^2 = 5^2$$

So we can can get the area.
What do you think guys.

Good luck noburu.. u threw some good CR questions in the forum. Cheers!

i am definetly with you, but OA seems to be A

As i have already said, if 2 points are fixed, and you wanna draw a right angle with the other vertex, there´s only one option (ok, yezz, u win, actually 2 taking into account its reflection).

Think of line joining two given vertices as diameter of a triangle, then if third angle is 90, it will be point lying on circle. Hust draw various third vertix and see that as base (diameter) remains same, height may vary.
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Re: vertices of a triangle  [#permalink]

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01 Sep 2010, 11:41
1
The answer is A. If you imagine the line joining the 2 points in question as the diameter then there are infinite triangles one can draw whose vertex is on the semicircle with different heights. Not suff. Hence A.

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Re: vertices of a triangle  [#permalink]

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02 Sep 2010, 00:42
D

From st-1 (x,y) could be (x,3) or (x,1).
Other two points are located in the line Y=2 you know height and base. Suff.

From st-2: xy is right angle.. so there will be only two such points... however this is out of GMAT scope but should be sufficient.
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Re: vertices of a triangle  [#permalink]

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07 Jun 2013, 23:42
Bunuel wrote:

(1) |y-2|=1 --> $$y=3$$ or $$y=1$$ --> vertex C could be anywhere on the blue line $$y=3$$ or anywhere on the red line $$y=1$$. But in ANY case the are of ABC will be the same --> $$area=\frac{1}{2}*base*height$$ so $$base=AB=5$$ and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> $$area=\frac{1}{2}*base*height=\frac{5}{2}$$. Sufficient.

Can someone please explain the Red Part. Why height of C2 will be 1.
Thanks
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Re: vertices of a triangle  [#permalink]

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07 Jun 2013, 23:50
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imhimanshu wrote:

Can someone please explain the Red Part. Why height of C2 will be 1.
Thanks

If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

We have the three points (-2,2), (3,2) and (x,y).

A) |y-2|=1

This tells us that the y coordinate of $$(x,y)$$ is 1 away from $$y=2$$. What does it mean? It means that the point will have a coordinate in the form of $$(x,3)$$ or $$(x,1)$$.

Given the base from the points (-2,2), (3,2), that equals $$5$$, in both cases the heigth will be 1. The first two points are on the $$y=2$$ line, and the third will be 1 "away" from it, on $$y=1$$ or on $$y=3$$.
So the area will be the same in both cases $$\frac{base*height}{2}$$, the base will not change and the height is 1
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Re: vertices of a triangle  [#permalink]

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08 Jun 2013, 02:16
imhimanshu wrote:
Bunuel wrote:

(1) |y-2|=1 --> $$y=3$$ or $$y=1$$ --> vertex C could be anywhere on the blue line $$y=3$$ or anywhere on the red line $$y=1$$. But in ANY case the are of ABC will be the same --> $$area=\frac{1}{2}*base*height$$ so $$base=AB=5$$ and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> $$area=\frac{1}{2}*base*height=\frac{5}{2}$$. Sufficient.

Can someone please explain the Red Part. Why height of C2 will be 1.
Thanks

Consider triangle $$ABC_1$$: the height is the distance (perpendicular) from $$C_1$$ to AB, which is 1;
Consider triangle $$ABC_2$$: the height is the distance (perpendicular) from $$C_2$$ to AB, which is 1.

Hope it's clear.
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Re: vertices of a triangle  [#permalink]

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15 Jun 2013, 04:57
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Bunuel wrote:
imhimanshu wrote:
Bunuel wrote:

(1) |y-2|=1 --> $$y=3$$ or $$y=1$$ --> vertex C could be anywhere on the blue line $$y=3$$ or anywhere on the red line $$y=1$$. But in ANY case the are of ABC will be the same --> $$area=\frac{1}{2}*base*height$$ so $$base=AB=5$$ and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> $$area=\frac{1}{2}*base*height=\frac{5}{2}$$. Sufficient.

Can someone please explain the Red Part. Why height of C2 will be 1.
Thanks

Consider triangle $$ABC_1$$: the height is the distance (perpendicular) from $$C_1$$ to AB, which is 1;
Consider triangle $$ABC_2$$: the height is the distance (perpendicular) from $$C_2$$ to AB, which is 1.

Hope it's clear.

Nothing to add after Bunuel's explanation. This link might help as a visual aid.
Any number of triangles with the same base and between 2 parallel lines will always have the same area.

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Re: If vertices of a triangle have coordinates (-2,2), (3,2) and  [#permalink]

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18 Jun 2013, 09:04
noboru wrote:
If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

A) |y-2|=1
B) angle at the vertex (x,y) equals to 90 degrees

dude here's my take and this contradicts ur QA

s1: this gives two values for Y i.e. 3 and -1 so its insufficient as we dont know any info other than this we need to know either value or x or the nature of triange.
s2: now given angle is 90, here if the angle is 90 then we get a equation from slope, two variables and one equation so again NS

now consider s1 and s2, we have y two values and each of which will yield 2 different values for x : for example

use slope form y2-y1/ x2-x1 => u get equation as x2 - x-5= 0
so again NS
please correct me if im wrong
Re: If vertices of a triangle have coordinates (-2,2), (3,2) and &nbs [#permalink] 18 Jun 2013, 09:04

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