Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

(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.

Re: Co-ordinate Geometry- Area of a Triangle [#permalink]
18 Feb 2011, 09:37

1

This post received KUDOS

Area of triangle= 1/2*base*height

Consider line segment joining (-2,1) and (3,1) as Base. Base=5. It is a line parallel to x-axis that intercepts at y=1.

Now; the y-coordinate of (x,y) can't be 1.

1) |y-1|=1 y-1=1 y=2 or y-1=-1 y=0

We know the y-coordinate of the third vertex. It's either 1 or 0.

Say y=1; Irrespective of what x is; the distance between base and the vertex will be 1.

Distance is always the perpendicular distance. from the base i.e. make a line of infinite length from vertex(x,y) parallel to the base. The perpendicular line from base to this line will be the distance.

So; height=1

Area = 1/2*base*height = 1/2*5*1=5/2

Likewise for y=0; base=5; height=1 Sufficient.

2) Knowing that it is a right triangle won't give us the area. We don't know the length of the sides. Neither do we know anything about the other two angles. Not sufficient.

Re: What is the area of the triangle? [#permalink]
08 Mar 2012, 07:16

1

This post received KUDOS

Expert's post

babusona wrote:

The vertices of triangle are (-2,1), (3,1) and (x,y) on a plane. What is the area of the triangle? a)|y-1|=1 b)The angle at vertex (x,y) is 90 degrees.

Can this be considered as a 700+ or 600-700 level question? I am new to GMAT and had no clue on this one.

Merging similar topics. Please refer to the solutions above and ask if anything remains unclear.

As for difficulty level: I'd say since the problem involves several concepts (geometry, coordinate geometry, absolute value) then it can be considered 650+ question. _________________

What is the area of the triangle? [#permalink]
08 Mar 2012, 06:25

The vertices of triangle are (-2,1), (3,1) and (x,y) on a plane. What is the area of the triangle? a)|y-1|=1 b)The angle at vertex (x,y) is 90 degrees.

Can this be considered as a 700+ or 600-700 level question? I am new to GMAT and had no clue on this one.

Re: What is the area of the triangle? [#permalink]
08 Mar 2012, 06:59

babusona wrote:

The vertices of triangle are (-2,1), (3,1) and (x,y) on a plane. What is the area of the triangle? a)|y-1|=1 b)The angle at vertex (x,y) is 90 degrees.

Can this be considered as a 700+ or 600-700 level question? I am new to GMAT and had no clue on this one.

Plot the given points on an xy graph. S1) You will see that y=0 or 2 and this means the height of the triangle is 1 for either case. Since we know the length of the base from points (-2,1) and (3,1), we can thus find the area of the triangle

S2) The value of angle at point (x,y) is not enough determine the height.

If vertices of a triangle have coordinates [#permalink]
17 Sep 2012, 14:05

swatirpr wrote:

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

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

Sorry to revive this old post.

So 1. \(|y - 2| = 1\) ===>\(y-2=1\)or\(y-2=-1\) ===> two solutions \(y=1\) or\(y=3\). Notice that that regardless of what value of x, the base will always be the distance between \((-2, 2) and (3, 2)\) and the height will always be 1. Therefore the area will always be the same.

2) We know the base and that the third vertex will form a right angle. Well if that is the case, the only right angles that can formed for a given line is one in which it is the diameter of a circle. The third point on this line will always form a right angled triangle. All right angled triangles have their hypotenuse as the diameter of a circle and the third vertex a point on the circle. As you can see, while we know that this third, unknown, vertex will be on this circle, we dont't know where. Therefore the right triangle will have different areas as the third point glides across the circle. The area will reach a maximum when the point forms a isoclese right triangle.

Attachments

gmat_delete2.PNG [ 10.16 KiB | Viewed 2983 times ]

Re: Vertices of a traingle are at (-2,1) , (3,1) & (x,y) on a [#permalink]
19 Jun 2014, 07:51

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Vertices of a traingle are at (-2,1) , (3,1) & (x,y) on a [#permalink]
03 Aug 2014, 05:34

Bunuel wrote:

sandhyash wrote:

Q - Vertices of a traingle are at (-2,1) , (3,1) & (x,y) on a coordinate plane. What is the area of the triangle ?

1) |y-1| =1 2) Angle at Vertex (x,y) = 90 degrees

Source : MGMAT Test

Solution given : (A)

The explanation given in the solution is that :

From Statement (1) either y=0 or y=2, either way base is 5 and height =1. Hence area can be computed

Statement (2) is not enough

I did not understand this explanation. Please help !! Thanks in advance.

Below is almost identical question:

Vertices of a triangle have coordinates (-2, 2), (3, 2), (x, y). What is the area of the triangle?

(1) |y - 2| = 1 (2) angle at the vertex \((x, y)\) equals 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.

Answer: A.

Hope it heps.

Hey bunuel, brilliant explanation.

But, i have a silly doubt in this problem. arent we assuming that the base is fixed at (-2, 2), (3, 2). why cant the base be on the unknown point.

from 1st statement we get the value of y as 3 or 1. why cant the base be on (x,1) and (3,2) or (x,3) and (3,2)?

Re: Vertices of a traingle are at (-2,1) , (3,1) & (x,y) on a [#permalink]
12 Aug 2014, 01:44

Expert's post

saggii27 wrote:

Bunuel wrote:

sandhyash wrote:

Q - Vertices of a traingle are at (-2,1) , (3,1) & (x,y) on a coordinate plane. What is the area of the triangle ?

1) |y-1| =1 2) Angle at Vertex (x,y) = 90 degrees

Source : MGMAT Test

Solution given : (A)

The explanation given in the solution is that :

From Statement (1) either y=0 or y=2, either way base is 5 and height =1. Hence area can be computed

Statement (2) is not enough

I did not understand this explanation. Please help !! Thanks in advance.

Below is almost identical question:

Vertices of a triangle have coordinates (-2, 2), (3, 2), (x, y). What is the area of the triangle?

(1) |y - 2| = 1 (2) angle at the vertex \((x, y)\) equals 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.

Answer: A.

Hope it heps.

Hey bunuel, brilliant explanation.

But, i have a silly doubt in this problem. arent we assuming that the base is fixed at (-2, 2), (3, 2). why cant the base be on the unknown point.

from 1st statement we get the value of y as 3 or 1. why cant the base be on (x,1) and (3,2) or (x,3) and (3,2)?

We can consider any side of the triangle to be the base. _________________

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

McCombs Acceptance Rate Analysis McCombs School of Business is a top MBA program and part of University of Texas Austin. The full-time program is small; the class of 2017...